Goyal Brothers Solutions for ICSE Class 9 Physics Chapter 2 Motion In One Dimension

Exercise 1

Question 1. What do you understand by the terms
1. rest
2. motion? Support your answer by giving two examples each
Answer:
(i) Rest : A body is said to be at rest if it does not change its position with respect to its immediate surroundings.
For example :
(a) A chair lying in a room.
(b) A stone lying on the ground.
(ii) Motion : A body is said to be in motion if it changes its position with respect to its immediate surroundings.
For example :
(a) A car running on the road.
(b) A football moving on the ground.
In simple words: Rest is when an object stays in the same spot, like a book on a table. Motion is when an object moves from one place to another, like a ball being kicked.

📝 Teacher's Note: Use classroom objects to demonstrate that "surroundings" are the key to deciding if something is moving or not.

🎯 Exam Tip: Always mention "with respect to its immediate surroundings" in your definitions of rest and motion to score full marks.

Question 2. By giving an example, prove that rest and motion are relative terms.
Answer: Rest and motion are relative terms. It means a body at rest in one situation at certain time can be in motion in another situation at the same time.
A person sitting in the compartment of a moving train is in the state of rest with respect to the surrounding of the compartment yet he is in state of motion, if he compare himself with the surroundings outside the compartment.
In simple words: Imagine you are sitting in a moving car. To your friend sitting next to you, you aren't moving. But to someone standing on the sidewalk, you are zooming past!

📝 Teacher's Note: Use the "Bus/Train" analogy as it is the most intuitive way for students to understand frames of reference.

🎯 Exam Tip: Keywords like "relative terms" and "frame of reference" are what examiners look for in this explanation.

Question 3. Define :
1. Scalar quantities.
2. Vector quantities. Give two differences between scalar and vector quantities.
Answer:
(i) Scalar quantities : The physical quantities which are expressed in magnitude only are called scalar quantities.
For example: Mass, length, time, distance, density, energy etc.
(ii) Vector quantities : The physical quantities which are expressed in magnitude as well as direction are called vector quantities.
For example : Displacement, velocity, acceleration, momentum, force etc.
Differences between scalar and vector quantities:

In simple words: A scalar is just a number, like "5 kg". A vector is a number with a direction, like "5 km North".

📝 Teacher's Note: Explain that "magnitude" simply means the "size" or "amount" of something.

🎯 Exam Tip: Remember that vectors can change even if the speed stays the same, as long as the direction changes!

Question 4. Pick out the scalar and vector quantities from the following list:
1. mass
2. density
3. displacement
4. distance
5. momentum
6. acceleration
7. temperature
8. time
Answer:
1. Scalar quantities : Mass, density, distance, temperature, time.
2. Vector quantities : Displacement, momentum, acceleration.
In simple words: Things like time and mass don't have a "direction", so they are scalars. Things like acceleration do have a direction, so they are vectors.

📝 Teacher's Note: Ask students: "Does it make sense to say 5 o'clock North?" If not, it's a scalar.

🎯 Exam Tip: Distance and displacement are often confused. Remember: Distance is scalar, Displacement is vector.

Question 5. Define : (i) Speed (ii) Velocity. Give two differences between speed and velocity.
Answer:
(i) Speed : The distance covered by a body in a unit time is called its speed. It is also defined as the rate of change position of a body in any direction.
\[ \text{speed} = \frac{\text{distance}}{\text{time}} \]
(ii) Velocity : "Rate of change of displacement with time" is called velocity, or "The time rate of change of displacement of an object" is called the velocity.
\[ V = \frac{\text{distance}}{\text{time}} \]
(Note: Velocity is Displacement/Time).
Differences between speed and velocity:
Speed :
1. The rate of change of position of a body in any direction is known as its speed.
2. It is a scalar quantity.
3. It can be positive or zero.
Velocity :
1. The rate of change of position of a body in a particular direction is known velocity.
2. It is a vector quantity.
3. It can be positive, negative or zero.
In simple words: Speed is how fast you are going. Velocity is how fast you are going AND in which direction.

📝 Teacher's Note: Illustrate that a car driving at 60 km/h in a circle has a constant speed but a changing velocity.

🎯 Exam Tip: While calculating, always check if units are consistent (e.g., all in m/s or all in km/h).

Question 6. Define (i) Distance (ii) Displacement. Give two differences between displacement and distance.
Answer:
(i) Distance : The length of path travelled by a body in certain interval of time is called distance.
(ii) Displacement : of an object between two points is the shortest distance between these two points. "It is the unique path which can take the body from its initial to final position."
Differences between displacement and distance:
Distance :
1. It is a scalar quantity.
2. Distance travelled is always positive.
3. The distance travelled by a moving body is the actual length of path.
4. Distance travelled is always greater than or equal to the displacement.
Displacement :
1. It is a vector quantity.
2. Displacement may be positive negative or zero.
3. The displacement of a body is the shortest distance between the initial and final positions of the body.
4. Displacement is always less than or equal to the distance travelled.
In simple words: Distance is the total ground you covered. Displacement is how far out of place you are from where you started.

📝 Teacher's Note: Draw a zig-zag path on the board. The zig-zag length is distance; the straight line from start to finish is displacement.

🎯 Exam Tip: Remember: Displacement can be zero if you return to your starting point, but distance cannot be zero if you moved.

Question 7. By giving one example each, define
1. variable velocity.
2. average velocity.
3. uniform velocity.
Answer:
(i) Variable velocity : When a body covers unequal distances in equal intervals of time in a specified direction, the body is said to be moving with a variable velocity.
Example : A rotating fan at a constant speed has variable velocity, because of continuous change in direction.
(ii) Average velocity : The ratio of the total distance travelled in a specified direction to the total time taken by the body to travel the distance is called average velocity.
Example : If you walk to a campsite 1 km away and then back to your starting point with in 1 hour, then your average velocity will be zero because your initial and final position is same.
(iii) Uniform velocity : When a body covers equal distances in equal intervals of time (however small may be the time interval) in a specified direction, the body is laid to be moving with uniform velocity.
Example : A car moving on a straight road with constant speed has uniform velocity.
In simple words: Uniform velocity is like a car on a highway with cruise control on. Variable velocity is like a car in city traffic. Average velocity looks at your whole trip from start to finish.

📝 Teacher's Note: Emphasize that "uniform" requires both speed and direction to be constant.

🎯 Exam Tip: For average velocity, always use the formula \( \frac{\text{Total Displacement}}{\text{Total Time}} \).

Question 8. What do you understand by the term acceleration? When is the acceleration
1. positive
2. negative?
Answer:
• Acceleration : "Rate of change of velocity with respect to time" is called acceleration. S.I. unit of acceleration is \( \text{m s}^{-2} \).
• Positive acceleration : When velocity of a body increases with time, then acceleration of the body is said to be positive acceleration.
• Negative acceleration : When the velocity of a body decreases with time, then acceleration of the body is said to be negative acceleration.
In simple words: Acceleration is when you speed up (positive) or slow down (negative).

📝 Teacher's Note: Negative acceleration is also called "retardation" or "deceleration".

🎯 Exam Tip: Always include the S.I. unit \( \text{m/s}^2 \) or \( \text{ms}^{-2} \) when defining acceleration.

Question 9. Define the term acceleration due to gravity. State its value in C.GS. as well as in S.I. system. When is acceleration due to gravity
1. positive
2. negative?
Answer:
Acceleration due to gravity : The acceleration of a freely falling body, under the action of gravity of earth, is called acceleration due to gravity. It is represented by ‘g’.
In S.I. system, \( g = 9.8 \text{ ms}^{-2} \); In C.GS. system, \( g = 980 \text{ cms}^{-2} \).
If a body falls towards the earth, then value of acceleration due to gravity is positive.
If the body rises vertically upwards, then value of acceleration due to gravity is negative.
In simple words: When you drop a ball, gravity pulls it down and makes it faster and faster. This "pulling power" is \( g \). It is negative when you throw something up because it slows down.

📝 Teacher's Note: Remind students that \( g \) is approximately \( 10 \text{ m/s}^2 \) for easier calculations, but they must use 9.8 in exams unless told otherwise.

🎯 Exam Tip: Don't forget that \( g \) acts downwards. Upward motion = negative \( g \), Downward motion = positive \( g \).

Question 10. Give an example of a body which covers a certain distance, but its displacement is zero.
Answer: Yes, displacement of a body can be zero even if the distance covered by it is not zero.
For example, if a body moves in a circle, then displacement of the body in one rotation is zero but the distance covered by it in one rotation = \( 2\pi r \), where r is the radius of the circle in which the body is moving.
In simple words: If you run one full lap on a circular track, you've run a long distance, but you ended up exactly where you started, so your displacement is zero.

📝 Teacher's Note: This is a classic concept test. Ask students about their daily trip to school and back home to verify displacement vs distance.

🎯 Exam Tip: The circular path is the most standard example for this question. Make sure to mention "one complete rotation".

Question 11. Give an example of an accelerated body, moving with a uniform speed.
Answer: When a body moves over a circular path with constant speed, then body is said to be accelerated. Acceleration is produced in the body due to continuous change in its direction.
In simple words: Imagine a car going around a roundabout at exactly 20 mph. Even though the speedometer doesn't change, the car is accelerating because it's constantly turning.

📝 Teacher's Note: This helps clarify that velocity is a vector. Changing direction = changing velocity = acceleration.

🎯 Exam Tip: Uniform circular motion is the only case where speed is constant but acceleration is present.

Question 12. What is the relation between distance and time when
1. body is moving with uniform velocity
2. body is moving with variable velocity?
Answer:
1. When body moves with uniform velocity, then body covers equal distance in equal interval of time in a specified direction.
2. When a body moves with variable velocity, then body covers unequal distance in equal interval of time in a specified direction.
In simple words: Uniform means steady timing. Variable means you might go fast for one second and slow for the next.

📝 Teacher's Note: Relate this to a distance-time graph: uniform velocity is a straight line, variable velocity is a curve.

🎯 Exam Tip: "Equal distance in equal intervals of time" is the standard phrase for uniform motion.

Question 13. (a) Distinguish between scalar and vector quantities,
(b) State whether following are scalar or vector quantities
1. speed
2. force
3. acceleration
4. energy
Answer:
(a) See Question 3 of Exercise-1
(b)
1. Speed is a scalar quantity because it has magnitude and no direction.
2. Force is a vector quantity because it has both magnitude as well as direction.
3. Acceleration is a vector quantity.
4. Energy is a scalar quantity because it has magnitude and no direction.
In simple words: If you need to know "which way" it's going (like Force), it's a vector. If direction doesn't matter (like your age or energy), it's a scalar.

📝 Teacher's Note: Energy can be confusing for students. Explain that energy is the *capacity* to do work and doesn't point anywhere.

🎯 Exam Tip: Force is always a vector. If you push an object, you must push it *somewhere*.

Question 14. Copy the following table and fill in the blank spaces.

Answer: (Values filled in the table above).
In simple words: Displacement is measured in metres (like length) but is a vector. Density is how heavy something is for its size and is a scalar.

📝 Teacher's Note: Density is a ratio of two scalars (mass and volume), so it must be a scalar.

🎯 Exam Tip: S.I. units are crucial. Remember for density it is \( \text{kg/m}^3 \).

Question 15. Draw a diagram to show the motion of a body whose speed remains constant, but velocity changes continuously.
Answer: A body moving in a circle with constant speed but variable velocity as the velocity of the body at any point is along the tangent to the circle at that point as shown in the figure.
In simple words: Think of a stone tied to a string being spun around. It moves at the same speed, but it's always changing its direction.

📝 Teacher's Note: This is a standard diagram for "uniform circular motion". The velocity is always perpendicular to the radius.

🎯 Exam Tip: Use the word "tangent" to describe the direction of velocity at any point on the circle.

Practice Problems 1

Question 1. A car covers 90 km in 1 1/2 hours towards east. Calculate
(i) displacement of car,
(ii) its velocity in (a) \( \text{kmh}^{-1} \)
(b) \( \text{ms}^{-1} \).
Answer:
(i) A car covers 90 km in \( 1 \frac{1}{2} \) hour i.e. \( \frac{3}{2} \) hour towards east.
\( \therefore \text{Displacement} = 90 \text{ km} - \text{east} \)
(ii) Time \( = 1 \frac{1}{2} \text{ hour} = \frac{3}{2} \text{ hour} \)
\[ \text{Velocity} = \frac{\text{Displacement}}{\text{Time}} \]
\( v = \frac{90}{3/2} \)
\( v = \frac{90 \times 2}{3} = 60 \text{ km/h} \)
\( v = 60 \times \frac{1000}{3600} \text{ m/s} \)
\( v = \frac{100}{6} \text{ m/s} = 16.67 \text{ ms}^{-1} \)
In simple words: The car travels 90 km East. Its velocity is 60 km per hour East. To get the speed in meters per second, we use a conversion trick.

📝 Teacher's Note: Teach the shortcut: multiply by \( \frac{5}{18} \) to convert km/h to m/s.

🎯 Exam Tip: Don't forget to write "East" in your velocity answer—velocity must have a direction!

Question 2. A race horse runs straight towards north and covers 540 m in one minute. Calculate (i) displacement of horse, (ii) its velocity in (a) \( \text{ms}^{-1} \) (b) \( \text{kmh}^{-1} \).
Answer: A race horse runs straight towards north and covers 540 m in one minute.
1. Displacement = 540 m – north
2. Time = 1 minute = 60 s
\[ \text{Velocity} = \frac{\text{Displacement}}{\text{Time}} \]
\( v = \frac{540}{60} = 9 \text{ ms}^{-1} \)
\( v = 9 \times \frac{3600}{1000} \text{ km/h} = \frac{324}{10} \text{ km/h} \)
\( v = 32.4 \text{ km/h or kmh}^{-1} \)
In simple words: The horse moves 9 meters every second toward the North. This is the same as going 32.4 kilometers in an hour.

📝 Teacher's Note: Remind students to convert time (minutes) into S.I. units (seconds) before starting the calculation.

🎯 Exam Tip: To convert m/s back to km/h, multiply by \( \frac{18}{5} \) or 3.6.

Practice Problems 2

Question 1. The change in velocity of a motor bike is \( 54 \text{ kmh}^{-1} \) in one minute. Calculate its acceleration in (a) \( \text{ms}^{-2} \) (b) \( \text{kmh}^{-2} \).
Answer:
Change in velocity \( = 54 \text{ kmh}^{-1} = 54 \times \frac{1000}{3600} \text{ ms}^{-1} = 15 \text{ ms}^{-1} \)
Time \( = 1 \text{ minute} = 60 \text{s} \)
\[ \text{Acceleration (a)} = \frac{\text{Change in velocity}}{\text{Time}} \]
\( a = \frac{15}{60} \text{ ms}^{-2} \)
\( a = 0.25 \text{ ms}^{-2} \)
\( a = 0.25 \times \frac{3600 \times 3600}{1000} \text{ kmh}^{-2} = 3240 \text{ kmh}^{-2} \)
In simple words: The bike is picking up speed at a rate of 0.25 meters per second every second.

📝 Teacher's Note: "Change in velocity" is \( v - u \). Since it's given directly here, we don't need to subtract.

🎯 Exam Tip: \( \text{ms}^{-2} \) is the same as \( \text{m/s}^2 \). Both notations are acceptable.

Question 2. A speeding car changes its velocity from \( 108 \text{ kmh}^{-1} \) to \( 36 \text{ kmh}^{-1} \) in 4 s. Calculate its deceleration in
1. \( \text{ms}^{-2} \)
2. \( \text{kmh}^{-2} \).
Answer:
Initial velocity \( = u = 108 \text{ kmh}^{-1} = 108 \times \frac{5}{18} \text{ m/s} = 30 \text{ ms}^{-1} \)
Final velocity \( = v = 36 \text{ kmh}^{-1} = 36 \times \frac{5}{18} \text{ m/s} = 10 \text{ ms}^{-1} \)
Time \( = t = 4 \text{ S} \)
\[ \text{Deceleration} = a = \frac{v - u}{t} \]
\( a = \frac{10 - 30}{4} \)
\( a = -\frac{20}{4} = -5 \text{ ms}^{-2} \)
Deceleration is 5 \( \text{ms}^{-2} \).
\( a = \frac{-5 \times 3600 \times 3600}{1000} \text{ kmh}^{-2} = -64800 \text{ kmh}^{-2} \)
In simple words: The car is slamming on the brakes, slowing down by 5 meters per second every second.

📝 Teacher's Note: Deceleration is just negative acceleration. If the question asks for "deceleration", give the positive magnitude (5). If it asks for "acceleration", give the negative sign (-5).

🎯 Exam Tip: Always subtract Initial from Final (\( v - u \)). If velocity decreases, your answer *must* come out negative.

UNIT II - EXERCISE 2

(A) Objective Questions

Question 1. A graph is a straight line parallel to the time axis in a distance – time graph. From the graph, it implies:
(a) body is stationary
(b) body is moving with a uniform speed
(c) body is moving with a variable speed
(d) None of the options
Answer: (a) body is stationary
In simple words: If the graph line is flat, the distance isn't changing as time passes. This means the object is just sitting still.

📝 Teacher's Note: Draw a flat line on a distance-time graph to show that at \( t=1, t=2, t=3 \), the distance stays at, say, 5 meters.

🎯 Exam Tip: Parallel to the time axis on a *distance-time* graph means speed is zero. On a *velocity-time* graph, it would mean uniform speed.

Question 2. The slope of a displacement – time graph represents :
(a) uniform speed
(b) non-uniform speed
(c) uniform velocity
(d) uniform acceleration
Answer: (c) uniform velocity
In simple words: "Slope" means steepness. A steeper line on this graph means the object is covering more distance in less time, which means more velocity.

📝 Teacher's Note: Remind students: Slope of displacement-time = Velocity. Slope of velocity-time = Acceleration.

🎯 Exam Tip: If the graph is a straight line, the slope is constant, so the velocity is uniform.

Question 3. A body dropped from the top of a tower reaches the ground in 4s. Height of the tower is :
(a) 39.2 m
(b) 44.1 m
(c) 58.8 m
(d) 78.4 m
Answer: (d) 78.4 m
Explanation :
Initial velocity \( = u = 0 \)
Time \( = t = 4 \text{s} \)
As body falls downwards, so acceleration \( = a = +g = +9.8 \text{ ms}^{-2} \)
(Distance) \( S = ut + \frac{1}{2} at^2 \)
\( S = (0)(4) + \frac{1}{2} \times 9.8 \times (4)^2 \)
\( s = 0 + \frac{1}{2} \times 9.8 \times 16 \)
\( S = 9.8 \times 8 = 78.4 \text{ m} \)
In simple words: Since the body was dropped, it started with zero speed. We use the gravity formula to find how far it fell in 4 seconds.

📝 Teacher's Note: "Dropped" always implies that initial velocity \( u = 0 \).

🎯 Exam Tip: Memorize the equations of motion; they are the foundation for almost every mechanics problem.

Question 4. The speed of a car reduces from 15 m/s to 5 m/s over a displacement of 10 m. The uniform acceleration of the car is :
(a) \( -10 \text{ m/s}^2 \)
(b) \( +10 \text{ m/s}^2 \)
(c) \( 2 \text{ m/s}^2 \)
(d) \( 0.5 \text{ m/s}^2 \)
Answer: (a) -10 m/s²
Explanation :
Initial velocity \( = u = 15 \text{ m/s} \)
Final velocity \( = v = 5 \text{ m/s} \)
Displacement \( = S = 10 \text{ m} \)
Acceleration \( = a = ? \)
\( v^2 - u^2 = 2aS \); \( (5)^2 - (15)^2 = 2a (10) \)
\( 25 - 225 = 20a \)
\( a = -220/20 = -10 \text{ ms}^{-2} \)
In simple words: The car slowed down, so the acceleration is negative. We use the third equation of motion because we don't know the time.

📝 Teacher's Note: Choosing the right equation of motion (the one without 't') is the first step to solving this quickly.

🎯 Exam Tip: Be careful with the squares: \( 15^2 \) is 225. A common mistake is using \( 15 \times 2 = 30 \).

Question 5. A body projected vertically up with a velocity 10 m/s reaches a height of 20 m. If it is projected with a velocity of 20 m/s, then the maximum height reached by the body is :
(a) 20 m
(b) 10 m
(c) 80 m
(d) 40 m
Answer: (c) 80 m
Explanation :
Case-I:
Initial velocity \( = u = 10 \text{ m/s} \)
Height \( = \text{Distance} = S = 20 \text{ m} \)
At highest point, Final velocity \( = v = 0 \)
\[ v^2 - u^2 = 2aS \); \( (0)^2 - (10)^2 = 2a (20) \)
\( a = -\frac{100}{40} = -2.5 \text{ ms}^{-2} \)
Case-II:
\( u = 20 \text{ m/s} \); \( v = 0 \)
\( a = -2.5 \text{ ms}^{-2} \); \( S = h = ? \)
\( v^2 - u^2 = 2aS \)
\( (0)^2 - (20)^2 = 2(-2.5)h \)
\( h = \frac{400}{5} = 80 \text{ m} \)
In simple words: Height depends on the square of the speed. If you double the speed (from 10 to 20), the height becomes 4 times greater (\( 2 \times 2 = 4 \)), so \( 20 \times 4 = 80 \).

📝 Teacher's Note: This is a ratio problem. \( H \propto u^2 \). Using ratios is often faster than calculating 'a'.

🎯 Exam Tip: Remember that for vertical motion, velocity at the very top is always zero.

Question 6. What does the area of an acceleration – time graph represent?
(a) Uniform velocity
(b) Displacement
(c) Distance
(d) Change in velocity
Answer: (d) Change in velocity
In simple words: Area usually means multiplication. Acceleration times time equals change in velocity.

📝 Teacher's Note: Just like Area under Velocity-Time graph is Displacement, Area under Accel-Time graph is Velocity.

🎯 Exam Tip: This is a direct factual question. Memorize: Area under V-T = Displacement; Area under A-T = Velocity change.

Question 7. A driver applies brakes when he sees a child on the railway track, the speed of the train reduces from 54 km/h to 18 km/h in 5 s. What is the distance travelled by the train during this interval of time?
(a) 52 m
(b) 50 m
(c) 25 m
(d) 80 m
Answer: (b) 50 m
Explanation :
\( u = 54 \text{ km/h} = 54 \times \frac{5}{18} \text{ m/s} = 15 \text{ m/s} \)
\( v = 18 \text{ km/h} = 18 \times \frac{5}{18} \text{ m/s} = 5 \text{ m/s} \)
\( t = 5 \text{ S} \); \( S = ? \)
\( a = \frac{v - u}{t} = \frac{5 - 15}{5} = -2 \text{ m/s}^2 \)
\( v^2 - u^2 = 2aS \)
\( (5)^2 - (15)^2 = 2(-2)S \)
\( S = \frac{25 - 225}{-4} = \frac{-200}{-4} = 50 \text{ m} \)
In simple words: We first turn the speeds into meters per second. Then we find out how hard the train slowed down and use that to find the distance.

📝 Teacher's Note: You can also use \( S = \frac{(u+v)}{2} \times t \) for a quicker answer if acceleration is uniform.

🎯 Exam Tip: Always convert km/h to m/s *immediately* if time is given in seconds.

Question 8. In velocity time graph, the acceleration is :
(a) – 4 \( \text{m/s}^2 \)
(b) 4 \( \text{m/s}^2 \)
(c) 10 \( \text{m/s}^2 \)
(d) zero+
Answer: (a) -4 m/s²
Explanation :
Acceleration = Slope of velocity time graph
\( a = -\frac{20}{5} \) [Velocity is decreasing with time]
\( a = -4 \text{ m/s}^2 \) [Acceleration will be negative]
In simple words: The graph shows velocity dropping from 20 to 0 in 5 seconds. So it loses 4 meters per second every second.

📝 Teacher's Note: A downward-sloping line on a V-T graph always means negative acceleration.

🎯 Exam Tip: Slope is \( \frac{\text{Rise}}{\text{Run}} \). Here, Rise is -20 and Run is 5.

Question 9. The distance covered in adjoining velocity – time graph is :
(a) 25 m
(b) 40 m
(c) 50 m
(d) 45 m
Answer: (c) 50 m
Explanation :
Distance covered = Area under velocity – time graph
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)
\( \text{Area} = \frac{1}{2} \times 5 \times 20 = 50 \text{ m} \)
In simple words: To find the distance from a speed graph, you calculate the area of the shape under the line. Here it's a triangle.

📝 Teacher's Note: The area of the triangle represents the total ground covered during the 5 seconds of braking.

🎯 Exam Tip: Don't forget the \( \frac{1}{2} \) in the triangle area formula!

Question 10. At the maximum height, a body thrown vertically upwards has :
(a) velocity not zero but acceleration zero.
(b) acceleration not zero but velocity zero.
(c) both acceleration and velocity are zero.
(d) both acceleration and velocity are not zero.
Answer: (b) acceleration not zero but velocity zero.
In simple words: At the very top, the ball stops for a split second (velocity = 0), but gravity is still pulling it down (acceleration = 9.8). If gravity were zero, the ball would float away!

📝 Teacher's Note: This is a very important conceptual trap. Acceleration due to gravity is constant throughout the flight.

🎯 Exam Tip: Velocity can be zero while acceleration is non-zero (turning point). Acceleration can be zero while velocity is non-zero (uniform motion).

(B) Subjective Questions

Question 1. Draw displacement – time graphs for the following situations :
1. When a body is stationary.
2. When a body is moving with uniform velocity.
3. When a body is moving with variable velocity.
Answer:
(i) Displacement – time graph when body is stationary:
(ii) Displacement – time graph when body is moving with uniform velocity:
(iii) Displacement – time graph when body is moving with variable velocity:
In simple words: A flat line means not moving. A straight sloped line means moving at a steady pace. A curved line means your speed is changing.

📝 Teacher's Note: Show that the steeper the straight line, the faster the uniform velocity is.

🎯 Exam Tip: Always label your axes with quantity and units, even in sketches.

Question 2. Draw velocity – time graphs for the following situations :
1. When a body is moving with uniform velocity.
2. When a body is moving with variable velocity, but uniform acceleration.
3. When a body is moving with variable velocity, but uniform retardation.
4. When a body is moving with variable velocity and variable acceleration.
Answer:
Following are the velocity – time graph :
(i) When a body is moving with uniform velocity : Then velocity time graph is AB is a straight line parallel to time axis.
(ii) When a body is moving with variable velocity, but uniform acceleration : Then velocity – time graph OA is a straight line inclined to time axis.
(iii) When a body is moving with variable velocity, but uniform retardation :
(iv) When a body is moving with variable velocity and variable acceleration :
In simple words: For speed graphs, a flat line means you aren't speeding up. A line going up means steady acceleration. A line going down means steady slowing down. A curve means the acceleration itself is changing.

📝 Teacher's Note: Compare these to the displacement-time graphs. A flat line means something very different in each!

🎯 Exam Tip: Uniform acceleration = Straight line with positive slope. Uniform retardation = Straight line with negative slope.

Question 3. How can you find the following?
1. Velocity from a displacement – time graph.
2. Acceleration from velocity – time graph.
3. Displacement from velocity – time graph.
4. Velocity from acceleration – time graph.
Answer:
(i) Velocity from a displacement – time graph : We can find the velocity by knowing the slope of the displacement-time graph. For uniform motion (straight line OP):
\[ \text{Slope} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1} = \text{Velocity of the body} \]
(ii) Acceleration from velocity – time graph : We can find the acceleration of the body by knowing the slope of the velocity-time graph.
\[ \text{Slope} = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1} = \text{Acceleration of the body} \]
(iii) Displacement from velocity-time graph : Displacement covered by a body is equal to the area under the velocity – time graph.
\[ \text{Area of rectangle ABCD} = \text{Length} \times \text{Breadth} = v \times (t_2 - t_1) = \text{Displacement} \]
(iv) Velocity from acceleration – time graph : Area under the acceleration – time graph gives the change in velocity of the body.
\[ \text{Area} = a \times (t_2 - t_1) = v - u \]
If \( u = 0 \), then \( \text{Area} = v = \text{velocity} \).
In simple words: To find a "rate" (velocity/acceleration), find the steepness (slope). To find a "total" (distance/velocity), find the area under the line.

📝 Teacher's Note: Teach the "Graph Ladder": Displacement \( \xrightarrow{\text{slope}} \) Velocity \( \xrightarrow{\text{slope}} \) Acceleration. Going down the ladder (Acceleration \( \xrightarrow{\text{area}} \) Velocity \( \xrightarrow{\text{area}} \) Displacement) uses area.

🎯 Exam Tip: Always state "Slope represents..." or "Area represents..." before starting a graphical calculation.

Question 4. What do you understand by the term acceleration due to gravity? What is its value in C.G.S. and S.I. systems?
Answer:
Acceleration due to gravity : The acceleration of a freely falling body, under the action of gravity of earth, is called acceleration due to gravity. It is represented by ‘g’.
In S.I. system, \( g = 9.8 \text{ ms}^{-2} \); In C.G.S. system, \( g = 980 \text{ cms}^{-2} \).
If a body falls towards the earth, then value of acceleration due to gravity is positive. If the body rises vertically upwards, then value of acceleration due to gravity is negative.
In simple words: This is the speed a falling object gains every second just because Earth is pulling it.

📝 Teacher's Note: Use the "coin and feather" vacuum experiment video link to show that \( g \) is the same for all masses.

🎯 Exam Tip: Mention the direction of motion relative to Earth when discussing the sign (+ or -) of \( g \).

Question 5. Can you suggest about the kind of motion of a body from the following distance – time graphs?
Answer:
(a) Distance – time graph in figure (a) shows that body is stationary.
(b) Distance – time graph figure (b) shows that body is in uniform motion.
(c) Distance – time graph in figure (c) shows that initially body is in uniform motion i.e. it covers equal distance in equal intervals of time. From O to A, body is in uniform motion and from A to B, body is at rest.
In simple words: (a) stays at one distance, (b) moves steadily away, and (c) moves steadily then stops.

📝 Teacher's Note: Point out that in graph (c), the "kink" at point A means a sudden change in state from moving to stopped.

🎯 Exam Tip: For graph (b), specify "uniform motion" as it's a straight line starting from the origin.

Question 6. Can you suggest real life examples about the motion of a body from the following velocity – time graphs?
Answer:
(a) A car running on a road at constant velocity.
(b) A car starting from rest and moving with uniform acceleration (velocity increases equally in equal intervals).
(c) A train starting from rest, accelerating, moving at constant velocity, and then slowing down to a stop.
(d) A moving vehicle coming to a stop in front of a house (velocity decreases with time).
(e) A ball is thrown vertically upward and it returns back to earth after sometime.
In simple words: (a) cruise control, (b) flooring the gas, (c) a normal subway trip, (d) hitting the brakes, (e) throwing a ball in the air.

📝 Teacher's Note: Graph (e) is special—the velocity hits zero at the top and then becomes negative as the ball falls back down.

🎯 Exam Tip: For graph (c), break it into three parts: acceleration, constant velocity, and retardation.

Question 7. Diagram shows a velocity – time graph for a car starting from rest. The graph has three section AB, BC and CD.
1. From a study of this graph, state how the distance travelled in any section is determined.
2. Compare the distance travelled in section BC with the distance travelled in section AB.
3. In which section, car has a zero acceleration?
4. Is the magnitude of acceleration higher or lower than that of retardation? Give a reason.
Answer:
(i) The area under that part above the x-axis gives distance travelled.
(ii) \[ \frac{\text{Distance travelled in part BC}}{\text{Distance travelled in part AB}} = \frac{v_0 \times t}{\frac{1}{2} \times t \times v_0} = \frac{2}{1} = 2:1 \]
(iii) Car has zero acceleration in section BC because there is no change in velocity from B to C.
(iv) \[ \text{Acceleration} = \text{slope of AB} = \frac{v_0}{t} \]
\[ \text{Retardation} = \text{slope of CD} = \frac{v_0}{t/2} = 2\frac{v_0}{t} \]
\( \because \) Magnitude of retardation is twice the magnitude of acceleration.
\( \therefore \) Magnitude of acceleration is lower.
In simple words: The car takes longer to speed up than to slow down. The steady middle part covers twice as much ground as the speeding-up part.

📝 Teacher's Note: Use the "slope" calculation to prove that a shorter time for the same velocity change means higher acceleration/retardation.

🎯 Exam Tip: To compare distances, find the area of the triangle (AB) and rectangle (BC) and divide them.

Question 8. Write down the type of motion of a body along the A – O – B in each of the following distance – time graphs.
Answer:
1. Body is stationary.
2. Body is in uniform motion i.e. it covers equal distance in equal intervals of time.
3. From A to O, body is in uniform motion having positive slope and from O to B, body is in uniform motion having negative slope (returning to start).
In simple words: 1 is still, 2 goes away, 3 goes away and then comes back.

📝 Teacher's Note: Graph 3 is a common way to show a round trip. Point O is the furthest point from the start.

🎯 Exam Tip: Negative slope on a distance-time graph means the object is moving back towards the reference point.

Practice Problems 1

Question 1. From the displacement – time graph shown given below calculate : –
1. Average velocity in first three seconds.
2. Displacement from initial position at the end of 13 s.
3. Time after which the body is at the initial position,
4. Average velocity after 8 s.
Answer:
(i) In first three seconds:
Total displacement = 8m; Total time = 3 s
Average velocity = \( \frac{8}{3} = 2.67 \text{ ms}^{-1} \)
(ii) Displacement at the end of 13s = -8m.
(iii) Body is at initial position after 8s and 17s (where displacement is zero).
(iv) Average velocity after 8s is zero because at 8s, the displacement is zero.
In simple words: Look at where the line is on the Y-axis at each time to find the answers. If the line crosses the zero mark, the object is back home.

📝 Teacher's Note: This problem teaches students how to read coordinates directly from a physical graph.

🎯 Exam Tip: For "initial position," look for where the graph line intersects the time (X) axis.

Question 2. From the displacement – time graph shown given below calculate :
1. Velocity between 0 – 2 s.
2. Velocity between 8 s – 12 s.
3. Average velocity between 5 s – 12 s.
Answer:
(i) Velocity between 0 – 2 s \( = \frac{-(25 - 10)}{2} = -7.5 \text{ ms}^{-1} \) [Note: calculation from graph values].
(ii) Velocity between 8 s – 12 s \( = \frac{25 - 20}{12 - 8} = \frac{5}{4} = 1.25 \text{ ms}^{-1} \)
(iii) Average velocity between 5 s – 12 s \( = \frac{\text{Total displacement}}{\text{Time}} = \frac{25 - 10}{12 - 5} = \frac{15}{7} = 2.1 \text{ ms}^{-1} \)
In simple words: We pick the start and end points for each time gap and find the slope.

📝 Teacher's Note: Remind students that average velocity for a specific interval only depends on the endpoints, not what happened in between.

🎯 Exam Tip: Always show the subtraction of coordinates (\( y_2-y_1 / x_2-x_1 \)) clearly.

Practice Problems 2

Question 1. A train starting from rest, picks up a speed of \( 20 \text{ ms}^{-1} \) in 200 s. It continues to move at the same rate for next 500s, and is then brought to rest in another 100 s.
(i) Plot a speed-time graph.
(ii) From graph calculate
(a) uniform rate of acceleration
(b) uniform rate of retardation
(c) total distance covered before stopping
(d) average speed.
Answer:
(i)
(ii) (a) Acceleration \( = \frac{20}{200} = 0.1 \text{ ms}^{-2} \)
(b) Retardation \( = \frac{20}{100} = 0.2 \text{ ms}^{-2} \)
(c) Distance \( = \text{Area of Trapezium} = \frac{1}{2} \times (500 + 800) \times 20 = 13000 \text{ m} = 13 \text{ km} \)
(d) Average speed \( = \frac{13000}{800} = 16.25 \text{ ms}^{-1} \)
In simple words: The train speeds up slowly (0.1), stays steady, then slows down faster (0.2). It covers 13 kilometers total.

📝 Teacher's Note: The "next 500s" means the time becomes 200 + 500 = 700s. Many students incorrectly label the x-axis as 200, 500, 100.

🎯 Exam Tip: For distance, the area of a trapezium is easier than adding the triangle and rectangle separately: \( \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} \).

Question 2. A ball is thrown up vertically, and returns back to thrower in 6 s. Assuming there is no air friction, plot a graph between velocity and time. From the graph calculate :
1. deceleration
2. acceleration
3. total distance covered by ball
4. average velocity.
Answer: Total time = 6s. Upward trip = 3s, Downward trip = 3s.
For upward motion: \( v = u + at \implies 0 = u - 10(3) \implies u = 30 \text{ m/s} \).
(i) Deceleration (upwards) \( = -\text{slope} = \frac{30}{3} = 10 \text{ ms}^{-2} \).
(ii) Acceleration (downwards) \( = \text{slope} = \frac{30}{3} = 10 \text{ ms}^{-2} \).
(iii) Distance \( = \text{Area of two triangles} = (\frac{1}{2} \times 3 \times 30) \times 2 = 45 + 45 = 90 \text{ m} \).
(iv) Average velocity \( = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{0}{6} = 0 \).
In simple words: The ball goes up 45m and falls back 45m. Its total travel is 90m, but it ends where it started, so the average velocity is zero.

📝 Teacher's Note: Explain that since air friction is ignored, the motion is perfectly symmetrical.

🎯 Exam Tip: Total distance is the *sum* of upward and downward trips, but total displacement for a return trip is always zero.

Question 3. A racing car is moving with a velocity of 50 m/s. On applying brakes, it is uniformly retarded and comes to rest in 20 seconds. Calculate its acceleration.
Answer:
\( u = 50 \text{ m/s} \); \( v = 0 \); \( t = 20 \text{s} \)
\( v = u + at \)
\( 0 = 50 + a(20) \)
\( -20a = 50 \)
\( a = \frac{-50}{20} = -2.5 \text{ ms}^{-2} \)
In simple words: The car slows down by 2.5 meters per second every single second until it stops.

📝 Teacher's Note: The negative sign confirms the car is slowing down (retardation).

🎯 Exam Tip: Always state your givens (\( u, v, t \)) before picking an equation.

Question 4. A body falls freely downward from a certain height. Show graphically the relation between the distance fallen and square of time. How will you determine ‘g’ from the graph?
Answer: For a freely falling body (\( u=0 \)), \( S = \frac{1}{2} gt^2 \). This means \( S \propto t^2 \).
The graph of \( S \) vs \( t^2 \) is a straight line.
By knowing the slope of the displacement (\( S \)) Vs. square of time (\( t^2 \)) graph, we can find \( g \):
\( \text{Slope} = \frac{g}{2} \implies g = 2 \times \text{Slope} \).
In simple words: If you plot distance against time-squared, you get a straight line. Gravity is just twice the steepness of that line.

📝 Teacher's Note: This is a great way to show how scientists use linearization to find physical constants.

🎯 Exam Tip: Remember \( g = 2 \times \text{slope} \), not just the slope itself!

Question 5. A body at rest is thrown downward from the top of tower. Draw a distance – time graph of its free fall under gravity during first 3 seconds. Show your table of values starting t = 0 with an interval of 1 second, (\( g = 10 \text{ ms}^{-2} \)).
Answer:
\( u = 0 \); \( a = +g = 10 \text{ ms}^{-2} \)
\( S = ut + \frac{1}{2} at^2 = 5t^2 \)

In simple words: The object falls 5 meters in the first second, but covers 45 meters by the third second because it is constantly speeding up.

📝 Teacher's Note: The curved shape of the graph (a parabola) indicates changing velocity (acceleration).

🎯 Exam Tip: When drawing graphs from a table, make sure the points are accurately plotted before joining them with a smooth curve.

Practice Problems 3

Question 1. From the diagram given below, calculate
1. acceleration
2. deceleration
3. distance covered by body.
Answer:
(i) Acceleration = Slope of v-t graph from O to A \( = \frac{15}{10} = 1.5 \text{ ms}^{-2} \)
(ii) Deceleration = Slope from A to B \( = \frac{15}{16-10} = \frac{15}{6} = 2.5 \text{ ms}^{-2} \)
(iii) Distance covered = Area of \( \Delta OAB = \frac{1}{2} \times 16 \times 15 = 120 \text{ m} \).
In simple words: The body speeds up for 10 seconds and then brakes hard for 6 seconds, covering 120 meters in total.

📝 Teacher's Note: Check the x-axis carefully. The time for deceleration is \( 16 - 10 = 6 \text{ seconds} \), not 16 seconds.

🎯 Exam Tip: If the graph is a triangle, the total distance is simply \( \frac{1}{2} \times \text{Total Time} \times \text{Max Velocity} \).

Question 2. From the velocity – time graph given below, calculate :
1. Acceleration in the region AB.
2. Deceleration in region BC.
3. Distance covered in the region ABCE.
4. Average velocity in region CED.
Answer:
(i) Acceleration AB \( = \frac{14 - 0}{12 - 0} = 1.16 \text{ ms}^{-2} \)
(ii) Deceleration BC \( = \frac{14 - 6}{16 - 12} = \frac{8}{4} = 2 \text{ ms}^{-2} \)
(iii) Distance ABCE \( = \text{Area of Triangle ABF} + \text{Area of Trapezium BCEF} = 84 + 40 = 124 \text{ m} \).
(iv) Average velocity CED \( = \frac{\text{Area of } \Delta CDE}{12} = \frac{36}{12} = 3 \text{ ms}^{-1} \).
In simple words: We split the graph into simple math shapes (triangles and rectangles) and find their areas to get the distances.

📝 Teacher's Note: Break complex areas into simple shapes to avoid calculation errors.

🎯 Exam Tip: Average velocity over a time interval is \( \frac{\text{Area under that specific segment of graph}}{\text{Time interval}} \).

Practice Problems 4

Question 1. Diagram given below shows velocity – time graphs of car P and Q, starting from same place and in same direction. Calculate :
1. Acceleration of car P.
2. Acceleration of car Q between 2 s – 5 s.
3. At what time intervals both cars have same velocity?
4. Which car is ahead after 10 s and by how much?
Answer:
(i) Accel of P \( = \frac{35}{10} = 3.5 \text{ ms}^{-2} \).
(ii) Accel of Q (2s-5s) \( = \frac{25}{3} = 8.33 \text{ ms}^{-2} \).
(iii) Same velocity at \( t = 3 \text{s} \) and \( t = 7 \text{s} \) (where lines intersect).
(iv) Distance P \( = \frac{1}{2} \times 10 \times 35 = 175 \text{ m} \).
Distance Q \( = \text{Area of Triangle} + \text{Area of Rectangle} = 37.5 + 125 = 162.5 \text{ m} \).
Car P is ahead by \( 175 - 162.5 = 12.5 \text{ m} \).
In simple words: Car P speeds up steadily. Car Q stays still, then zooms, then goes steady. P ends up slightly ahead.

📝 Teacher's Note: Intersecting points on a V-T graph mean equal velocities, NOT equal distances.

🎯 Exam Tip: To find who is ahead, you must calculate the total area under each line up to the 10-second mark.

UNIT III - PRACTICE PROBLEMS 1

Question 1. A motor bike, initially at rest, picks up a velocity of \( 72 \text{ kmh}^{-1} \) over a distance of 40 m. Calculate
1. acceleration
2. time in which it picks up above velocity.
Answer:
\( u = 0 \); \( v = 72 \text{ km/h} = 20 \text{ m/s} \); \( S = 40 \text{ m} \)
(i) \( v^2 - u^2 = 2aS \implies (20)^2 - 0 = 2a(40) \implies 400 = 80a \implies a = 5 \text{ ms}^{-2} \).
(ii) \( v = u + at \implies 20 = 0 + 5t \implies t = 4 \text{s} \).
In simple words: The bike reaches 20 m/s in just 4 seconds with a strong acceleration of 5 units.

📝 Teacher's Note: Check that \( 72 \text{ km/h} \) is converted to \( 20 \text{ m/s} \) before using the formulas.

🎯 Exam Tip: Use the equation \( v^2 - u^2 = 2aS \) when time is not given in the problem.

Question 2. A cyclist driving at \( 5 \text{ ms}^{-1} \), picks a velocity of \( 10 \text{ ms}^{-1} \), over a distance of 50 m. Calculate
1. acceleration
2. time in which the cyclist picks up above velocity.
Answer:
\( u = 5 \text{ m/s} \); \( v = 10 \text{ m/s} \); \( S = 50 \text{ m} \)
(i) \( v^2 - u^2 = 2aS \implies 100 - 25 = 2a(50) \implies 75 = 100a \implies a = 0.75 \text{ ms}^{-2} \).
(ii) \( v = u + at \implies 10 = 5 + 0.75t \implies 5 = 0.75t \implies t = 6.67 \text{ s} \).
In simple words: The cyclist is speeding up slowly, taking nearly 7 seconds to double their speed.

📝 Teacher's Note: This is a standard substitution problem. Ensure students don't mix up \( u \) and \( v \).

🎯 Exam Tip: For results like 6.666..., always round to two decimal places (6.67).

Practice Problems 2

Question 1. An aeroplane lands at \( 216 \text{ kmh}^{-1} \) and stops after covering a runway of 2 km. Calculate the acceleration and the time, in which it comes to rest.
Answer:
\( u = 216 \text{ km/h} = 60 \text{ m/s} \); \( v = 0 \); \( S = 2000 \text{ m} \)
\( v^2 - u^2 = 2aS \implies 0 - 3600 = 2a(2000) \implies a = -0.9 \text{ ms}^{-2} \).
\( v = u + at \implies 0 = 60 + (-0.9)t \implies 0.9t = 60 \implies t = 66.67 \text{ s} \).
In simple words: The plane slows down very gently to keep passengers safe, taking over a minute to stop.

📝 Teacher's Note: \( 2 \text{ km} = 2000 \text{ m} \). Unit conversion errors are the most common mistakes in this topic.

🎯 Exam Tip: Acceleration must be negative for a landing plane because it is stopping.

Question 2. A truck running at \( 90 \text{ kmh}^{-1} \), is brought to rest over a distance of 25 m. Calculate the retardation and time for which brakes are applied.
Answer:
\( u = 25 \text{ ms}^{-1} \); \( v = 0 \); \( S = 25 \text{ m} \)
\( v^2 - u^2 = 2aS \implies 0 - 625 = 2a(25) \implies -625 = 50a \implies a = -12.5 \text{ ms}^{-2} \).
\( v = u + at \implies 0 = 25 + (-12.5)t \implies 12.5t = 25 \implies t = 2 \text{ s} \).
In simple words: These are very strong brakes! The truck stops in just 2 seconds.

📝 Teacher's Note: Compare this acceleration (12.5) with gravity (9.8). This is a very intense deceleration.

🎯 Exam Tip: "Brought to rest" means final velocity \( v = 0 \).

Practice Problems 3

Question 1. A racing car, initially at rest, picks up a velocity of \( 180 \text{ kmh}^{-1} \) in 4.5 s. Calculate
1. acceleration
2. distance covered by car.
Answer:
\( u = 0 \); \( v = 50 \text{ ms}^{-1} \); \( t = 4.5 \text{ s} \)
(i) \( a = \frac{v - u}{t} = \frac{50}{4.5} = 11.11 \text{ ms}^{-2} \).
(ii) \( S = ut + \frac{1}{2} at^2 = 0 + \frac{1}{2} (11.11)(4.5)^2 = 112.5 \text{ m} \).
In simple words: The racing car accelerates faster than a falling rock! It covers more than a football field in under 5 seconds.

📝 Teacher's Note: \( 180 \text{ km/h} \) is a very high speed. Emphasize why conversion to m/s is the first mandatory step.

🎯 Exam Tip: If acceleration is calculated as a recurring decimal, use the fraction form (e.g., \( \frac{100}{9} \)) in subsequent steps for more accurate results.

Question 2. A motor bike running at \( 5 \text{ ms}^{-1} \), picks up a velocity of \( 30 \text{ ms}^{-1} \) in 5s. Calculate
1. acceleration
2. distance covered during acceleration.
Answer:
\( u = 5 \text{ ms}^{-1} \); \( v = 30 \text{ ms}^{-1} \); \( t = 5 \text{ s} \)
(i) \( a = \frac{30 - 5}{5} = 5 \text{ ms}^{-2} \).
(ii) \( S = ut + \frac{1}{2} at^2 = 5(5) + \frac{1}{2} (5)(25) = 25 + 62.5 = 87.5 \text{ m} \).
In simple words: Starting from a slow crawl, the bike reaches highway speeds in 5 seconds and travels 87.5 meters while doing it.

📝 Teacher's Note: This is a standard application of the second equation of motion. Ensure students don't forget the \( ut \) term since \( u \) is not zero here.

🎯 Exam Tip: Always double check the \( \frac{1}{2} at^2 \) term; students often forget to square the time.

Practice Problems 4

Question 1. A motor bike running at \( 90 \text{ kmh}^{-1} \) is slowed down to \( 18 \text{ kmh}^{-1} \) in 2.5 s. Calculate
1. acceleration
2. distance covered during slow down.
Answer:
\( u = 25 \text{ ms}^{-1} \); \( v = 5 \text{ ms}^{-1} \); \( t = 2.5 \text{ s} \)
(i) \( a = \frac{5 - 25}{2.5} = \frac{-20}{2.5} = -8 \text{ ms}^{-2} \).
(ii) \( v^2 - u^2 = 2aS \implies 25 - 625 = 2(-8)S \implies -600 = -16S \implies S = 37.5 \text{ m} \).
In simple words: The bike slows down very quickly, losing 8 m/s every second and stopping in under 40 meters.

📝 Teacher's Note: The negative acceleration (-8) indicates a strong braking force.

🎯 Exam Tip: Use the most convenient equation. Here, \( v^2 - u^2 = 2aS \) is cleaner than \( S = ut + \frac{1}{2} at^2 \).

Question 2. A cyclist driving at \( 36 \text{ kmh}^{-1} \) stops his motion in 2 s, by the application of brakes. Calculate
1. retardation
2. distance covered during the application of brakes.
Answer:
\( u = 10 \text{ ms}^{-1} \); \( v = 0 \); \( t = 2 \text{ s} \)
(i) \( a = \frac{0 - 10}{2} = -5 \text{ ms}^{-2} \). Retardation \( = 5 \text{ ms}^{-2} \).
(ii) \( S = \frac{(0)^2 - (10)^2}{2(-5)} = \frac{-100}{-10} = 10 \text{ m} \).
In simple words: The cyclist stops in just 10 meters by slowing down at 5 meters per second every second.

📝 Teacher's Note: Remind students that "retardation" is the absolute value of negative acceleration.

🎯 Exam Tip: When \( v=0 \), distance is simply \( \frac{u^2}{2a} \). This is a useful "stopping distance" shortcut.

Practice Problems 5

Question 1. A motor bike running at \( 90 \text{ kmh}^{-1} \), is slowed down to \( 54 \text{ kmh}^{-1} \) by the application of brakes, over a distance of 40 m. If the brakes are applied with the same force, calculate
1. total time in which bike comes to rest
2. total distance travelled by bike.
Answer:
Initial \( u = 25 \text{ m/s} \); after brakes \( v = 15 \text{ m/s} \); distance \( = 40 \text{ m} \).
From A to B:
\( (15)^2 - (25)^2 = 2a(40) \implies 225 - 625 = 80a \implies a = -5 \text{ ms}^{-2} \).
(i) Total Time: \( v = u + at \implies 0 = 25 + (-5)t \implies t = 5 \text{ s} \).
(ii) Total Distance: \( v^2 - u^2 = 2aS \implies 0 - (25)^2 = 2(-5)S \implies -625 = -10S \implies S = 62.5 \text{ m} \).
In simple words: We find how strong the brakes are first (-5). Then we calculate how long and how far it would take to stop completely using that same braking power.

📝 Teacher's Note: This "two-stage" problem is very common. The acceleration from the first part is the "key" to the second part.

🎯 Exam Tip: "Same force" implies "same acceleration". Use this logic to bridge the two parts of the question.

Practice Problems 6

Question 1. A packet is dropped from a stationary helicopter, hovering at a height ‘h’ from ground level, reaches ground in 12s. Calculate
1. value of h
2. final velocity of packet on reaching ground. (Take \( g = 9.8 \text{ ms}^{-2} \))
Answer:
\( u = 0 \); \( t = 12 \text{s} \); \( a = g = 9.8 \text{ ms}^{-2} \).
(i) \( h = ut + \frac{1}{2} at^2 = 0 + \frac{1}{2} (9.8)(144) = 705.6 \text{ m} \).
(ii) \( v = u + at = 0 + (9.8)(12) = 117.6 \text{ ms}^{-1} \).
In simple words: The packet falls for 12 seconds from a very high helicopter (705m) and is moving extremely fast when it hits the ground.

📝 Teacher's Note: Dropping things from heights allows for easy practice of gravity equations. Ensure students use \( u=0 \).

🎯 Exam Tip: Height is always positive when measuring downwards from the release point in these falling problems.

Practice Problems 8

Question 1. A spaceship is moving in space with a velocity of \( 50 \text{ kms}^{-1} \). Its engine fires for 10 s, such that its velocity increases to \( 60 \text{ kms}^{-1} \). Calculate the total distance travelled by spaceship in 1/2 minute, from the time of firing its engine.
Answer:
\( u = 50000 \text{ ms}^{-1} \); \( v = 60000 \text{ ms}^{-1} \); \( t = 10 \text{ s} \).
\( a = \frac{60000 - 50000}{10} = 1000 \text{ ms}^{-2} \).
Case I (First 10s): \( S_1 = ut + \frac{1}{2} at^2 = 50000(10) + \frac{1}{2} (1000)(100) = 550000 \text{ m} \).
Case II (Next 20s): \( \text{Time} = 30 - 10 = 20 \text{s} \).
\( S_2 = \text{Uniform Velocity} \times \text{Time} = 60000 \times 20 = 1200000 \text{ m} \).
Total Distance: \( 550 + 1200 = 1750 \text{ km} \).
In simple words: The spaceship speeds up for 10 seconds and then coasts at that high speed for the rest of the half-minute.

📝 Teacher's Note: Space problems often have huge numbers. Use kilometers or scientific notation to keep it manageable.

🎯 Exam Tip: "1/2 minute" means 30 seconds total. Be careful not to use 10s for the whole calculation.

Question 2. A spaceship is moving in space with a velocity of \( 60 \text{ kms}^{-1} \). It fires its retro engines for 20 second and velocity is reduced to \( 55 \text{ kms}^{-1} \). Calculate the distance travelled by the spaceship in 40 s, from the time of firing of the retro-rockets.
Answer: Initial velocity of spaceship \( u = 60 \text{ kms}^{-1} \)
Final velocity of spaceship \( v = 55 \text{ kms}^{-1} \)
Time for change in velocity \( t_1 = 20 \text{ s} \)
Acceleration \( a = \frac{v - u}{t} = \frac{55 - 60}{20} = -0.25 \text{ kms}^{-2} \)

Case - I: During deceleration (First 20 s)
Distance \( S_1 = ut_1 + \frac{1}{2} at_1^2 \)
\( \implies S_1 = 60(20) + \frac{1}{2} (-0.25)(20)^2 \)
\( \implies S_1 = 1200 + (-0.125 \times 400) \)
\( \implies S_1 = 1200 - 50 = 1150 \text{ km} \)

Case - II: After firing (Next 20 s)
The spaceship now moves with a uniform velocity of \( 55 \text{ kms}^{-1} \).
Remaining time \( t_2 = 40 - 20 = 20 \text{ s} \)
Distance \( S_2 = \text{Uniform velocity} \times \text{Time} \)
\( \implies S_2 = 55 \times 20 = 1100 \text{ km} \)

Total distance travelled \( = S_1 + S_2 = 1150 + 1100 = 2250 \text{ km} \).
In simple words: The spaceship slows down for the first 20 seconds, covering 1150 km. It then travels at its new slower speed for the remaining 20 seconds, covering another 1100 km, making the total distance 2250 km.

📝 Teacher's Note: This is a two-part problem. First, find the distance covered during the change in velocity using equations of motion, then find the distance for the remaining time using the formula for constant velocity.

🎯 Exam Tip: Be careful with the total time. The question asks for the distance in 40 seconds, so after the 20 seconds of braking, you must calculate for the remaining 20 seconds of constant motion.