RBSE Solutions Class 11 Physics Chapter 3 Kinematics

RBSE Class 11 Physics Chapter 3 Very Short Answer Type Questions

Question 1. What is the name of branch of Physics related with the study of motion of particles?
Answer: The branch of Physics that studies the motion of particles is called Dynamics. It looks at how forces affect movement.
In simple words: Physics that studies how tiny things move is called dynamics.

🎯 Exam Tip: Define key terms precisely for full marks, showing clear understanding of the core concept.

Question 2. How many coordinates are required there in one, two and three dimensional motion?
Answer: For one-dimensional motion, 1 coordinate is needed. For two-dimensional motion, 2 coordinates are needed. For three-dimensional motion, 3 coordinates are needed to define a particle's position.
In simple words: Motion in one direction uses 1 number for position, in two directions uses 2 numbers, and in three directions uses 3 numbers.

🎯 Exam Tip: Understand how the number of coordinates directly relates to the spatial dimensions of motion.

Question 4. How much is the displacement in one cycle of a circular motion?
Answer: The displacement in one complete cycle of circular motion is zero. This is because the starting point and the ending point are the same, meaning there is no net change in position.
In simple words: If you go in a circle and end up where you started, your total change in position is zero.

🎯 Exam Tip: Remember that displacement considers only the initial and final positions, not the path taken.

Question 5. Write the formula of speed.
Answer: The formula for speed is:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]
In simple words: Speed is found by dividing the distance you travel by the time it takes.

🎯 Exam Tip: Clearly state the formula and define its components to avoid losing marks.

Question 6. How much would be the displacement if a person walks 4 m towards east then 3m towards the south and then 4m towards west.
Answer: Let's trace the path: First, 4m East. Then, 3m South. Finally, 4m West. The 4m East and 4m West movements cancel each other out as they are in opposite directions. So, the net movement is only 3m South. Therefore, the displacement is 3m towards south.
In simple words: The person walked 4m east and 4m west, which cancel out. So, only the 3m south movement remains, which is the total displacement.

🎯 Exam Tip: For displacement, visualize the start and end points directly, ignoring any back-and-forth movements that cancel each other.

Question 7. What is negative acceleration called?
Answer: Negative acceleration is commonly called retardation or deceleration. This term is used when an object is slowing down or its velocity is decreasing.
In simple words: When something slows down, it has negative acceleration, also known as retardation.

🎯 Exam Tip: Use the correct terminology like deceleration or retardation when describing negative acceleration.

Question 8. What is the rate of change of displacement called with respect to time?
Answer: The rate of change of displacement with respect to time is called velocity. Velocity is a vector quantity, meaning it has both magnitude (speed) and direction.
In simple words: How fast an object changes its position and in what direction is called its velocity.

🎯 Exam Tip: Always distinguish between speed (rate of distance) and velocity (rate of displacement), especially noting velocity's directional component.

Question 9. Name the physical quantity which obtained by the slope of velocity time graph.
Answer: The physical quantity obtained by the slope of a velocity-time graph is acceleration. A steeper slope indicates a greater acceleration.
In simple words: If you look at a graph where the vertical line is velocity and the horizontal line is time, how steep the line is tells you the acceleration.

🎯 Exam Tip: Understand the relationship between slopes of graphs and the physical quantities they represent in kinematics.

Question 11. Which physical quantity obtained by the area of velocity time graph.
Answer: The physical quantity obtained by the area under a velocity-time graph is distance (or displacement). This area represents the total ground covered by the object.
In simple words: If you calculate the area under the line on a velocity-time graph, that area tells you how far the object has travelled.

🎯 Exam Tip: Remember that area under velocity-time graph gives displacement/distance, while its slope gives acceleration.

Question 12. A particle is moving with a definite velocity then how much would be its acceleration?
Answer: If a particle is moving with a definite (constant) velocity, its acceleration would be zero. This is because acceleration is defined as the rate of change of velocity, and if velocity is constant, there is no change.
In simple words: If something moves at a steady speed in the same direction, it is not speeding up or slowing down, so its acceleration is zero.

🎯 Exam Tip: Constant velocity is a key indicator for zero acceleration; avoid confusing speed with velocity.

Question 13. How much should be the angle of a projection so that the body covers the maximum distance?
Answer: For a projectile to cover the maximum horizontal distance (range) when launched, the optimal angle of projection should be 45° from the horizontal. This angle provides the best balance between initial horizontal velocity and time in the air.
In simple words: To throw something the farthest, you should launch it at an angle of 45 degrees from the ground.

🎯 Exam Tip: Recall the optimal angle for maximum projectile range, which is a fundamental concept in projectile motion.

RBSE Class 11 Physics Chapter 3 Short Answer Type Questions

Question 1. Explain motion.
Answer: Motion is the phenomenon where a body changes its position over time. An object is considered to be in absolute motion if its speed is measured relative to a truly stable point in the universe. However, no such perfectly stable point is known in the world. Therefore, neither absolute rest nor absolute motion can be truly defined. Instead, motion and rest are relative concepts. An object is said to be moving relative to another object if its position changes with respect to that other object as time passes.
In simple words: Motion means something is changing its place over time. We usually talk about motion compared to something else, because there's no perfectly still point in the universe to compare it to.

🎯 Exam Tip: Emphasize that motion is a relative concept, not absolute, and requires a defined reference point for observation.

Question 2. Explain in short frame of reference and its importance.
Answer: A frame of reference is a coordinate system used to describe the position and movement of objects. It helps us understand how different observers might see the same motion differently. For example, if you throw a ball on a moving train, you see it go straight up and down. But an observer on the station platform sees the ball follow a curved path (a parabola). This difference arises because the observers are in different frames of reference, one moving and one stationary. Frames of reference are crucial for describing motion accurately and are classified into two types:
(a) **Inertial frame of reference:** This is a frame that is either at rest or moving at a constant velocity without acceleration. Newton's laws of motion are valid in these frames. Examples include the Earth's motion or a space shuttle moving at a steady speed.
(b) **Non-inertial frame of reference:** This is a frame that is accelerating. Newton's laws of motion are not valid in these frames without introducing fictitious forces. Examples include a car speeding up or turning a corner, or a rocket accelerating.
In simple words: A frame of reference is your viewpoint when watching something move. What you see depends on whether you are standing still or moving. It helps us understand how things move differently for different observers.

🎯 Exam Tip: Illustrate the concept of a frame of reference with a simple, clear example, and briefly explain the two main types: inertial and non-inertial.

Question 3. Explain and draw cartesian coordinate system.
Answer: A Cartesian coordinate system provides a convenient way to define a frame of reference. It involves three mutually perpendicular axes, typically named x, y, and z, all meeting at a common origin point O. The x, y, and z coordinates of a particle specify its exact position with respect to this frame. To measure time, a clock is assumed to be placed within the system. This system allows us to pinpoint any object's location and track its movement in space effectively.

          Y-axis ^
               |
               |
               |
               |
     Negative-X axis --- O --- X-axis
               |
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               |
               v Negative-Y axis
              /
             /
            Z-axis
  

🎯 Exam Tip: Clearly describe the axes and origin of the Cartesian system, and state its purpose in defining position and a reference point for time.

Question 4. On the basis of frame of reference in how many types is the motion divided? Write the names.
Answer: Based on the frame of reference, motion is divided into three main types:
(a) **One-dimensional motion:** This is motion along a single straight line, where only one coordinate changes. For example, a train moving on a straight track.
(b) **Two-dimensional motion:** This is motion in a plane, where two coordinates change. For example, a car moving on a flat road or a projectile in flight.
(c) **Three-dimensional motion:** This is motion in space, where all three coordinates (x, y, z) change. For example, a bird flying freely or an airplane in the sky.
In simple words: Motion can be in 3 ways: straight line (1D), flat surface (2D), or all around (3D), depending on how many directions the object moves in.

🎯 Exam Tip: List the types of motion clearly and provide a quick, simple example for each type to demonstrate understanding.

Question 5. Differentiate between distance and displacement.
Answer: Distance and displacement are two distinct concepts used to describe motion:
(a) **Distance:**
- **Definition:** It is the total path length covered by an object between its initial and final positions.
- **Nature:** It is a scalar quantity, meaning it only has magnitude (a numerical value) but no direction.
- **Path Dependency:** It depends on the actual path taken by the object.
- **Value:** Distance is always positive and can never be zero for a moving object.
- **Units:** Its unit is meter (m) in the M.K.S. system and centimeter (cm) in the C.G.S. system.
(b) **Displacement:**
- **Definition:** It is the shortest straight-line distance between the initial and final positions of an object.
- **Nature:** It is a vector quantity, meaning it has both magnitude and a specific direction (from initial to final position).
- **Path Dependency:** It does not depend on the actual path taken; only on the start and end points.
- **Value:** Displacement can be positive, negative, or zero (if the object returns to its starting point).
- **Units:** Its unit is meter (m) in the M.K.S. system and centimeter (cm) in the C.G.S. system.
In simple words: Distance is how much ground you cover totally, no matter the direction. Displacement is how far you are from where you started, in a straight line, and in what direction.

🎯 Exam Tip: Clearly highlight that distance is a scalar (path-dependent, always positive) while displacement is a vector (path-independent, can be zero or negative).

Question 6. Explain translatory motion with examples.
Answer: Translatory motion occurs when an object moves from one place to another without rotating or changing its orientation. In this type of motion, all particles within the object move through the same distance in the same direction over the same time interval. This means the object simply shifts its position.
**Example:**
- The motion of a car or bus on a straight road. All parts of the vehicle move together from one point to another.
- A book sliding across a table.
In simple words: Translatory motion is when a whole object moves from one spot to another without spinning. Like a car driving straight.

🎯 Exam Tip: Define translatory motion clearly and provide a simple, relatable example to show understanding.

Question 7. A man walks 4m east, then 5m north, then taking a right turn walks 8m straight. Calculate the distance travelled by the man and its displacement.
Answer: Let's break down the man's journey to calculate the distance and displacement:
**1. Distance Travelled:**
The total distance travelled is the sum of all the path lengths covered, regardless of direction.
Distance = 4m (East) + 5m (North) + 8m (East, as a right turn after North is East)
Distance = 4m + 5m + 8m = 17m

**2. Displacement:**
Displacement is the shortest straight-line distance from the starting point to the final point, along with the direction.
- Net movement in the East direction: 4m (East) + 8m (East) = 12m East
- Net movement in the North direction: 5m North
We can visualize this as a right-angled triangle where the legs are 12m East and 5m North. The hypotenuse will be the displacement.
Using the Pythagorean theorem:
\[ \text{Displacement}^2 = (\text{East displacement})^2 + (\text{North displacement})^2 \]
\[ \text{Displacement}^2 = (12m)^2 + (5m)^2 \]
\[ \text{Displacement}^2 = 144m^2 + 25m^2 \]
\[ \text{Displacement}^2 = 169m^2 \]
\[ \text{Displacement} = \sqrt{169m^2} \]
\[ \text{Displacement} = 13m \]
The direction of the displacement would be North-East from the starting point.

🎯 Exam Tip: Clearly differentiate between scalar distance (total path length) and vector displacement (straight line from start to end) and use vector addition or Pythagoras theorem for displacement calculations involving perpendicular movements.

Question 8. Explain the differences between average and instantaneous velocity with examples.
Answer: Average velocity and instantaneous velocity both describe how an object's position changes, but they do so over different time scales:
(a) **Average Velocity:**
- **Definition:** It is the total displacement \( (\Delta \vec{x}) \) of an object divided by the total time interval \( (\Delta t) \) over which that displacement occurred. It represents the overall rate and direction of change in position.
- **Formula:** \[ \overrightarrow{v_{av}} = \frac{\Delta \vec{x}}{\Delta t} = \frac{\overrightarrow{x_2} - \overrightarrow{x_1}}{t_2 - t_1} \]
- **Example:** If a car travels 100 km East in 2 hours, its average velocity is 50 km/h East.
(b) **Instantaneous Velocity:**
- **Definition:** It is the velocity of an object at a specific moment in time or at a particular point in its path. It is found by considering the limit of average velocity as the time interval approaches zero.
- **Formula:** \[ \vec{v} = \lim_{\Delta t \rightarrow 0} \frac{\Delta \vec{x}}{\Delta t} = \frac{d\vec{x}}{dt} \]
- **Example:** The reading on a car's speedometer at any given instant shows its instantaneous speed, and if combined with direction, it gives instantaneous velocity.
Velocity (both average and instantaneous) can be positive, negative, or zero, depending on the direction of motion relative to the chosen reference. Instantaneous speed is always equal to the magnitude of instantaneous velocity, but average speed can be greater than or equal to the magnitude of average velocity.
In simple words: Average velocity is your overall speed and direction for a whole trip. Instantaneous velocity is your exact speed and direction at one specific moment.

🎯 Exam Tip: Clearly state the definitions and formulas for both, emphasizing the "overall" versus "specific moment" distinction, and provide simple examples.

Question 9. A runner completes a circular path of 1000 m in 2 min 5 s. What is its average speed and average velocity?
Answer: Let's calculate the average speed and average velocity for the runner:
**Given:**
- Total distance travelled = 1000 m
- Total time taken = 2 min 5 s
**Step 1: Convert total time to seconds:**
\[ t = (2 \times 60 \text{ s}) + 5 \text{ s} = 120 \text{ s} + 5 \text{ s} = 125 \text{ s} \]
**Step 2: Calculate Average Speed:**
Average Speed = \( \frac{\text{Total Distance}}{\text{Total Time}} \)
Average Speed = \( \frac{1000 \text{ m}}{125 \text{ s}} = 8 \text{ m/s} \)
**Step 3: Calculate Average Velocity:**
Since the runner completes one full circular path, their final position is the same as their initial position. This means the total displacement is zero.
Average Velocity = \( \frac{\text{Total Displacement}}{\text{Total Time}} \)
Average Velocity = \( \frac{0 \text{ m}}{125 \text{ s}} = 0 \text{ m/s} \)
Therefore, the runner's average speed is 8 m/s, and their average velocity is 0 m/s.
In simple words: The runner's average speed was 8 m/s because they covered 1000 meters in 125 seconds. But since they finished back where they started on the circular path, their average velocity (change in position) is 0 m/s.

🎯 Exam Tip: Remember to differentiate clearly between distance (scalar) and displacement (vector) in problems, especially for circular motion where displacement after a full cycle is zero.

Question 11. Draw a velocity-time graph for a uniform accelerated motion. What does its slope depicts?
Answer: A velocity-time graph for uniform accelerated motion shows velocity increasing steadily over time. This is represented by a straight line with a positive slope, indicating that the velocity changes by the same amount in equal time intervals.
The slope of this velocity-time graph directly depicts the constant uniform acceleration of the moving body.
\[ \text{Acceleration} = \text{Slope} = \frac{\text{Change in Velocity}}{\text{Change in Time}} = \tan \theta \]

🎯 Exam Tip: For a velocity-time graph, remember that the slope represents acceleration, and a straight line indicates constant acceleration.

Question 12. A ball when thrown vertically upwards with a velocity u reaches height h. If velocity is increased 2 times (i. e. 2 u) then what would be the effect on height?
Answer: When a ball is thrown vertically upwards, its velocity at the highest point becomes zero. We use the third equation of motion:
\[ v^2 = u^2 + 2as \]
Here, \( v = 0 \) (final velocity at max height), \( u \) (initial velocity), \( a = -g \) (acceleration due to gravity, negative as it acts downwards), and \( s = h \) (maximum height).
Substituting these values:
\[ 0^2 = u^2 + 2(-g)h \]
\[ 0 = u^2 - 2gh \]
Rearranging the equation to solve for \( h \):
\[ 2gh = u^2 \]
\[ h = \frac{u^2}{2g} \]
Now, if the initial velocity is doubled (i.e., \( u_{\text{new}} = 2u \)), let the new maximum height be \( h' \).
Using the same formula with the new velocity:
\[ h' = \frac{(u_{\text{new}})^2}{2g} = \frac{(2u)^2}{2g} = \frac{4u^2}{2g} \]
We can see that \( \frac{u^2}{2g} \) is equal to the original height \( h \). So,
\[ h' = 4 \times \left(\frac{u^2}{2g}\right) = 4h \]
Therefore, if the initial velocity is doubled, the maximum height reached will increase by 4 times.
In simple words: When you throw a ball up, its height depends on how fast you throw it initially. If you throw it twice as fast, it will go four times higher because the initial speed is squared in the height formula.

🎯 Exam Tip: Remember the relationship \( h \propto u^2 \) for maximum height in projectile motion. This means doubling the initial velocity quadruples the height, and similar exponential scaling applies.

Question 13. What is projectile motion?
Answer: Projectile motion describes the movement of an object (a projectile) that is launched near the Earth's surface and then follows a curved path under the sole influence of gravity. In this type of motion, air resistance is usually ignored for simplicity. To analyze projectile motion, the velocity and acceleration components acting on the object are typically split into horizontal and vertical directions and studied independently. This allows for a clear understanding of the projectile's trajectory, range, and maximum height.
In simple words: Projectile motion is when you throw something, and it flies in a curved path because only gravity is pulling it down. We study its up-and-down movement and its forward movement separately.

🎯 Exam Tip: Define projectile motion by highlighting that gravity is the only acting force (neglecting air resistance), leading to a characteristic parabolic trajectory.

RBSE Class 11 Physics Chapter 3 Long Answer Type Questions

Question 1. Describe the three equations for uniform accelerated motion.
Answer: The three fundamental equations of motion describe the relationship between an object's velocity, displacement, acceleration, and time when it moves with uniform (constant) acceleration. These equations can be derived using the calculus method:
**(i) First Equation of Motion (Velocity-Time Relation):**
From the definition of acceleration, it is the rate of change of velocity:
\[ a = \frac{dv}{dt} \]
Rearranging this, we get \( dv = a\,dt \).
To find the total change in velocity, we integrate both sides. Assume initial velocity is \( u \) at time \( t=0 \), and final velocity is \( v \) at time \( t \):
\[ \int_{u}^{v} dv = \int_{0}^{t} a\,dt \]
Since \( a \) is constant (uniform acceleration), it can be moved outside the integral:
\[ [v]_{u}^{v} = a[t]_{0}^{t} \]
\[ v - u = a(t - 0) \]
Thus, the first equation of motion is:
\[ v = u + at \]
This equation relates the final velocity, initial velocity, acceleration, and time.

**(ii) Second Equation of Motion (Position-Time Relation):**
From the definition of instantaneous velocity, it is the rate of change of displacement:
\[ v = \frac{dx}{dt} \]
Rearranging gives \( dx = v\,dt \).
We know from the first equation that \( v = u + at \). Substitute this into the displacement equation:
\[ dx = (u + at)\,dt \]
To find the total displacement, we integrate both sides. Assume initial position is \( x_0 \) at time \( t=0 \), and final position is \( x \) at time \( t \):
\[ \int_{x_0}^{x} dx = \int_{0}^{t} (u + at)\,dt \]
Separate the integral on the right side:
\[ [x]_{x_0}^{x} = \int_{0}^{t} u\,dt + \int_{0}^{t} at\,dt \]
\[ x - x_0 = u[t]_{0}^{t} + a\left[\frac{t^2}{2}\right]_{0}^{t} \]
\[ x - x_0 = u(t - 0) + a\left(\frac{t^2}{2} - 0\right) \]
If \( S \) represents the total displacement \( (x - x_0) \), then the second equation of motion is:
\[ S = ut + \frac{1}{2}at^2 \]
This equation relates displacement, initial velocity, acceleration, and time.

**(iii) Third Equation of Motion (Velocity-Displacement Relation):**
We start with the definition of acceleration: \( a = \frac{dv}{dt} \).
Using the chain rule, we can express acceleration in terms of displacement:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} \]
Since \( \frac{dx}{dt} = v \) (instantaneous velocity), we have:
\[ a = v\frac{dv}{dx} \]
Rearranging this equation, we get \( v\,dv = a\,dx \).
To find the relationship between velocity and displacement, we integrate both sides. Assume initial velocity is \( u \) at position \( x_0 \), and final velocity is \( v \) at position \( x \):
\[ \int_{u}^{v} v\,dv = \int_{x_0}^{x} a\,dx \]
Since \( a \) is constant, it can be moved outside the integral:
\[ \left[\frac{v^2}{2}\right]_{u}^{v} = a[x]_{x_0}^{x} \]
\[ \frac{v^2}{2} - \frac{u^2}{2} = a(x - x_0) \]
Multiplying the entire equation by 2:
\[ v^2 - u^2 = 2a(x - x_0) \]
If \( S \) represents the total displacement \( (x - x_0) \), then the third equation of motion is:
\[ v^2 = u^2 + 2aS \]
This equation relates final velocity, initial velocity, acceleration, and displacement.
In simple words: These three equations help us understand how objects move when they speed up or slow down steadily. The first one tells us the final speed. The second one tells us how far it travels. The third one connects final speed, initial speed, and distance travelled. All are derived using small changes in velocity and time.

🎯 Exam Tip: Understand the physical meaning of each equation and be able to derive them using the calculus method, clearly showing each integration step and limits.

Question 3. Derive the equations of motion with the help of the graphical method.
Answer: We can derive the three equations of motion for uniformly accelerated motion using a velocity-time graph.
Consider an object moving along a straight line with uniform acceleration \( a \).
- Let \( u \) be the initial velocity of the object at time \( t=0 \). (Represented by OP)
- Let \( v \) be the final velocity of the object at time \( t \). (Represented by SQ)
- Let \( t \) be the time interval. (Represented by OS or PR)
- Let \( S \) be the distance travelled by the object in time \( t \).
The velocity-time graph for this motion is a straight line PQ, as shown in the figure below.

🎯 Exam Tip: Clearly label the graph and explain how its slope and area relate to kinematic quantities. Demonstrate each derivation step logically, connecting it back to the graph's features.

Question 13. What is projectile motion?
Answer: Projectile motion describes how an object moves when it is thrown near Earth's surface. The object, called a projectile, follows a curved path only under the influence of gravity. To understand this motion, the velocity and acceleration are separated into horizontal and vertical parts and studied independently.
In simple words: Projectile motion is the curved path an object takes when thrown, influenced only by gravity. We study its horizontal and vertical movements separately.

🎯 Exam Tip: Remember that in projectile motion, horizontal velocity is constant (neglecting air resistance), while vertical velocity changes due to gravity.

RBSE Class 11 Physics Chapter 3 Long Answer Type Questions

Question 1. Describe the three equations for uniform accelerated motion.
Answer:
The three equations for uniformly accelerated motion describe how velocity, displacement, and time are related when an object moves with constant acceleration. These can be derived using calculus methods.

(i) Velocity-time relation:
From the definition of acceleration, which is the rate of change of velocity:
\( a = \frac{dv}{dt} \)
This means \( dv = a \, dt \)
If we integrate both sides from an initial velocity \( u \) at time \( t=0 \) to a final velocity \( v \) at time \( t \), and since acceleration \( a \) is uniform (constant):
\[ \int_{u}^{v} dv = \int_{0}^{t} a \, dt \]
\[ [v]_{u}^{v} = a[t]_{0}^{t} \]
This gives \( v - u = a(t - 0) \)
So, \( v - u = at \)
Or,
\( v = u + at \quad ...(vi) \)
This is the first equation of motion.
In simple words: This equation shows that your final speed is your starting speed plus how much your speed changed due to constant acceleration over time.

(ii) Distance-time relation:
The instantaneous velocity of an object in uniformly accelerated motion is given by:
\( v = \frac{dx}{dt} \)
So, \( dx = v \, dt \)
Since \( v = u + at \) (from the first equation), we can substitute this into the equation for \( dx \):
\( dx = (u + at) \, dt \quad ...(vii) \)
If we integrate both sides from an initial position \( x_0 \) at time \( t=0 \) to a final position \( x \) at time \( t \):
\[ \int_{x_0}^{x} dx = \int_{0}^{t} (u + at) \, dt \]
\[ [x]_{x_0}^{x} = \int_{0}^{t} u \, dt + \int_{0}^{t} at \, dt \]
\[ x - x_0 = u[t]_{0}^{t} + a\left[\frac{t^2}{2}\right]_{0}^{t} \]
\[ x - x_0 = u(t - 0) + a\left(\frac{t^2}{2} - 0\right) \quad ...(viii) \]
This simplifies to:
\( x - x_0 = ut + \frac{1}{2}at^2 \)
If \( S \) is the distance covered, then \( S = x - x_0 \). So, the equation becomes:
\( S = ut + \frac{1}{2}at^2 \quad ...(ix) \)
This is the second equation of motion.
In simple words: This equation helps you find the total distance an object travels. It considers its starting speed, the time it moves, and how much it speeds up or slows down consistently.

(iii) Velocity-displacement relation:
Instantaneous acceleration is given by:
\( a = \frac{dv}{dt} \)
We can also write this as:
\( a = \frac{dv}{dx} \cdot \frac{dx}{dt} \)
Since \( v = \frac{dx}{dt} \), we can substitute it:
\( a = v \frac{dv}{dx} \)
This means \( v \, dv = a \, dx \quad ...(x) \)
If we integrate both sides from an initial velocity \( u \) at position \( x_0 \) to a final velocity \( v \) at position \( x \):
\[ \int_{u}^{v} v \, dv = \int_{x_0}^{x} a \, dx \]
\[ \left[\frac{v^2}{2}\right]_{u}^{v} = a[x]_{x_0}^{x} \]
\[ \frac{v^2}{2} - \frac{u^2}{2} = a(x - x_0) \quad ...(xi) \]
Multiply by 2:
\( v^2 - u^2 = 2a(x - x_0) \)
If \( S \) is the distance covered, then \( S = x - x_0 \). So, the equation becomes:
\( v^2 - u^2 = 2aS \)
Or,
\( v^2 = u^2 + 2aS \quad ...(xii) \)
This is the third equation of uniform acceleration motion.
In simple words: This equation connects how fast something is going, where it started, and how far it moved, without needing to know the time taken.

🎯 Exam Tip: When solving problems, choose the equation that directly relates the given quantities to the unknown, minimizing calculation steps. Remember these equations are only valid for constant acceleration.

Question 2. Explain in short frame of reference and its importance.
Answer:
A frame of reference is a coordinate system used to describe the position and movement of an object. Imagine watching a ball on a moving train. To someone on the train, the ball goes straight up and down. But to someone standing on the platform, the ball follows a curved path. This difference happens because the observers are in different frames of reference, which are moving relative to each other.

Frames of reference help us understand motion from different viewpoints. They are crucial because motion is relative; an object's movement depends on where the observer is. There are two main types:

(a) Inertial frame of reference: This is a frame that is either standing still or moving at a constant speed in a straight line. Newton's laws of motion work perfectly in these frames. For example, the Earth's motion, a space shuttle moving at a constant velocity, or a rocket doing the same, can be considered inertial frames.
(b) Non-inertial frame of reference: This frame is accelerating, meaning its velocity is changing. Newton's laws of motion do not directly apply in these frames without adding "fictitious" forces. Examples include a car speeding up, slowing down, or turning a corner, or any frame moving along a curved path, even at a steady speed.
In simple words: A frame of reference is like your viewpoint for observing motion. It's important because how something moves can look different depending on your own movement or stillness.

🎯 Exam Tip: Clearly distinguish between inertial and non-inertial frames, as the validity of Newton's laws depends on this distinction. Remember that motion is always relative to a chosen frame.

Question 3. Explain and draw cartesian coordinate system.
Answer:
A Cartesian coordinate system is a convenient way to define a frame of reference. It uses three mutually perpendicular axes, usually labeled X, Y, and Z, which all meet at a common point called the origin (O). The coordinates (x, y, z) of a particle tell us its exact position relative to this origin within the frame. To measure time, a clock is also placed within this system. The figure below illustrates a Cartesian system.
In simple words: A Cartesian system uses three straight lines (X, Y, Z) that cross at one point (origin) to pinpoint an object's exact location, like a map for physics.

🎯 Exam Tip: When drawing, ensure the three axes are clearly perpendicular, the origin is labeled, and positive/negative directions are indicated. This visual representation is key to understanding position in 3D space.

Question 4. On the basis of frame of reference in how many types is the motion divided? Write the names.
Answer: On the basis of the frame of reference, motion is divided into three main types:

• One dimensional motion.
• Two dimensional motion.
• Three dimensional motion.

🎯 Exam Tip: Understand that these types refer to the minimum number of coordinates needed to describe the motion, not necessarily the actual path taken by the object.

Question 5. Differentiate between distance and displacement.
Answer: Distance and displacement are two important quantities in physics that describe motion, but they have distinct meanings:

(i) Distance: Distance is a scalar quantity, meaning it only has magnitude (a value) and no direction. It is the total length of the path covered by an object between its initial and final positions, regardless of the path's twists and turns. For example, if you walk 5 meters east and then 3 meters west, the total distance travelled is 8 meters.
Units: Meter (m) in M.K.S. system, Centimeter (cm) in C.G.S. system.
(ii) Displacement: Displacement is a vector quantity, meaning it has both magnitude and direction. It refers to the shortest straight-line distance between the object's initial and final positions. Its direction always points from the starting point to the ending point. In the previous example, if you walk 5 meters east and then 3 meters west, your final position is 2 meters east of your starting point, so your displacement is 2 meters east.
Units: Meter (m) in M.K.S. system, Centimeter (cm) in C.G.S. system.
In simple words: Distance is how much ground you covered in total, no matter the path. Displacement is the straight-line distance from where you started to where you ended, including the direction.

🎯 Exam Tip: Remember that distance can never be negative, while displacement can be negative (indicating movement in the opposite direction). Distance is always greater than or equal to the magnitude of displacement.

Question 6. Explain translatory motion with examples.
Answer: Translatory motion occurs when an object moves from one place to another relative to a fixed frame of reference, without any rotation. In this type of motion, all points on the object move with the same velocity in parallel paths. Essentially, the object shifts its position without changing its orientation.
Example: A car or a bus moving along a straight road is an example of translatory motion. The entire vehicle moves forward, and every part of it moves by the same amount in the same direction.
In simple words: Translatory motion is when an object moves from one spot to another without spinning. Like a car driving straight, all its parts move together.

🎯 Exam Tip: For translatory motion, remember that the object moves as a whole, meaning all its constituent particles have the same velocity and displacement at any given instant.

Question 7. A man walks 4m east, then 5m north, then taking a right turn walks 8m straight. Calculate the distance travelled by the man and its displacement.
Answer:
Distance travelled:
The total distance travelled by the man is the sum of all the individual paths he walked:
Distance = 4m (east) + 5m (north) + 8m (straight, which implies east based on a right turn after north)
Distance = 4 + 5 + 8 = 17 m

Displacement:
To find the displacement, we need to draw a vector diagram. Let's assume "taking a right turn walks 8m straight" means he turns right after walking North, so he walks 8m East.
Let's set up a coordinate system where East is the positive X-axis and North is the positive Y-axis.

• Starting point A (0,0)
• Walks 4m East: Reaches B (4,0)
• Walks 5m North: Reaches C (4,5)
• Turns right (East) and walks 8m: Reaches D (4+8, 5) = D (12,5)

The displacement is the straight-line distance from the initial position A (0,0) to the final position D (12,5).

Using the Pythagorean theorem for the right-angled triangle formed by the total eastward movement (4m + 8m = 12m) and the total northward movement (5m):
\( AD^2 = (Total \, East \, distance)^2 + (Total \, North \, distance)^2 \)
\( AD^2 = (4 + 8)^2 + (5)^2 \)
\( AD^2 = (12)^2 + (5)^2 \)
\( AD^2 = 144 + 25 \)
\( AD^2 = 169 \)
\( AD = \sqrt{169} \)
\( AD = 13 \, m \)
So, the displacement is 13 m. The direction can be found using trigonometry, but the question only asks for the magnitude.
In simple words: The man walked a total of 17 meters. His displacement, which is the shortest straight line from start to end, is 13 meters.

🎯 Exam Tip: Always differentiate between scalar distance and vector displacement. For displacement in multiple directions, use vector addition or the Pythagorean theorem for perpendicular movements.

Question 8. Explain the differences between average and instantaneous velocity with examples.
Answer:
Average velocity: Average velocity is defined as the total change in position (displacement) divided by the total time taken for that displacement to occur. It gives an overall idea of how fast an object moved and in what direction over a period of time. It doesn't tell us the speed or direction at any specific moment.
Formula: \( \overrightarrow{v_{av}} = \frac{\Delta \vec{x}}{\Delta t} = \frac{\overrightarrow{x_2} - \overrightarrow{x_1}}{t_2 - t_1} \)
Example: If a car travels 100 km east in 2 hours, its average velocity is 50 km/h east. Even if it stopped or sped up during that time, the average remains 50 km/h east.

Instantaneous velocity: Instantaneous velocity is the velocity of an object at a very specific instant in time or at a particular point along its path. It is found by taking the average velocity over an extremely small time interval that approaches zero. This tells us exactly how fast and in what direction the object is moving right now.
Formula: \( \vec{v} = \lim_{\Delta t \rightarrow 0} \frac{\Delta \vec{x}}{\Delta t} = \frac{d\vec{x}}{dt} \)
Example: When you look at your car's speedometer, it shows your instantaneous speed (and implied velocity if you know your direction) at that exact moment. If the car's speedometer reads 60 km/h east, that's its instantaneous velocity.

Velocity can be positive, negative, or zero, depending on the direction of motion relative to the chosen reference direction. An important result from studying speed and velocity is that while average speed can be equal to or greater than the magnitude of average velocity, instantaneous speed is always equal to the magnitude of instantaneous velocity.
In simple words: Average velocity is your overall speed and direction over a whole trip. Instantaneous velocity is your exact speed and direction at one specific moment, like what your speedometer shows.

🎯 Exam Tip: Remember that velocity is a vector quantity (has direction), while speed is a scalar quantity. The key difference lies in the "time interval" – for average velocity, it's a measurable duration; for instantaneous, it's an infinitesimally small moment.

Question 9. A runner completes a circular path of 1000 m in 2 min 5 s. What is its average speed and average velocity?
Answer:
Given:
Total distance travelled by the runner = 1000 m
Total time of the whole journey \( t = 2 \, \text{min} \, 5 \, \text{s} \)
First, convert the time to seconds:
\( t = (2 \times 60) \, \text{s} + 5 \, \text{s} \)
\( t = 120 \, \text{s} + 5 \, \text{s} = 125 \, \text{s} \)

Average speed:
Average speed is total distance divided by total time.
\( \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} \)
\( \text{Average speed} = \frac{1000 \, \text{m}}{125 \, \text{s}} \)
\( \text{Average speed} = 8 \, \text{m/s} \)

Average velocity:
Since the runner completes one full circular path, the initial and final positions are the same. This means the displacement is zero.
\( \text{Displacement} = 0 \, \text{m} \)
Average velocity is total displacement divided by total time.
\( \text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}} \)
\( \text{Average velocity} = \frac{0 \, \text{m}}{125 \, \text{s}} \)
\( \text{Average velocity} = 0 \, \text{m/s} \)
In simple words: The runner's average speed is 8 meters per second because they covered 1000 meters in 125 seconds. However, their average velocity is 0 meters per second because they ended up exactly where they started on the circular track.

🎯 Exam Tip: For circular motion problems, always remember that after a complete cycle, the displacement is zero, even if a significant distance has been covered.

Question 11. Draw a velocity-time graph for a uniform accelerated motion. What does its slope depicts?
Answer:
For uniform accelerated motion, the velocity-time graph is a straight line with a positive slope (if acceleration is positive). This indicates that the velocity changes at a constant rate over time. The slope of this straight-line graph represents the constant uniform acceleration of the moving body.

Below is a typical velocity-time graph for uniformly accelerated motion:

🎯 Exam Tip: Remember that the area under a velocity-time graph represents displacement, while its slope represents acceleration.

Question 12. A ball when thrown vertically upwards with a velocity u reaches height h. If velocity is increased 2 times (i.e. 2u) then what would be the effect on height?
Answer:
When a ball is thrown vertically upwards, its velocity becomes zero at the highest point (maximum height). We can use the third equation of motion: \( v^2 = u^2 + 2aS \)
Here:
- Final velocity \( v = 0 \) (at maximum height)
- Initial velocity \( u \)
- Acceleration \( a = -g \) (due to gravity, acting downwards)
- Displacement \( S = h \) (maximum height)

Substitute these values into the equation:
\( 0^2 = u^2 + 2(-g)h \)
\( 0 = u^2 - 2gh \)
Rearranging this equation to find \( h \):
\( 2gh = u^2 \)
\( h = \frac{u^2}{2g} \)

Now, if the initial velocity is doubled (i.e., \( u' = 2u \)), let the new maximum height be \( h' \). Using the same formula:
\( h' = \frac{(u')^2}{2g} \)
Substitute \( u' = 2u \):
\( h' = \frac{(2u)^2}{2g} \)
\( h' = \frac{4u^2}{2g} \)
We know that \( h = \frac{u^2}{2g} \), so we can substitute \( h \) into the equation for \( h' \):
\( h' = 4 \left(\frac{u^2}{2g}\right) \)
\( h' = 4h \)

Therefore, if the initial upward velocity is doubled, the maximum height reached by the ball will increase by 4 times.
In simple words: If you throw a ball up twice as fast, it will go four times higher. This is because the height depends on the square of the initial speed.

🎯 Exam Tip: Remember that maximum height is directly proportional to the square of the initial velocity. Neglecting air resistance, the only acceleration acting on the ball during its flight is due to gravity.

Question 13. How much should be the angle of a projection so that the body covers the maximum distance?
Answer: The angle of projection required for a body to cover the maximum horizontal distance (range) is 45°. At this angle, the horizontal component of velocity is large enough to carry the projectile a good distance, while the vertical component is sufficient to keep it in the air for an adequate time. This combination results in the longest possible horizontal range.
In simple words: To throw something the farthest horizontally, you should launch it at an angle of 45 degrees.

🎯 Exam Tip: Maximum range is achieved at 45 degrees. Angles above or below 45 degrees (e.g., 30 and 60 degrees, or 20 and 70 degrees) that are complementary will have the same range, but different maximum heights and times of flight.

RBSE Class 11 Physics Chapter 3 Numerical Problems

Question 1. Time taken from Jaipur to Ajmer by car is 1.5 h. If the average velocity of car is 80 km/h then how much is the distance between Jaipur and Ajmer?
Answer:
Given:
Time taken \( t = 1.5 \, \text{hr} \)
Average velocity \( \overline{v} = 80 \, \text{km/hr} \)

We know that average velocity is defined as total displacement (distance in this case, as it's a straight path) divided by time:
\( \overline{v} = \frac{S}{t} \)
To find the distance \( S \), we can rearrange the formula:
\( S = \overline{v} \times t \)
Substitute the given values:
\( S = 80 \, \text{km/hr} \times 1.5 \, \text{hr} \)
\( S = 120 \, \text{km} \)
Therefore, the distance between Jaipur and Ajmer is 120 km.
In simple words: The car travels 80 kilometers every hour. Since it drove for 1.5 hours, the total distance between the two cities is 120 kilometers.

🎯 Exam Tip: Always ensure units are consistent before performing calculations. In this case, both velocity and time were in terms of hours, so no conversion was needed.

Question 2. In 5s the speed of bus increases from 25km/h to 70km/h. What is the average acceleration of the bus?
Answer:
Given:
Time interval \( \Delta t = 5 \, \text{s} \)
Initial speed \( v_1 = 25 \, \text{km/h} \)
Final speed \( v_2 = 70 \, \text{km/h} \)

First, we need to convert the speeds from km/h to m/s to match the time unit (seconds). To convert km/h to m/s, multiply by \( \frac{5}{18} \).
\( v_1 = 25 \, \text{km/h} \times \frac{5}{18} \, \text{m/s per km/h} = \frac{125}{18} \, \text{m/s} \approx 6.94 \, \text{m/s} \)
\( v_2 = 70 \, \text{km/h} \times \frac{5}{18} \, \text{m/s per km/h} = \frac{350}{18} \, \text{m/s} = \frac{175}{9} \, \text{m/s} \approx 19.44 \, \text{m/s} \)

Next, calculate the change in speed \( \Delta v \):
\( \Delta v = v_2 - v_1 = \frac{175}{9} \, \text{m/s} - \frac{125}{18} \, \text{m/s} \)
\( \Delta v = \frac{350}{18} \, \text{m/s} - \frac{125}{18} \, \text{m/s} = \frac{225}{18} \, \text{m/s} = \frac{25}{2} \, \text{m/s} = 12.5 \, \text{m/s} \)

Now, calculate the average acceleration \( a \):
\( a = \frac{\Delta v}{\Delta t} \)
\( a = \frac{12.5 \, \text{m/s}}{5 \, \text{s}} \)
\( a = 2.5 \, \text{m/s}^2 \)
The average acceleration of the bus is 2.5 m/s².
In simple words: The bus speeds up by 12.5 meters per second over 5 seconds. This means its acceleration is 2.5 meters per second squared.

🎯 Exam Tip: Always pay attention to unit consistency. If different units are provided (e.g., km/h and seconds), convert them to a consistent system (like SI units: meters and seconds) before calculation.

Question 3. Car A and B travels 100 km journey together. Car A travels with the uniform speed of 40 km/h. Car B travels with 60 km/h. But after half an hour it stops for 15 min due to some problem and when it moves its speed remains 50 km/h.
(i) Draw a speed-time graph of the journey.
(ii) Tell which car will complete the journey first and how much time before?
Answer:
Given total journey distance = 100 km.

Car A:
Uniform speed \( V_A = 40 \, \text{km/h} \)
Time taken by Car A \( t_A = \frac{\text{Distance}}{\text{Speed}} = \frac{100 \, \text{km}}{40 \, \text{km/h}} = 2.5 \, \text{hr} \)

Car B:
1. Travels first part: Speed \( V_B = 60 \, \text{km/h} \) for half an hour \( t_1 = 0.5 \, \text{hr} \)
Distance covered \( l_1 = V_B \times t_1 = 60 \, \text{km/h} \times 0.5 \, \text{hr} = 30 \, \text{km} \)
2. Stops for 15 min: \( t' = 15 \, \text{min} = \frac{15}{60} \, \text{hr} = 0.25 \, \text{hr} \)
3. Travels remaining distance: Remaining distance \( l_2 = 100 \, \text{km} - 30 \, \text{km} = 70 \, \text{km} \)
Speed for remaining distance \( V'_B = 50 \, \text{km/h} \)
Time taken for remaining distance \( t_2 = \frac{l_2}{V'_B} = \frac{70 \, \text{km}}{50 \, \text{km/h}} = 1.4 \, \text{hr} \)
Total time taken by Car B \( t_B = t_1 + t' + t_2 = 0.5 \, \text{hr} + 0.25 \, \text{hr} + 1.4 \, \text{hr} = 2.15 \, \text{hr} \)

(i) Speed-time graph of the journey:
The graph below visually represents the speed of Car A and Car B over time. Car A maintains a constant speed of 40 km/h for 2.5 hours. Car B starts at 60 km/h for 0.5 hours, then drops to 0 km/h for 0.25 hours (15 minutes), and then continues at 50 km/h until it completes the 100 km journey at 2.15 hours.

(ii) Which car will complete the journey first and how much time before?
Car A's total time \( t_A = 2.5 \, \text{hr} \)
Car B's total time \( t_B = 2.15 \, \text{hr} \)

Since \( t_B < t_A \), Car B will complete the journey first.
Time difference \( \Delta t = t_A - t_B = 2.5 \, \text{hr} - 2.15 \, \text{hr} = 0.35 \, \text{hr} \)
Convert the time difference to minutes:
\( \Delta t = 0.35 \, \text{hr} \times 60 \, \text{min/hr} = 21 \, \text{min} \)
Car B will reach the destination 21 minutes before Car A.
In simple words: Car B finishes the 100 km trip in 2.15 hours, while Car A takes 2.5 hours. So, Car B arrives first, 21 minutes earlier than Car A.

🎯 Exam Tip: When analyzing multi-stage journeys, break down each car's travel into segments and calculate time for each, then sum them up for the total. Always convert units to be consistent.

Question 5. A particle is dropped from a 200 m tall tower. At the same time a second particle is thrown upwards with velocity 50 m/s. Calculate when and where would these two particles (bodies) meet.
Answer:
Let's assume the particles meet at a height \( h \) from the ground after time \( t \). The total height of the tower is 200 m.
Let's use the second equation of motion: \( S = ut + \frac{1}{2}at^2 \). We will take upward direction as positive and downward direction as negative, and the ground as the reference point (height = 0).

For the particle dropped from the tower (Body A):
- Initial velocity \( u_A = 0 \, \text{m/s} \) (dropped from rest)
- Acceleration \( a_A = -g \) (downwards, so -9.8 m/s² or use 10 m/s² if simplified, let's assume \( g \) is positive, so \( a = -g \))
- Displacement \( S_A \) (distance covered downwards) \( = -(200 - h) \). (If h is height from ground, it covers 200-h downwards).
So, \( -(200 - h) = (0)t + \frac{1}{2}(-g)t^2 \)
\( -(200 - h) = -\frac{1}{2}gt^2 \)
\( 200 - h = \frac{1}{2}gt^2 \quad ...(1) \)

For the particle thrown upwards (Body B):
- Initial velocity \( u_B = 50 \, \text{m/s} \) (upwards)
- Acceleration \( a_B = -g \) (downwards)
- Displacement \( S_B = h \) (height from the ground)
So, \( h = u_Bt + \frac{1}{2}(-g)t^2 \)
\( h = 50t - \frac{1}{2}gt^2 \quad ...(2) \)

Now, we have two equations. Let's add equation (1) and equation (2):
\( (200 - h) + h = \frac{1}{2}gt^2 + (50t - \frac{1}{2}gt^2) \)
\( 200 = 50t \)
\( t = \frac{200}{50} \)
\( t = 4 \, \text{s} \)
So, the two particles will meet after 4 seconds.

Now, substitute \( t = 4 \, \text{s} \) into equation (2) to find the height \( h \):
\( h = 50(4) - \frac{1}{2}g(4)^2 \)
Assuming \( g = 9.8 \, \text{m/s}^2 \):
\( h = 200 - \frac{1}{2}(9.8)(16) \)
\( h = 200 - (4.9)(16) \)
\( h = 200 - 78.4 \)
\( h = 121.6 \, \text{m} \)

So, the particles will meet after 4 seconds at a height of 121.6 m from the ground.
In simple words: The two particles will cross paths 4 seconds after they are released. They will meet at a height of 121.6 meters above the ground.

🎯 Exam Tip: When dealing with objects moving under gravity, consistently define a positive direction (e.g., upwards) and apply the sign of acceleration due to gravity (g) accordingly (e.g., -g if upward is positive).

Question 6. A 100 g mass particle falls 490 cm downwards from rest position and comes to rest after moving 70 cm down in the sand. Calculate the acceleration imposed by the sand on the particle. (g = 9.8m/s²)
Answer:
Given:
Mass \( m = 100 \, \text{g} = 0.1 \, \text{kg} \)
Initial fall distance (in air) \( h = 490 \, \text{cm} = 4.90 \, \text{m} \)
Distance of penetration in sand \( s = 70 \, \text{cm} = 0.70 \, \text{m} \)
Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)

First, calculate the velocity of the particle just before it hits the sand. It falls from rest, so initial velocity \( u_1 = 0 \).
Using \( v^2 = u^2 + 2aS \):
\( v^2 = u_1^2 + 2gh \)
\( v^2 = 0^2 + 2(9.8 \, \text{m/s}^2)(4.90 \, \text{m}) \)
\( v^2 = 19.6 \times 4.90 \)
\( v^2 = 96.04 \, (\text{m/s})^2 \)
\( v = \sqrt{96.04} \, \text{m/s} = 9.8 \, \text{m/s} \)
So, the velocity of the particle just before hitting the sand is 9.8 m/s.

Now, consider the motion of the particle inside the sand:
- Initial velocity \( u_2 = 9.8 \, \text{m/s} \) (this is the velocity \( v \) from the previous step)
- Final velocity \( v_2 = 0 \, \text{m/s} \) (comes to rest)
- Displacement \( s = 0.70 \, \text{m} \)
- Acceleration in sand \( a_{sand} = ? \)

Using \( v_2^2 = u_2^2 + 2a_{sand}s \):
\( 0^2 = (9.8)^2 + 2(a_{sand})(0.70) \)
\( 0 = 96.04 + 1.4 \, a_{sand} \)
\( -1.4 \, a_{sand} = 96.04 \)
\( a_{sand} = -\frac{96.04}{1.4} \)
\( a_{sand} = -68.6 \, \text{m/s}^2 \)
The negative sign indicates that the acceleration is in the opposite direction to the particle's motion, meaning it is a deceleration. The magnitude of the acceleration imposed by the sand is 68.6 m/s².
In simple words: The particle hits the sand at 9.8 m/s and stops after digging 70 cm. The sand applies a strong stopping force, causing a deceleration of 68.6 m/s².

🎯 Exam Tip: Break complex problems into stages. First, calculate the velocity at the interface between two mediums (air and sand), then use that as the initial velocity for the next stage of motion.

Question 7. A stone is dropped from a helicopter moving in upward direction at 500 m height. It reaches the earth after 6 seconds. What was the velocity of the helicopter when the stone was dropped? (g = 10m/s²)
Answer:
Given:
Height from which stone is dropped \( h = 500 \, \text{m} \)
Time taken to reach earth \( t = 6 \, \text{s} \)
Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

When the stone is dropped from the helicopter, it initially has the same velocity as the helicopter. Since the helicopter is moving upwards, the stone's initial velocity (\( u \)) is also upwards. We will take the upward direction as positive and downward as negative. The displacement of the stone is -500 m (since it moves downwards from its initial position).

Using the second equation of motion: \( S = ut + \frac{1}{2}at^2 \)
Here, \( S = -500 \, \text{m} \)
\( a = -g = -10 \, \text{m/s}^2 \)
\( -500 = u(6) + \frac{1}{2}(-10)(6)^2 \)
\( -500 = 6u - 5(36) \)
\( -500 = 6u - 180 \)
Add 180 to both sides:
\( -500 + 180 = 6u \)
\( -320 = 6u \)
\( u = -\frac{320}{6} \)
\( u = -53.33 \, \text{m/s} \)

The negative sign indicates that the initial velocity of the stone (and thus the helicopter) was actually downwards in order to reach the ground in 6 seconds, despite the problem stating the helicopter was moving in an upward direction at the moment of drop. This usually means the helicopter was slowing down while moving up, or it was actually descending when the stone was released. However, if we strictly follow the 'upward direction' wording for initial velocity, it means there's a contradiction with the result. Let's re-read the question carefully: "A stone is dropped from a helicopter moving in upward direction...". This implies \(u\) is initially upwards. If the result is negative, it means the initial velocity was actually downwards. Let's assume the question implied that the *direction* was upward but the *magnitude* is what we need to find, and the sign will tell us the true direction if our assumption was wrong. Or, the question intends for us to find the helicopter's velocity at the moment of drop, given the conditions. If the calculated initial velocity is negative, it means the helicopter was actually moving downwards when the stone was dropped.

Let's assume the question implicitly asks for the magnitude of the velocity. The initial velocity of the helicopter was 53.33 m/s downwards (since the sign is negative). If the question intends "moving in upward direction" as the *assumed* direction of \(u\), then the physics dictates that for the stone to reach the ground in 6 seconds from 500m with \( g = 10 \, \text{m/s}^2 \), the initial velocity must be downwards. If we must interpret "moving in upward direction" as a constraint that \(u\) must be positive, then the stone would take longer to reach the ground (it would first go up, then come down). Given the numerical result, the initial velocity must be downwards.
The magnitude of the velocity of the helicopter when the stone was dropped was 53.33 m/s.
In simple words: To fall 500 meters in 6 seconds with gravity, the stone must have been moving downwards at 53.33 m/s when it was dropped from the helicopter.

🎯 Exam Tip: Pay close attention to the sign conventions for displacement, velocity, and acceleration. A negative result for initial velocity when an upward direction was assumed means the actual initial velocity was in the opposite (downward) direction.

Question 8. A man walks with a speed of 5 km/h on a straight road and reaches his office which is 2.5 km away. Since, the office was closed, he at the same time returns back to his house with a speed of 7.5 km/h. Calculate his average speed and average velocity for following time intervals.
(i) 0 to 30 min
(ii) 0 to 50 min
(iii) 0 to 40 min.
Answer:
Given:
Distance to office \( D = 2.5 \, \text{km} \)
Speed to office \( V_1 = 5 \, \text{km/h} \)
Time to office \( t_{to \, office} = \frac{D}{V_1} = \frac{2.5 \, \text{km}}{5 \, \text{km/h}} = 0.5 \, \text{hr} = 30 \, \text{min} \)

Speed returning home \( V_2 = 7.5 \, \text{km/h} \)
Time returning home \( t_{to \, home} = \frac{D}{V_2} = \frac{2.5 \, \text{km}}{7.5 \, \text{km/h}} = \frac{1}{3} \, \text{hr} = 20 \, \text{min} \)

(i) For time interval 0 to 30 min:
In this interval, the man walks to his office (2.5 km) and reaches there exactly at 30 min.
Total distance travelled = 2.5 km
Total displacement = 2.5 km (from home to office)

Average speed \( = \frac{\text{Total distance}}{\text{Total time}} = \frac{2.5 \, \text{km}}{0.5 \, \text{hr}} = 5 \, \text{km/h} \)
Average velocity \( = \frac{\text{Total displacement}}{\text{Total time}} = \frac{2.5 \, \text{km}}{0.5 \, \text{hr}} = 5 \, \text{km/h} \) (in the direction of office)
In simple words: For the first 30 minutes, he just walks to his office. So his average speed and average velocity are both 5 km/h.

(ii) For time interval 0 to 50 min:
The man reaches office in 30 min (0.5 hr). He then returns home for the remaining time \( 50 \, \text{min} - 30 \, \text{min} = 20 \, \text{min} \).
Time returning home for 20 min \( = \frac{20}{60} \, \text{hr} = \frac{1}{3} \, \text{hr} \)
Distance travelled during return = \( V_2 \times (\text{time for return}) = 7.5 \, \text{km/h} \times \frac{1}{3} \, \text{hr} = 2.5 \, \text{km} \)
Since the time to return home fully is 20 min, by 50 min the man has returned all the way to his house (2.5 km back).
Total distance travelled = (Distance to office) + (Distance back home) = 2.5 km + 2.5 km = 5 km
Total displacement = 0 km (He started from home and returned to home).
Total time = 50 min \( = \frac{50}{60} \, \text{hr} = \frac{5}{6} \, \text{hr} \)

Average speed \( = \frac{\text{Total distance}}{\text{Total time}} = \frac{5 \, \text{km}}{5/6 \, \text{hr}} = 5 \times \frac{6}{5} \, \text{km/h} = 6 \, \text{km/h} \)
Average velocity \( = \frac{\text{Total displacement}}{\text{Total time}} = \frac{0 \, \text{km}}{5/6 \, \text{hr}} = 0 \, \text{km/h} \)
In simple words: After 50 minutes, he has walked to the office and back home. His average speed is 6 km/h, but his average velocity is 0 km/h because he's back at his starting point.

(iii) For time interval 0 to 40 min:
The man reaches office in 30 min (0.5 hr). He then returns home for the remaining time \( 40 \, \text{min} - 30 \, \text{min} = 10 \, \text{min} \).
Time returning home for 10 min \( = \frac{10}{60} \, \text{hr} = \frac{1}{6} \, \text{hr} \)
Distance travelled during return = \( V_2 \times (\text{time for return}) = 7.5 \, \text{km/h} \times \frac{1}{6} \, \text{hr} = 1.25 \, \text{km} \)
Total distance travelled = (Distance to office) + (Distance travelled back) = 2.5 km + 1.25 km = 3.75 km
Total displacement = (Distance to office) - (Distance travelled back) = 2.5 km - 1.25 km = 1.25 km (towards office)
Total time = 40 min \( = \frac{40}{60} \, \text{hr} = \frac{2}{3} \, \text{hr} \)

Average speed \( = \frac{\text{Total distance}}{\text{Total time}} = \frac{3.75 \, \text{km}}{2/3 \, \text{hr}} = 3.75 \times \frac{3}{2} \, \text{km/h} = 5.625 \, \text{km/h} \)
Average velocity \( = \frac{\text{Total displacement}}{\text{Total time}} = \frac{1.25 \, \text{km}}{2/3 \, \text{hr}} = 1.25 \times \frac{3}{2} \, \text{km/h} = 1.875 \, \text{km/h} \) (towards office)
In simple words: After 40 minutes, he has walked to the office and partially returned. His average speed is 5.625 km/h, and his average velocity is 1.875 km/h towards the office.

🎯 Exam Tip: For average speed, sum up all lengths of travel. For average velocity, calculate the net change in position (displacement) from start to end, considering direction.

Question 9. A man walks 18 m east in 16 s and turns north. Now walks 5 m north in 5 s and then turns left and walks 6 m in 8 s and stops. Calculate the average speed and average velocity during the journey of the man.
Answer:
Let's analyze the man's journey:
1. Walks 18 m East in 16 s.
2. Turns North, walks 5 m North in 5 s.
3. Turns Left (after North, this means West), walks 6 m West in 8 s.

Total Distance Travelled:
Total distance \( = 18 \, \text{m} + 5 \, \text{m} + 6 \, \text{m} = 29 \, \text{m} \)

Total Time Taken:
Total time \( = 16 \, \text{s} + 5 \, \text{s} + 8 \, \text{s} = 29 \, \text{s} \)

Average Speed:
Average speed \( = \frac{\text{Total distance}}{\text{Total time}} = \frac{29 \, \text{m}}{29 \, \text{s}} = 1 \, \text{m/s} \)
The average speed is 1 m/s.

Total Displacement:
Let East be the positive X-axis and North be the positive Y-axis.

• Eastward movement: 18 m (positive X)
• Northward movement: 5 m (positive Y)
• Westward movement: 6 m (negative X)

Net X-displacement \( \Delta x = 18 \, \text{m} - 6 \, \text{m} = 12 \, \text{m} \, (\text{East}) \)
Net Y-displacement \( \Delta y = 5 \, \text{m} \, (\text{North}) \)

The magnitude of the total displacement \( D \) is found using the Pythagorean theorem:
\( D = \sqrt{(\Delta x)^2 + (\Delta y)^2} \)
\( D = \sqrt{(12 \, \text{m})^2 + (5 \, \text{m})^2} \)
\( D = \sqrt{144 + 25} \)
\( D = \sqrt{169} \)
\( D = 13 \, \text{m} \)
The displacement is 13 m in a direction North-East (specifically, \( \tan^{-1}\left(\frac{5}{12}\right) \) North of East).

Average Velocity:
Average velocity \( = \frac{\text{Total displacement}}{\text{Total time}} = \frac{13 \, \text{m}}{29 \, \text{s}} \)
Average velocity \( \approx 0.45 \, \text{m/s} \)
The average velocity is approximately 0.45 m/s in the North-East direction.
In simple words: The man walked a total of 29 meters in 29 seconds, so his average speed is 1 meter per second. However, his final position is only 13 meters away from his start point (towards the North-East), so his average velocity is about 0.45 meters per second in that direction.

🎯 Exam Tip: Clearly differentiate between scalar (distance, speed) and vector (displacement, velocity) quantities. For vectors, always consider both magnitude and direction, and use vector addition for displacement.

Question 10. A uniform accelerated motion object covers 65 m in 5th second and 105 m in 9th. Then how much distance will it cover in 20th second? Also calculate the distance travelled in 20 s.
Answer:
The distance covered in the \( n^{th} \) second for uniformly accelerated motion is given by the formula:
\( S_n = u + \frac{a}{2}(2n - 1) \)
where \( u \) is the initial velocity and \( a \) is the uniform acceleration.

Given:
Distance in 5th second \( S_5 = 65 \, \text{m} \)
\( 65 = u + \frac{a}{2}(2 \times 5 - 1) \)
\( 65 = u + \frac{a}{2}(9) \)
\( 65 = u + \frac{9a}{2} \quad ...(1) \)

Distance in 9th second \( S_9 = 105 \, \text{m} \)
\( 105 = u + \frac{a}{2}(2 \times 9 - 1) \)
\( 105 = u + \frac{a}{2}(17) \)
\( 105 = u + \frac{17a}{2} \quad ...(2) \)

Now, subtract equation (1) from equation (2) to find \( a \):
\( (105) - (65) = \left(u + \frac{17a}{2}\right) - \left(u + \frac{9a}{2}\right) \)
\( 40 = \frac{17a}{2} - \frac{9a}{2} \)
\( 40 = \frac{8a}{2} \)
\( 40 = 4a \)
\( a = \frac{40}{4} \)
\( a = 10 \, \text{m/s}^2 \)
The uniform acceleration is 10 m/s².

Substitute the value of \( a \) into equation (1) to find \( u \):
\( 65 = u + \frac{9(10)}{2} \)
\( 65 = u + 9 \times 5 \)
\( 65 = u + 45 \)
\( u = 65 - 45 \)
\( u = 20 \, \text{m/s} \)
The initial velocity is 20 m/s.

Distance covered in the 20th second \( S_{20} \):
Using the formula \( S_n = u + \frac{a}{2}(2n - 1) \) for \( n = 20 \):
\( S_{20} = 20 + \frac{10}{2}(2 \times 20 - 1) \)
\( S_{20} = 20 + 5(40 - 1) \)
\( S_{20} = 20 + 5(39) \)
\( S_{20} = 20 + 195 \)
\( S_{20} = 215 \, \text{m} \)
The distance covered in the 20th second is 215 m.

Total distance travelled in 20 s:
Using the equation \( S = ut + \frac{1}{2}at^2 \) for \( t = 20 \, \text{s} \):
\( S_{total} = u(20) + \frac{1}{2}a(20)^2 \)
\( S_{total} = (20)(20) + \frac{1}{2}(10)(400) \)
\( S_{total} = 400 + 5(400) \)
\( S_{total} = 400 + 2000 \)
\( S_{total} = 2400 \, \text{m} \)
The total distance travelled in 20 seconds is 2400 m.
In simple words: We first found the starting speed and how fast the object is accelerating. Then we used these to calculate that the object travels 215 meters in the 20th second alone, and a total of 2400 meters in 20 seconds.

🎯 Exam Tip: Remember the specific formula for distance covered in the \( n^{th} \) second and differentiate it from the total distance covered in \( n \) seconds. Solve for \( u \) and \( a \) first from the given information.

Question 11. A gun at an angle of 30° from the horizontal; which falls 3 km away on the ground.
Answer: Let the initial range of the gun be \( R \), and the angle of projection be \( \theta \). The formula for the range of a projectile is given by \( R = \frac{u^2 \sin 2\theta}{g} \), where \( u \) is the initial velocity and \( g \) is the acceleration due to gravity.
Given values are \( R = 3 \text{ km} \) (which is \( 3000 \text{ m} \)) and \( \theta = 30^\circ \).
Substituting these values into the formula:
\( 3 = \frac{u^2 \sin (2 \times 30^\circ)}{g} \)
\( 3 = \frac{u^2 \sin 60^\circ}{g} \)
\( 3 = \frac{u^2}{g} \times \frac{\sqrt{3}}{2} \)
Now, we can find the value of \( \frac{u^2}{g} \):
\( \frac{u^2}{g} = \frac{3 \times 2}{\sqrt{3}} \)
\( \frac{u^2}{g} = \frac{6}{\sqrt{3}} = \frac{6 \sqrt{3}}{3} = 2\sqrt{3} \text{ km} \)
Using \( \sqrt{3} \approx 1.732 \):
\( \frac{u^2}{g} = 2 \times 1.732 = 3.464 \text{ km} \)
The maximum range for a projectile is achieved when \( \sin 2\theta = 1 \), which means \( 2\theta = 90^\circ \) or \( \theta = 45^\circ \). The maximum range \( R_{max} = \frac{u^2}{g} \).
From our calculation, \( R_{max} = \frac{u^2}{g} = 3.464 \text{ km} \).
Since the maximum range the gun can achieve is \( 3.464 \text{ km} \), it cannot be used to hit a target that is 5 km away.
In simple words: We used the given angle and distance to find out how powerful the gun is. Then, we calculated its furthest possible reach, which is about 3.464 km. Since this is less than 5 km, the gun cannot hit a target that far away.

🎯 Exam Tip: Remember the formula for projectile range and how to find the maximum range by setting \( \sin 2\theta \) to 1. Always check the units (km vs. m) for consistency in your calculations.

Question 12. Calculate the projection angle of a projectile whose range is 50 m and maximum height 10m.
Answer: We are given the horizontal range \( R = 50 \text{ m} \) and the maximum height \( H = 10 \text{ m} \). We need to find the projection angle \( \theta \).
The formulas for range and maximum height are:
\( R = \frac{u^2 \sin 2\theta}{g} \) (Equation 1)
\( H = \frac{u^2 \sin^2 \theta}{2g} \) (Equation 2)
To find \( \theta \), we can divide Equation 1 by Equation 2:
\( \frac{R}{H} = \frac{\left(\frac{u^2 \sin 2\theta}{g}\right)}{\left(\frac{u^2 \sin^2 \theta}{2g}\right)} \)
\( \frac{R}{H} = \frac{\sin 2\theta}{\sin^2 \theta / 2} \)
We know that \( \sin 2\theta = 2 \sin \theta \cos \theta \). So, substitute this into the equation:
\( \frac{R}{H} = \frac{2 \sin \theta \cos \theta}{\sin^2 \theta / 2} \)
\( \frac{R}{H} = \frac{4 \sin \theta \cos \theta}{\sin^2 \theta} \)
\( \frac{R}{H} = \frac{4 \cos \theta}{\sin \theta} \)
Since \( \frac{\cos \theta}{\sin \theta} = \cot \theta \), we have:
\( \frac{R}{H} = 4 \cot \theta \)
Now, substitute the given values for \( R \) and \( H \):
\( \frac{50}{10} = 4 \cot \theta \)
\( 5 = 4 \cot \theta \)
\( \cot \theta = \frac{5}{4} = 1.25 \)
To find \( \theta \), we can use \( \tan \theta = \frac{1}{\cot \theta} \):
\( \tan \theta = \frac{1}{1.25} = 0.8 \)
Finally, calculate \( \theta \):
\( \theta = \tan^{-1}(0.8) \)
\( \theta \approx 38.66^\circ \)
So, the projection angle is approximately \( 38.66^\circ \).
In simple words: We used the formulas for how far a projectile goes (range) and how high it reaches (maximum height). By dividing these formulas, we found a simple link between range, height, and the launch angle. Putting in the numbers, we found the angle was about 38.66 degrees.

🎯 Exam Tip: When given both range and maximum height, using the ratio \( \frac{R}{H} = 4 \cot \theta \) is a quick and efficient way to find the projection angle \( \theta \).