NCERT Solutions Class 9 Science Exploration Chapter 04 Describing Motion Around Us

 

Question 1. My father went to a shop from home which is located at a distance of 250 m on a straight road. On reaching there he discovered that he forgot to carry a cloth bag. He came home to take it, went to the shop again, bought provisions and came back home. How much was the total distance travelled by him? What was his displacement from home?
Answer: Your father's journey involved four separate trips. First, he traveled from home to the shop (250 m), then back home again (250 m), then to the shop once more (250 m), and finally returned home (250 m). Adding all these segments together gives a total distance of 1000 m or 1 km. However, since he began at home and ended at home, his net change in position equals zero. This means his displacement from home is 0 m.
In simple words: Even though your father walked a total of 1 km, he ended up exactly where he started, so his displacement is zero.

Exam Tip: Always remember that distance counts every step taken, but displacement only measures the straight-line difference between start and end positions.

 

Question 2. A student runs from ground floor to the fourth floor of a school building to collect a book and then comes down their classroom on the second floor. If the height of each floor is 3m, find: (i) the total vertical distance travelled, and (ii) their displacement from the starting point.
Answer: Setting up the floor heights: Ground floor is at 0 m, the 2nd floor is at 6 m, and the 4th floor is at 12 m. The student moves upward from ground to 4th floor (a rise of 12 m), then downward from 4th floor to 2nd floor (a drop of 6 m). Adding these movements gives a total vertical distance of 12 + 6 = 18 m. For displacement, we measure only the vertical separation between the starting point (ground floor at 0 m) and the ending point (2nd floor at 6 m), which is 6 m upward.
In simple words: The student travels 18 m up and down in total, but ends up only 6 m above where they started.

Exam Tip: In vertical motion problems, sketch a diagram showing each floor level - this makes it much easier to track both distance and displacement correctly.

 

Question 3. A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating and if so, how?
Answer: Yes, the scooter can be accelerating even with a steady speedometer reading. The speedometer only shows the magnitude of velocity, which we call speed. However, acceleration occurs whenever velocity changes - and velocity is more than just speed, it also includes direction. If the girl rides her scooter along a curved path, such as turning a corner or going around a roundabout, the direction of motion keeps changing at every instant. Even though the speedometer shows the same number, the direction is constantly shifting. This change in direction means the velocity is changing, which is the definition of acceleration. In curved motion like this, the scooter experiences what we call centripetal acceleration - the speed stays constant, but the direction changes continuously.
In simple words: Turning a corner at a steady speed still counts as acceleration because you are changing direction, not just how fast you are going.

Exam Tip: Remember that acceleration depends on changes in velocity (a vector), not just changes in speed (a scalar) - direction matters just as much as magnitude.

 

Question 4. A car starts from rest and its velocity reaches 24 m s⁻¹ in 6 s. Find the average acceleration and the distance travelled in these 6 s.
Answer: Given: Initial velocity (u) = 0 m s⁻¹, Final velocity (v) = 24 m s⁻¹, Time (t) = 6 s

For average acceleration, use the formula a = (v - u) / t. Substituting the values: a = (24 - 0) / 6 = 4 m s⁻²

For distance, use the kinematic equation s = ut + ½at²: s = (0)(6) + ½ × 4 × (6)² = 0 + ½ × 4 × 36 = 72 m

We can verify this answer using v² = u² + 2as: (24)² = (0)² + 2 × 4 × s, which gives 576 = 8s, so s = 72 m
In simple words: The car speeds up at a rate of 4 metres per second each second, and covers 72 metres while doing so.

Exam Tip: Always verify your answer using a different kinematic equation - if you get the same result, you know your calculation is correct.

 

Question 5. A motorbike moving with initial velocity 28 m s⁻¹ and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and time taken to come to a stop.
Answer: Given: Initial velocity (u) = 28 m s⁻¹, Final velocity (v) = 0 m s⁻¹ (comes to rest), Distance (s) = 98 m

Step 1 - Finding acceleration: Use v² = u² + 2as: (0)² = (28)² + 2 × a × 98, which gives 0 = 784 + 196a. Solving for a: a = -784 / 196 = -4 m s⁻². The negative sign indicates that the motorbike is slowing down (deceleration).

Step 2 - Finding time: Use v = u + at: 0 = 28 + (-4) × t, which simplifies to 4t = 28, so t = 7 s
In simple words: The motorbike slows down at a rate of 4 metres per second per second and takes 7 seconds to come to a complete stop.

Exam Tip: When an object is slowing down, acceleration is negative - this is called deceleration or negative acceleration, and it is just as important to include the negative sign.

 

Question 6. Fig. 4.27 shows a position-time graph of two objects A and B moving along parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer.
Answer: No, objects A and B never have equal velocity. On a position-time graph, the velocity of an object is shown by the slope (steepness) of the line. A line that is more tilted indicates a higher velocity. Both objects A and B travel in straight lines with constant velocities, shown by their straight-line paths. However, the two lines have different slopes - one is steeper than the other. Two straight lines with different slopes will never become parallel to each other, which means their slopes remain different at all times. Equal velocity would require both lines to have identical slopes (be parallel), which is not the case here. While the two lines may cross at a specific point in time, at that crossing point the two objects are at the same position but still have different velocities.
In simple words: Different slopes on a graph mean different speeds. Two lines with different angles can meet at a point, but they never have the same tilt.

Exam Tip: Always check whether two position-time lines are parallel - if they are, the objects have equal velocity; if they are not, velocities are different at all times.

 

Question 7. A graph in Fig. 4.28 shows the change in position with time for two objects A and B moving in a straight line from 0 to 10 seconds. Choose the correct option(s).
(i) The average velocity of both over the 10 s time interval is equal since they have the same initial and final positions.
(ii) The average speeds of both over the 10s time interval is equal since they have the same initial and final positions.
(iii) The average speed of A over the 10 s time interval is lower than that of B since it covers a shorter distance than B in 10 seconds.
(iv) The average speed of A over the 10 s time interval is greater than that of B since B's speed is lower than A's in some segments.
Answer: Looking at Fig. 4.28, both A and B start at the same position and end at the same position at t = 10 s. Since the starting and ending positions are identical, the displacement is the same for both objects. Since average velocity depends only on displacement, both objects have the same average velocity. Object A follows a curved path that bends more, meaning it takes a longer route to reach the final position, covering more total distance. Object B travels in a straight line, taking the shortest route and covering less total distance. Because object A covers more total distance in the same time period (10 seconds), its average speed is greater than that of object B.

Therefore: (i) TRUE - Same starting and ending positions mean the same displacement, which gives the same average velocity. (iv) TRUE - A's curved path means it covers more total distance, so its average speed is greater than B's.
In simple words: Both reach the same final spot, so they have the same average velocity. But A took a longer route, so A moved faster on average.

Exam Tip: Never confuse average velocity (depends on displacement) with average speed (depends on total distance) - they are not the same thing unless the path is a straight line.

 

Question 8. A truck driver driving at the speed of 54 km h⁻¹ notices a road sign with a speed limit of 40 km h⁻¹ (Fig. 4.29) for trucks. He slows down to 36 km h⁻¹ in 36 s. What was the distance travelled by him during his time? Assume the acceleration to be constant while slowing down.
Answer: First, convert the speeds to m s⁻¹: Initial velocity u = 54 km h⁻¹ = 54 × (1000/3600) = 15 m s⁻¹, Final velocity v = 36 km h⁻¹ = 36 × (1000/3600) = 10 m s⁻¹, Time t = 36 s

Finding acceleration: a = (v - u) / t = (10 - 15) / 36 = -5 / 36 m s⁻²

Finding distance using s = ut + ½at²: s = 15 × 36 + ½ × (-5/36) × (36)² = 540 + ½ × (-5/36) × 1296 = 540 + (-5 × 1296 / 72) = 540 - 90 = 450 m

Alternative method using average velocity: s = ((u + v) / 2) × t = ((15 + 10) / 2) × 36 = 12.5 × 36 = 450 m
In simple words: The truck slowed down gradually over 36 seconds while covering a distance of 450 metres.

Exam Tip: Always convert units to m s⁻¹ and seconds before using kinematic equations - mixing units is a common source of errors.

 

Question 9. A car starts from rest and accelerates uniformly to 20 m s⁻¹ in 5 seconds. It then travels at 20 m s⁻¹ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.
Answer: This motion has three distinct phases.

Acceleration phase (0 to 5 s): u = 0, v = 20 m s⁻¹, t = 5 s. Using s = (u + v)/2 × t: s₁ = (0 + 20)/2 × 5 = 10 × 5 = 50 m

Constant velocity phase (5 to 15 s): v = 20 m s⁻¹, t = 10 s. Using s = v × t: s₂ = 20 × 10 = 200 m

Braking phase (15 to 21 s): u = 20 m s⁻¹, v = 0, t = 6 s. Using s = (u + v)/2 × t: s₃ = (20 + 0)/2 × 6 = 10 × 6 = 60 m

Total distance: s₁ + s₂ + s₃ = 50 + 200 + 60 = 310 m
In simple words: The car speeds up for 50 m, cruises steadily for 200 m, then slows down over 60 m, for a total of 310 m.

Exam Tip: Always break multi-phase motion problems into separate phases with constant acceleration - handle each phase independently, then add the results.

 

Question 10. A bus is travelling at 36 km h⁻¹ when the driver sees an obstacle 30 m ahead. The driver takes 0.5 s to react before pressing the brake. Once the brake is applied, the velocity of the bus reduces with constant acceleration of 2.5 m s⁻². Will the bus be able to stop before reaching the obstacle?
Answer: Convert initial speed: 36 km h⁻¹ = 10 m s⁻¹. Distance to obstacle = 30 m.

Distance covered during reaction time: During the 0.5 s reaction period, the driver has not yet pressed the brake, so the bus moves at constant speed: s₁ = u × t = 10 × 0.5 = 5 m

Distance covered during braking: Using v² = u² + 2as, where v = 0 (bus comes to rest), u = 10 m s⁻¹, a = -2.5 m s⁻²: 0 = (10)² + 2 × (-2.5) × s₂, which gives 0 = 100 - 5s₂, so s₂ = 20 m

Total stopping distance: s₁ + s₂ = 5 + 20 = 25 m

Since the total stopping distance (25 m) is less than the distance to the obstacle (30 m), yes, the bus will stop safely before hitting the obstacle.
In simple words: The bus needs 25 metres to come to a complete stop, but the obstacle is 30 metres away, so the bus stops safely.

Exam Tip: Reaction time is crucial in real-world braking problems - during this time, the vehicle continues at constant speed before deceleration begins.

 

Question 11. A student said, "The Earth moves around the Sun". In the context, discuss whether an object kept on the Earth can be considered to be at rest.
Answer: Whether an object is at rest depends entirely on which reference point we choose.

With respect to Earth as the reference point: An object placed on Earth, such as a book resting on a table, does not change its position relative to Earth. From Earth's perspective, the book is at rest. The book and Earth move together as one unit.

With respect to the Sun as the reference point: Since Earth itself is moving in an orbit around the Sun, everything on Earth - including our book - is also moving in that same orbit. From the Sun's viewpoint, the book is in motion because it travels with Earth around the Sun.

With respect to distant stars: The Sun itself is moving within the Milky Way galaxy. This means the book is also moving relative to distant stars in the galaxy.

The key lesson is that motion and rest are relative concepts - they depend on the observer and the reference point chosen. There is no absolute rest or absolute motion in the universe.
In simple words: An object can be at rest from one viewpoint but moving from another - it all depends on what you compare it to.

Exam Tip: Always state your reference point when describing whether something is at rest or moving - the same object can have different answers depending on the reference frame.

 

Question. Shade the area (in different colours) representing the displacement of the cyclist (i) while cyclist is moving with constant velocity. (ii) when the velocity of cyclist is decreasing. Also, calculate the displacement and average acceleration in the 120 s time interval.
Answer: (i) Displacement during constant velocity (40-80 s) - shaded rectangle: During this 40-second interval at a steady velocity of 3 m s⁻¹, the displacement is s₂ = v × t = 3 × (80 - 40) = 3 × 40 = 120 m

(ii) Displacement during decreasing velocity (80-120 s) - shaded triangle: As the velocity drops from 3 m s⁻¹ to 0, this triangular region represents: s₃ = ½ × base × height = ½ × 40 × 3 = 60 m

Total displacement over the full 120-second interval: The first phase (0-40 s) also forms a triangle with s₁ = ½ × 40 × 3 = 60 m. Adding all three sections: s_total = 60 + 120 + 60 = 240 m

Average acceleration over 120 s: a = (v_final - v_initial) / t = (0 - 0) / 120 = 0 m s⁻². The average acceleration is zero because the cyclist begins at rest and ends at rest, even though there were periods of positive and negative acceleration in between.
In simple words: The cyclist travels a total of 240 metres and the average acceleration over the whole period is zero because they start and stop at rest.

Exam Tip: When average acceleration is zero, it does not mean there was no acceleration during the motion - only that the total change in velocity from start to finish is zero.

 

Question 13. A girl is preparing for her first marathon by running on a straight road. She uses a smartwatch to calculate her running speed at different intervals. The graph (Fig. 4.31) depicts her velocity versus time. Estimate the running distance based on the graph.
Answer: The velocity-time graph shows two distinct phases of motion. The area under each section of the graph represents the distance covered in that phase.

Distance in Phase 1 (rectangular section): The runner maintains a constant velocity of 7.5 m h⁻¹ for 4 seconds. The area of this rectangle is s₁ = 7.5 × 4 = 30 km

Distance in Phase 2 (trapezoidal section): The runner's velocity decreases linearly from 7.5 m h⁻¹ to 5 m h⁻¹ over 2 seconds. The area of this trapezoid is s₂ = ½ × (7.5 + 5) × 2 = ½ × 12.5 × 2 = 12.5 km

Total estimated running distance: Total = 30 + 12.5 = 42.5 km, which we round to approximately 42 km. (Note: This is a marathon distance, which is indeed about 42 km.)
In simple words: The girl ran at full speed for the first part, then slowed down, covering a total distance of about 42 kilometres.

Exam Tip: The area under a velocity-time graph always equals displacement - use geometry (rectangles, triangles, trapezoids) to find areas rather than trying to calculate from the equation alone.

 

Question 14. On entering a state highway, a car continues to move with a constant velocity of 6 m s⁻¹ for 2 minutes and then accelerates with a constant acceleration 1 m s⁻² for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.
Answer: This motion consists of two phases. Phase 1: v = 6 m s⁻¹, t₁ = 2 min = 120 s. Phase 2: u = 6 m s⁻¹, a = 1 m s⁻², t₂ = 6 s

Displacement in Phase 1 (constant velocity section forms a rectangle on the v-t graph): s₁ = 6 × 120 = 720 m

Displacement in Phase 2 (acceleration section forms a trapezoid): First find the final velocity: v = u + at = 6 + 1 × 6 = 12 m s⁻¹. The displacement is the area of the trapezoid: s₂ = ½ × (6 + 12) × 6 = ½ × 18 × 6 = 54 m

Total displacement: s = s₁ + s₂ = 720 + 54 = 774 m
In simple words: The car travels at steady speed for 120 seconds covering 720 metres, then speeds up for 6 more seconds covering another 54 metres, for a total of 774 metres.

Exam Tip: When drawing a velocity-time graph, remember that constant velocity gives a horizontal line, and constant acceleration gives a sloped line - use the shapes formed to calculate displacements.

 

Question 15. Two cars A and B start moving with a constant acceleration from rest in a straight line. Car A attains a velocity of 5 m s⁻¹ in 5 s. Car B attains in a velocity of 3 m s⁻¹ in 10 s. Plot the velocity-time graphs for both the car in the same graph. Using the graph, calculate the displacement mentioned in the two time intervals.
Answer: First, find the acceleration for each car: Acceleration of A = (5 - 0) / 5 = 1 m s⁻², Acceleration of B = (3 - 0) / 10 = 0.3 m s⁻²

Velocity data for plotting:

Displacement of Car A (in 5 s - triangle area): s_A = ½ × base × height = ½ × 5 × 5 = 12.5 m

Displacement of Car B (in 10 s - triangle area): s_B = ½ × 10 × 3 = 15 m
In simple words: Car A accelerates faster and reaches 5 m/s in 5 seconds, covering 12.5 metres. Car B accelerates more slowly and reaches 3 m/s in 10 seconds, covering 15 metres.

Exam Tip: When plotting multiple objects on the same velocity-time graph, use different colours or line styles for each object - this makes comparisons much clearer and helps you visualize which car accelerates faster.

 

Question 16. Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minute's hand of the wall clock. During the given time interval, what is its (i) distance travelled, (ii) displacement, (iii) speed, and (iv) velocity. The length of the minute's hand is 7 cm (Fig. 4.32).
Answer: (i) Distance travelled by the tip: The tip of the minute hand traces out circular arcs as it moves around the clock face. From 6:00 PM to 7:30 PM is 1.5 hours, meaning the minute hand completes 1.5 full revolutions. The circumference of one revolution is 2πr = 2 × 3.14 × 7 = 43.96 cm. Total distance = 1.5 × 43.96 = 65.94 cm, approximately 66 cm

(ii) Displacement of the tip: At 6:00 PM, the minute hand points to 12 o'clock position (tip at the top of the clock). At 7:30 PM (after 1.5 revolutions), the minute hand points to the 6 o'clock position (tip at the bottom of the clock). The straight-line distance between these two positions is the diameter of the clock circle: Displacement = 2r = 2 × 7 = 14 cm, directed downward from the 12 position to the 6 position

(iii) Average speed: Average speed = Distance / Time. The time interval is 1.5 hours = 1.5 × 3600 = 5400 seconds. Average speed = 65.94 cm / 5400 s = 0.0122 cm s⁻¹, approximately 1.22 × 10⁻² cm s⁻¹

(iv) Average velocity: Average velocity = Displacement / Time = 14 cm / 5400 s = 0.0026 cm s⁻¹, approximately 2.6 × 10⁻³ cm s⁻¹
In simple words: The tip of the minute hand travels about 66 centimetres in a curved path but ends up only 14 centimetres away (straight line) from where it started.

Exam Tip: In circular motion, always remember that distance can be much larger than displacement - distance counts the entire curved path, while displacement measures only the straight-line separation between start and end points.

 

Very Short Answer Type Questions

 

Question 1. What is linear motion?
Answer: Linear motion occurs when an object travels along a straight path. Examples include a ball dropping straight down, a vehicle driving on a straight highway, and swimmers racing in a pool lane.
In simple words: Motion is linear when something moves in a straight line without turning.

Exam Tip: The defining feature of linear motion is a straight path - if the path curves, it is no longer linear motion.

 

Question 2. Define displacement.
Answer: Displacement measures the net change in position of an object between two moments in time. It has both magnitude and direction, making it a vector quantity. The SI unit is the metre (m).
In simple words: Displacement tells you how far an object is from where it started and in what direction.

Exam Tip: Never confuse displacement with distance - displacement has direction and can be zero, but distance is always positive and has no direction.

 

Question 3. What is the SI unit of average speed and average velocity?
Answer: The SI unit for both average speed and average velocity is metre per second, written as m s⁻¹ or m/s. In everyday use, speeds are also expressed in km h⁻¹ (kilometres per hour).
In simple words: Both speed and velocity are measured in metres per second.

Exam Tip: When solving problems with speeds in km/h, always convert to m/s using the factor 1000/3600 or approximately 5/18.

 

Question 4. When is the displacement of a moving object zero?
Answer: Displacement becomes zero when an object returns to its starting position. An example is an athlete who completes one full lap around a track - even though they travelled a large distance, they end up at the starting point, making displacement zero.
In simple words: Displacement is zero when you end up back where you started.

Exam Tip: Remember that non-zero distance with zero displacement is common - think of any motion where you return to your starting point.

 

Question 5. What is uniform motion in a straight line?
Answer: An object undergoes uniform motion in a straight line when it travels equal distances in equal time intervals while maintaining a straight path. This means the object moves at a constant speed in the same direction without changing either the speed or the direction.
In simple words: Uniform motion in a straight line means moving at the same speed in the same direction.

Exam Tip: For uniform motion, both the distance-time graph and velocity-time graph follow predictable patterns - distance increases linearly, and velocity remains horizontal.

 

Question 6. Define average acceleration.
Answer: Average acceleration is the rate of change of velocity with respect to time. It equals the change in velocity divided by the time interval during which the change takes place. The SI unit is m s⁻². Like velocity, it is a vector quantity and has both magnitude and direction.
In simple words: Average acceleration tells you how much faster or slower something gets moving per second.

Exam Tip: Negative acceleration (called deceleration) indicates slowing down - be sure to include the negative sign in your answer.

 

Question 7. What does the slope of a position-time graph represent?
Answer: The slope of a position-time graph represents the velocity of the object. A steeper slope indicates higher velocity, while a gentler slope indicates lower velocity. A horizontal line (zero slope) shows that the object is stationary or at rest.
In simple words: The steeper the line, the faster the object is moving.

Exam Tip: A curved position-time graph means the velocity is changing (acceleration is happening) - only straight lines indicate constant velocity.

 

Question 8. What does the area under a velocity-time graph represent?
Answer: The area enclosed between the velocity-time graph line and the time axis represents the displacement of the object during that time interval. Different shapes (rectangles, triangles, trapezoids) give the displacement for different types of motion.
In simple words: The area under a velocity-time graph tells you how far an object travelled.

Exam Tip: To find the area, use geometry formulas: rectangle (length × width), triangle (½ × base × height), trapezoid (½ × (parallel side 1 + parallel side 2) × height).

 

Question 9. State the three kinematic equations for motion with constant acceleration.
Answer: The three kinematic equations are:
(i) v = u + at
(ii) s = ut + ½at²
(iii) v² = u² + 2as

In these equations, u is the initial velocity, v is the final velocity, a is the acceleration, s is the displacement, and t is the time. These equations apply only when acceleration remains constant throughout the motion.
In simple words: These three formulas help you find missing values in motion problems when acceleration stays the same.

Exam Tip: Choose the equation that contains only the three quantities you know and the one you need to find - this avoids unnecessary calculations.

 

Question 10. What is uniform circular motion?
Answer: Uniform circular motion describes the motion of an object moving along a circular path at a constant speed. Although the speed remains unchanged, the velocity continuously changes direction at every point along the path, which means the object is constantly accelerating.
In simple words: Uniform circular motion is when something moves in a circle at a steady speed but is always turning.

Exam Tip: Do not mistake uniform circular motion for zero acceleration - the direction change creates acceleration even though speed is constant.

 

Question 11. What is the displacement of an object after one complete revolution in circular motion?
Answer: After one complete revolution, the displacement is zero because the object finishes at its starting position. However, the distance travelled is not zero - it equals the circumference of the circular path, which is 2πR, where R is the radius of the circular path.
In simple words: After going all the way around in a circle, you end up where you started, so displacement is zero.

Exam Tip: This is a key difference between distance and displacement - the distance around a circle is 2πR, but displacement after a complete revolution is always zero.

 

Question 12. What is the formula for average speed in uniform circular motion?
Answer: In uniform circular motion, average speed = 2πR / T, where R is the radius of the circular path and T is the time taken to complete one full revolution. This formula shows that average speed depends on both the size of the circle and how long one revolution takes.
In simple words: Average speed in circular motion equals the circle's circumference divided by the time for one complete trip around.

Exam Tip: This formula applies only to complete revolutions - do not use it for partial circles without adjusting the numerator accordingly.

 

Question 13. Can an object have zero velocity but non-zero acceleration?
Answer: Yes, this is absolutely possible. A classic example is a ball thrown vertically upward - at the highest point of its flight, the ball momentarily has zero velocity (it stops rising), but gravitational acceleration of 9.8 m s⁻² continues to act downward at that instant, pulling the ball back toward the ground.
In simple words: An object can be at rest (zero velocity) while still being pushed or pulled (non-zero acceleration).

Exam Tip: This concept shows that acceleration and velocity are independent - do not assume that if one is zero, the other must be zero too.

 

Question 14. What is the acceleration due to gravity as shown in Class 9 Science Exploration Chapter 4?
Answer: When an object is dropped from a height or thrown upward, it experiences a constant acceleration due to Earth's gravitational pull. This acceleration is 9.8 m s⁻² directed downward. It is denoted by the symbol g and is called acceleration due to gravitational force or acceleration due to gravity.
In simple words: Gravity pulls everything downward with an acceleration of 9.8 metres per second per second.

Exam Tip: When solving problems involving falling objects or projectiles, use g = 9.8 m s⁻² (or sometimes 10 m s⁻¹ for simpler calculations) and always include the downward direction.

 

Question 15. What does a straight line parallel to the time axis on a position-time graph indicate?
Answer: A straight line parallel to the time axis on a position-time graph indicates that the position of the object is not changing as time passes. This means the object is not moving - it is stationary or at rest at that location.
In simple words: A horizontal line on a position-time graph means the object is sitting still.

Exam Tip: Always look at whether a line is horizontal (zero velocity), slanted (constant velocity), or curved (acceleration) to understand the motion pattern.

 

Short Answer Type Questions

 

Question 1. Distinguish between distance and displacement with an example.
Answer: Distance measures the total path length that an object travels, counting every step or turn, without any regard to direction. Displacement, by contrast, measures only the net change in position from the starting point to the ending point and includes direction. For example, imagine an athlete running from point O to point A (covering 100 m) and then back to point B (covering 60 m from A). The total distance travelled is 100 + 60 = 160 m. However, the displacement is only 40 m in the positive direction (from the starting point O to the final position B).
In simple words: Distance is the total path travelled. Displacement is the straight-line distance from start to finish.

Exam Tip: In round-trip journeys, distance is always greater than or equal to displacement, and displacement can be zero if you return to your starting point.

 

Question 2. What is the difference between average speed and average velocity? When are they equal?
Answer: Average speed is calculated by dividing the total distance travelled by the total time taken. It has no direction and is a scalar quantity. Average velocity, on the other hand, is calculated by dividing the displacement by the time taken. It always includes direction and is a vector quantity. Average speed can never be zero for a moving object, but average velocity can be zero if an object returns to its starting position. The two are equal only when an object moves in a straight line without ever turning back, because in this case the distance travelled and the magnitude of displacement are identical.

Comparison table:

In simple words: Average speed counts every metre you travel. Average velocity only cares about how far you end up from where you started.

Exam Tip: In problems where an object changes direction (returns partway), these two quantities will be different - calculate both to show you understand the distinction.

 

Question 3. A swimmer completes one length of a 25 m pool and returns to the starting point in 50 seconds. Find average speed and average velocity.
Answer: Total distance travelled = 25 + 25 = 50 m (one length plus the return)

Displacement = 0 m (the swimmer returned to the starting point)

Average speed = 50 m ÷ 50 s = 1 m s⁻¹

Average velocity = 0 m ÷ 50 s = 0 m s⁻¹

This example demonstrates that average velocity can be zero even when average speed is clearly non-zero.
In simple words: The swimmer was moving and covered 50 metres, but ended up right back where they started.

Exam Tip: Whenever you see round-trip motion or any path that returns to the starting point, immediately recognize that displacement (and thus average velocity) will be zero.

 

Question 4. Explain what the slope of a velocity-time graph tells us about the motion of an object.
Answer: The slope of a velocity-time graph provides information about the acceleration of the object. When the slope is positive (the line goes upward as time increases), the velocity is increasing and the object is accelerating in the direction of motion. When the slope is negative (the line goes downward as time increases), the velocity is decreasing and the acceleration acts opposite to the direction of motion - this is called deceleration or negative acceleration. A zero slope (a horizontal line) means the velocity is constant, which indicates zero acceleration and uniform motion.
In simple words: A slanted line shows acceleration. A steep slope means large acceleration. A horizontal line means no acceleration.

Exam Tip: The steeper the slope (positive or negative), the larger the magnitude of acceleration - use the slope formula rise/run to calculate the actual acceleration value.

 

Question 5. How are the three kinematic equations derived? Name the two primary equations.
Answer: The two primary kinematic equations are fundamental and are derived from basic motion definitions:

The first primary equation v = u + at comes directly from the definition of average acceleration. Acceleration is the rate of change of velocity, so a = (v - u) / t, which rearranges to give this equation.

The second primary equation s = ut + ½at² is derived from the area under a velocity-time graph. For constant acceleration, this area represents the displacement.

The third equation v² = u² + 2as is derived by eliminating time (t) between the two primary equations. This allows us to relate velocity, acceleration, and displacement without needing the time value.

All three equations remain valid only when acceleration is constant throughout the entire motion.
In simple words: The first two equations are the foundations. The third equation is a shortcut that combines the first two.

Exam Tip: Know which equation to use based on which variables you have and which one you need - avoid using equations unnecessarily and keep your working clear and organized.

 

Question 6. Why is a moving object in uniform circular motion said to be accelerating, even though its speed is constant?
Answer: Acceleration is defined as the rate of change of velocity, not as the rate of change of speed. Velocity is a vector quantity that has both magnitude (the speed) and direction. In uniform circular motion, the speed (magnitude of velocity) remains constant, but the direction of motion changes continuously at every point along the circular path. Since the direction is constantly changing, the velocity itself is constantly changing. Whenever velocity changes - whether due to a change in speed, a change in direction, or both - there is acceleration. Therefore, an object moving in a circle at constant speed is indeed accelerating because its velocity is changing direction.
In simple words: Acceleration comes from any change in velocity, including direction changes - you can turn at the same speed and still be accelerating.

Exam Tip: This is a critical concept - memorize that acceleration depends on velocity (which includes direction), not just on speed (which is magnitude only).

 

Question 7. What is a reference point? Why is it important in describing motion?
Answer: A reference point is a fixed location or object chosen as a starting position from which we measure the position of another object. The position of any object is described by stating its distance and direction away from this reference point. The importance of a reference point is fundamental: without choosing a reference point, it is impossible to determine whether an object is at rest or in motion. An object may be stationary relative to one reference point but moving relative to another. For example, a person sitting in a moving train is at rest relative to the train but in motion relative to the ground. For motion along a straight line, the reference point is typically marked as the origin O, from which all distances are measured.
In simple words: A reference point is where you measure from - without it, you cannot say if something is moving or still.

Exam Tip: Always state your reference point clearly when describing motion - different observers using different reference points may describe the same object differently.

 

Question 8. In Example 4.8 of Class 9 Science Exploration Chapter 4, a car brakes with acceleration -4 m s⁻². Why does doubling the speed quadruple the stopping distance?
Answer: Using the kinematic equation v² = u² + 2as with v = 0 (the car comes to rest) and a = -4 m s⁻², we can solve for stopping distance s: 0 = u² + 2 × (-4) × s, which simplifies to s = u² / 8. This relationship shows that stopping distance is proportional to the square of the initial speed (u²). When you double the initial speed (u becomes 2u), the stopping distance becomes s = (2u)² / 8 = 4u² / 8, which is 4 times the original stopping distance. This quadratic relationship explains why high-speed collisions are so much more dangerous than low-speed ones - the stopping distance increases with the square of speed, not linearly.
In simple words: Doubling your speed quadruples your stopping distance - this is why speed limits are so important for safety.

Exam Tip: Recognize relationships between variables by rearranging kinematic equations - squared relationships like this one show why safety becomes critical at higher speeds.

 

Question 9. How can you find displacement from a velocity-time graph for an object moving with constant acceleration?
Answer: When an object moves with constant acceleration, the displacement over a time period equals the area trapped between the velocity-time graph line and the time axis during that interval. This area breaks down into two parts - a rectangle and a triangle. As an example, the displacement between 10 s and 20 s in Fig. 4.18b is found by adding the area of rectangle ACDE plus the area of triangle ABC, which equals 50 m + 25 m = 75 m.
In simple words: To find how far an object moved, measure the space under the line on a velocity-time graph. Add the rectangle part and the triangle part together.

Exam Tip: Always identify whether the graph region forms a rectangle, triangle, or trapezoid - this determines which area formula you use. Mark the time interval clearly on the graph before calculating.

 

Question 10. What is the difference between uniform and non-uniform motion in a straight line?
Answer:

In non-uniform motion, when the distances covered in successive equal time intervals are getting larger, the object is speeding up. When they are getting smaller, the object is slowing down.
In simple words: Uniform motion means going at the same speed the whole time. Non-uniform motion means the speed keeps changing - sometimes faster, sometimes slower.

Exam Tip: To identify motion type from a position-time graph, check if the line is straight (uniform) or curved (non-uniform) - curved lines always indicate acceleration or deceleration.

 

Question 1. Explain the concepts of position, distance travelled, and displacement using a suitable example. Under what conditions is the magnitude of displacement equal to the distance travelled?
Answer: An object's position is determined by stating how far it is from a fixed reference point (origin O) and in which direction. For straight-line motion, locations to the right of O count as positive, and locations to the left count as negative.Example: An athlete begins at O, reaches point A (100 m) at t = 10 s, then turns around and runs back to point B (40 m) at t = 16 s.

From the example: Total distance travelled = OA + AB = 100 m + 60 m = 160 m; Displacement = OB = 40 m in the positive direction. For straight-line motion, distance and displacement magnitude are equal only when the object travels in one direction without reversing. The moment an object changes direction, these two quantities become different, because distance keeps growing while displacement may shrink. A single instant (one moment in time) is distinct from a time interval (the span between two moments).
In simple words: Distance is the complete path you walk. Displacement is the shortest straight line from start to end. They match only if you never turn back.

Exam Tip: When drawing a motion diagram, always label start and end points clearly and calculate both quantities separately - they will differ whenever the path is not a straight line in one direction.

 

Question 2. Describe the three kinematic equations for motion in a straight line with constant acceleration. Derive the first and third equations and state the conditions under which these equations are valid.
Answer: For motion in a straight line at constant acceleration, five physical quantities interact - displacement (s), time interval (t), initial velocity (u), final velocity (v) and acceleration (a) - and they are linked by three kinematic equations.Equation 1: v = u + atStarting from the definition of average acceleration: a = (v - u) / t. Rearranging this gives: v = u + at. This equation finds the final velocity at any moment t if the initial velocity u and acceleration a are known.Equation 2: s = ut + ½at²This comes from the area under a velocity-time graph (sum of rectangle + triangle area): s = u × t + ½ × t × (v - u). Substituting (v - u) = at from Equation 1 gives: s = ut + ½at²Equation 3: v² = u² + 2asFrom Equation 1, we get t = (v - u)/a. Inserting this into Equation 2 and simplifying yields: v² = u² + 2as. This is useful when time is not given or not needed.

Conditions for validity:

• Acceleration must be constant throughout the motion.
• All equations apply to motion in a straight line.
• For motion in one direction, distance equals magnitude of displacement and speed equals magnitude of velocity.
• Signs of u, v, a, and s indicate direction - always assign negative sign when a quantity opposes the chosen positive direction.

In simple words: These three equations connect velocity, distance, time, and acceleration. Pick the one that has what you know and what you need to find - one quantity will always be missing from each equation.

Exam Tip: Always write down the given values and identify which equation has all known values and the unknown you seek - this prevents wrong formula selection and calculation errors.

 

Question 3. Describe position-time graphs and velocity-time graphs for different types of motion. What physical quantities can be calculated from each type of graph?
Answer: Graphs give a visual way to see how position and velocity shift over time. Class 9 Science Exploration Chapter 4 covers two main types of motion graphs.A. Position-Time Graphs

Calculation from a position-time graph:

• The object's position at any moment - read the y-axis value directly.
• Velocity - calculate the slope of the line using the formula: slope = (s₂ - s₁) / (t₂ - t₁) = BC/CA in the triangle method. A steeper slope indicates higher velocity; a gentler slope indicates lower velocity.

B. Velocity-Time GraphsCalculation from a velocity-time graph:

• Velocity at any moment - read the y-axis directly.
• Acceleration - calculate from the slope of the line: a = (v - u) / (t₂ - t₁) = BC/CA
• Displacement - calculate from the area between the graph line and the time axis:
  • For constant velocity: area = rectangle = velocity × time
  • For changing velocity: area = rectangle + triangle

A graph is not a map of the actual path. It shows how position or velocity changes relative to time, not the physical route the object takes. All graphs in Class 9 Science Exploration Chapter 4 apply to motion in a straight line in one direction only.
In simple words: A position-time graph's slope tells you speed. A velocity-time graph's slope tells you acceleration. The area under a velocity-time graph tells you how far something moved.

Exam Tip: Always label axes clearly with units and identify the graph type (position-time or velocity-time) before extracting information - mixing them up leads to wrong answers.

 

Question 4. What is uniform circular motion? Explain with examples and prove that an object in uniform circular motion is accelerated even though its speed is constant.
Answer:Definition: Uniform circular motion occurs when an object travels along a circular path at constant speed.Examples:

• A child on a merry-go-round moving at constant speed
• A satellite moving in a circular orbit around Earth
• A stone tied to a string and whirled in a horizontal circle at constant speed
• A car making a perfectly circular turn at constant speed

Distance and Displacement in Circular Motion: For a child moving from A to B to C on a merry-go-round:

• Distance = length of the arc ABC (curved path)
• Displacement = straight line AC

After completing one full circle:

• Distance = circumference = 2πR
• Displacement = 0 (returns to starting point)
• Average velocity = 0/T = 0 m s⁻¹
• Average speed = 2πR/T (non-zero)

Why is uniform circular motion accelerated? Velocity is a vector quantity - it has both magnitude (speed) and direction. In uniform circular motion:

• The speed (magnitude of velocity) stays constant at every point.
• The direction of velocity changes continuously - at every point on the circle, velocity points along the tangent to the circle at that point.

Since direction of velocity keeps changing, velocity itself is changing. Since acceleration is the rate of change of velocity, acceleration must be non-zero. This is proven by Activity 4.5 in the chapter - when a marble moving inside a ring is released by lifting the ring, it immediately travels in a straight line (the tangential direction at that instant), showing that the ring was continuously redirecting the marble's motion while it was inside.Key distinction: In everyday language, a vehicle is said to be accelerating only when speed increases. But in physics, acceleration takes place any time velocity changes - and velocity changes whenever direction changes, regardless of whether speed stays the same. Uniform circular motion perfectly demonstrates this principle.
In simple words: Even though speed stays the same, an object going in circles is always accelerating because it keeps turning. Turning is a change in direction, and any change in direction counts as acceleration.

Exam Tip: Emphasize that acceleration is about change in velocity (which includes both speed AND direction), not just change in speed - this conceptual clarity often separates marks on descriptive answers.

 

Question 5. A bus is travelling at 36 km h⁻¹ when the driver sees an obstacle 30 m ahead. The driver takes 0.5 s to react before pressing the brake. Once the brake is applied, velocity reduces with constant acceleration of 2.5 m s⁻². Will the bus stop before hitting the obstacle?
Answer: This problem from the Revise Reflect Refine section of Class 9 Science Exploration Chapter 4 nicely links kinematic equations to real-world road safety.Converting speed to SI units: u = 36 km h⁻¹ = 36 ÷ 3.6 = 10 m s⁻¹Calculating distance covered during reaction time: The driver takes 0.5 s to react. During this period, the bus keeps moving at 10 m s⁻¹ with no braking applied. Distance during reaction = speed × reaction time = 10 × 0.5 = 5 mCalculating braking distance after brake is applied: Given: u = 10 m s⁻¹, v = 0 m s⁻¹ (bus stops), a = -2.5 m s⁻² Using v² = u² + 2as:
0 = (10)² + 2 × (-2.5) × s
0 = 100 - 5s
s = 100/5 = 20 mCalculating total stopping distance: Total distance = reaction distance + braking distance = 5 + 20 = 25 mConclusion: The total stopping distance is 25 m, which is less than the 30 m distance to the obstacle. The bus will stop safely before reaching the obstacle - but with only 5 m to spare.
In simple words: The driver takes time to react (5 m traveled). Then the bus slows down (20 m more traveled). Together that is 25 m. Since the obstacle is 30 m away, the bus stops just in time.

Exam Tip: Always separate reaction distance and braking distance in real-world motion problems - many students forget the reaction time component, leading to an incorrect final answer.