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NCERT Class 9 Science Describing Motion Around Us: Exercise Solutions

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 9 Science book, chapter “Describing Motion Around Us”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Exercises Q1

   A father walks 250 m along a straight road from home to a shop, back home for a bag, to the shop again, and back home. Find the total distance and his displacement from home.

   1. The father walks 250 m from home to the shop.
   2. He walks 250 m back from the shop to home.
   3. He walks 250 m from home to the shop again.
   4. He walks 250 m back from the shop to home.
   5. Total distance is the sum of all lengths: 250 + 250 + 250 + 250 = 1000 m.
   6. Displacement is the shortest distance between the initial and final positions.
   7. Since he starts at home and ends at home, the displacement is 0 m.

   Answer: Total distance = 1000 m; Displacement = 0 m

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2. Exercises Q2

   A student runs from the ground floor up to the fourth floor and then down to a classroom on the second floor; each floor is 3 m high. Find (i) the total vertical distance and (ii) the displacement from the start.

   1. (i) The student starts at the ground floor (0 m).
   2. (i) He runs up to the fourth floor, which is 4 × 3 m = 12 m above the ground.
   3. (i) He then runs down to the second floor, which is 2 × 3 m = 6 m above the ground.
   4. (i) The distance covered while going up is 12 m.
   5. (i) The distance covered while going down is 12 m − 6 m = 6 m.
   6. (i) Total vertical distance = 12 m + 6 m = 18 m.
   7. (ii) Displacement is the shortest distance between the starting and final positions.
   8. (ii) The starting position is the ground floor (0 m).
   9. (ii) The final position is the second floor (6 m).
   10. (ii) Displacement = 6 m − 0 m = 6 m.

   Answer: (i) 18 m; (ii) 6 m

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3. Exercises Q3

   A scooter's speedometer reading stays constant. Can the scooter still be accelerating? How?

   1. The speedometer reading represents the magnitude of the velocity at that instant.
   2. A constant speedometer reading means the speed of the scooter is not changing.
   3. Acceleration is defined as the rate of change of velocity, which includes both speed and direction.
   4. If the scooter changes its direction of motion while maintaining the same speed, its velocity changes.
   5. Since the direction of velocity is changing, the scooter is undergoing acceleration.

   Answer: Yes, the scooter can be accelerating if it changes direction while maintaining a constant speed.

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4. Exercises Q4

   A car starts from rest and reaches 24 m s⁻¹ in 6 s. Find its average acceleration and the distance covered in those 6 s.

   1. Given: initial velocity u = 0 m s⁻¹, final velocity v = 24 m s⁻¹, time t = 6 s.
   2. Using the first equation of motion, v = u + at.
   3. Substituting the values, 24 = 0 + a × 6.
   4. Solving for acceleration, a = 24 ÷ 6 = 4 m s⁻².
   5. Using the second equation of motion, s = ut + ½at².
   6. Substituting the values, s = 0 × 6 + ½ × 4 × 6².
   7. Calculating the distance, s = 0 + 2 × 36 = 72 m.

   Answer: Average acceleration is 4 m s⁻² and distance covered is 72 m.

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5. Exercises Q5

   A motorbike moving at 28 m s⁻¹ with constant acceleration stops after 98 m. Find the acceleration and the time taken to stop.

   1. Use the equation v² = u² + 2as to find the acceleration.
   2. Substitute the values: 0² = 28² + 2 × a × 98.
   3. Simplify to get 0 = 784 + 196a.
   4. Solve for a: 196a = −784, so a = −4 m s⁻².
   5. Use the equation v = u + at to find the time taken.
   6. Substitute the values: 0 = 28 + (−4) × t.
   7. Solve for t: 4t = 28, so t = 7 s.

   Answer: Acceleration = −4 m s⁻²; Time taken = 7 s

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6. Exercises Q6

   Fig. 4.27 shows position-time graphs of objects A and B moving the same way on parallel tracks. Do they ever have equal velocity? Justify.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. Velocity is determined by the slope of a position-time graph, where a steeper slope indicates a higher velocity.
   2. Object A has a steeper slope than object B throughout the entire time interval shown in the graph.
   3. Since the slope of A is always greater than the slope of B, their velocities are never equal at any point in time.

   Answer: No, they do not have equal velocity because the slope of object A's graph is always steeper than that of object B.

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7. Exercises Q7

   Fig. 4.28: A and B start and end at the same positions in 10 s. Which are correct? (i) their average velocities are equal (same start and end) (ii) their average speeds are equal (equal distance in equal time) (iii) A's average speed is lower because it covers less distance (iv) A's average speed is higher because B is slower in some parts

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. (i) Average velocity depends on displacement, which is the change in position. Since both objects start and end at the same positions, their displacements are identical.
   2. (i) With equal displacement and the same time interval of 10 s, the average velocity for both objects is equal.
   3. (ii) Average speed is the total distance travelled divided by the time interval. The graph indicates that both objects cover the same total distance in the given time.
   4. (ii) Therefore, their average speeds are also equal because the distance and time are the same for both.
   5. (iii) and (iv) are incorrect because the total distances covered by A and B are equal, not different.

   Answer: (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 10 s time interval are equal since both cover equal distance in equal time.

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8. Exercises Q8

   A truck at 54 km h⁻¹ sees a 40 km h⁻¹ limit sign and slows uniformly to 36 km h⁻¹ in 36 s. How far does it travel in that time?

   1. Convert the initial speed from 54 km/h to m/s by multiplying by 5/18, which gives 15 m/s.
   2. Convert the final speed from 36 km/h to m/s by multiplying by 5/18, which gives 10 m/s.
   3. Calculate the acceleration using the formula a = (v - u) / t, which is (10 - 15) / 36 = -5/36 m/s².
   4. Use the equation of motion s = ut + 1/2 at² to find the distance.
   5. Substitute the values: s = 15 × 36 + 1/2 × (-5/36) × 36².
   6. Simplify the second term: 1/2 × (-5/36) × 1296 = -90 m.
   7. Add the terms: s = 540 - 90 = 450 m.

   Answer: 450 m

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9. Exercises Q9

   A car accelerates uniformly from rest to 20 m s⁻¹ in 5 s, moves at 20 m s⁻¹ for 10 s, then brakes uniformly to rest in 6 s. Find the total distance.

   1. The motion is divided into three parts: acceleration, constant speed, and deceleration.
   2. In the first part, the car accelerates from 0 to 20 m/s in 5 s, so the distance covered is s1 = (1/2) × 20 × 5 = 50 m.
   3. In the second part, the car moves at a constant speed of 20 m/s for 10 s, so the distance covered is s2 = 20 × 10 = 200 m.
   4. In the third part, the car decelerates from 20 m/s to 0 in 6 s, so the distance covered is s3 = (1/2) × 20 × 6 = 60 m.
   5. The total distance is the sum of the distances in all three parts: s = 50 + 200 + 60 = 310 m.

   Answer: 310 m

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10. Exercises Q10

    A bus at 36 km h⁻¹ sees an obstacle 30 m ahead; the driver reacts in 0.5 s, then brakes at 2.5 m s⁻². Will it stop before the obstacle?

    1. Convert the bus speed from 36 km/h to m/s by multiplying by 5/18, which gives 10 m/s.
    2. Calculate the distance covered during the 0.5 s reaction time using the formula distance = speed × time, which is 10 × 0.5 = 5 m.
    3. Use the equation of motion v² = u² + 2as to find the braking distance, where v = 0, u = 10 m/s, and a = -2.5 m/s².
    4. Substitute the values into the equation: 0 = (10)² + 2(-2.5)s, which simplifies to 0 = 100 - 5s.
    5. Solve for s to find the braking distance is 20 m.
    6. Add the reaction distance and braking distance to get the total stopping distance: 5 m + 20 m = 25 m.
    7. Compare the total stopping distance of 25 m with the obstacle distance of 30 m to see if the bus stops in time.

    Answer: Yes, the bus will stop before the obstacle as the total stopping distance is 25 m which is less than 30 m.

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11. Exercises Q11

    Someone says "the Earth moves around the Sun". Can an object kept on the Earth then be called at rest? Discuss.

    1. (i) An object is at rest if its position relative to a chosen reference point does not change with time.
    2. (ii) If the Sun is taken as the reference point, the object on Earth changes its position as the Earth revolves around the Sun, so it is in motion.
    3. (iii) If the Earth or a point on the Earth is taken as the reference point, the object's position does not change, so it is at rest.
    4. (iv) Thus, the state of rest or motion depends on the reference point chosen for observation.

    Answer: (i) Yes, if the reference point is on Earth; (ii) No, if the reference point is the Sun.

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12. Exercises Q12

    A cyclist's velocity-time graph (Fig. 4.30): 0 to 3 m s⁻¹ uniformly in the first 20 s, steady 3 m s⁻¹ from 20 s to 100 s, then down uniformly to 2 m s⁻¹ at 120 s. Which areas show the displacement (i) at constant velocity (ii) while slowing? Find the displacement and average acceleration over 120 s.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. The displacement at constant velocity is represented by the area of the rectangle from 20 s to 100 s.
    2. The displacement while slowing is represented by the area of the trapezium from 100 s to 120 s.
    3. Displacement in the first 20 s is the area of the triangle: (1/2) × 20 s × 3 m/s = 30 m.
    4. Displacement from 20 s to 100 s is the area of the rectangle: 3 m/s × (100 − 20) s = 240 m.
    5. Displacement from 100 s to 120 s is the area of the trapezium: (1/2) × (3 + 2) m/s × (120 − 100) s = 50 m.
    6. Total displacement is the sum of all parts: 30 m + 240 m + 50 m = 320 m.
    7. Average acceleration is the change in velocity divided by total time: (2 − 0) m/s ÷ 120 s = 1/60 m/s².

    Answer: 320 m, 1/60 m s⁻²

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13. Exercises Q13

    From the velocity-time graph of a girl's practice run (Fig. 4.31, speed about 7-7.5 km h⁻¹ for roughly 6.5 hours), estimate the distance she ran.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. On a velocity-time graph, the distance travelled is the area under the graph.
    2. Reading Fig. 4.31 approximately: about 7 km h⁻¹ for the first 0.6 h, rising to 7.5 km h⁻¹ by about 1.6 h, steady at 7.5 km h⁻¹ until about 3 h, then falling to about 6.5 km h⁻¹ by about 5.6 h, and steady at 6.5 km h⁻¹ until about 6.6 h.
    3. Split the area into strips: 7 × 0.6 ≈ 4.2 km; 7.25 × 1.0 ≈ 7.3 km; 7.5 × 1.4 ≈ 10.5 km; 7.25 × 1.6 ≈ 11.6 km; 6.75 × 1.0 ≈ 6.8 km; 6.5 × 1.0 ≈ 6.5 km.
    4. Adding the strips: about 47 km. Because the values are read off a graph, an answer between about 45 km and 48 km is a fair estimate.

    Answer: About 47 km (the area under the velocity-time graph; readings from the graph give roughly 45-48 km).

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14. Exercises Q14

    A car moves at a constant 6 m s⁻¹ for 2 minutes, then accelerates at 1 m s⁻² for 6 s. Using a velocity-time graph, find its displacement in the 2 min 6 s.

    1. The car moves at a constant velocity of 6 m/s for the first 2 minutes.
    2. Convert the time of 2 minutes into seconds: 2 × 60 = 120 s.
    3. Calculate the distance covered in the first part: 6 m/s × 120 s = 720 m.
    4. The car then accelerates at 1 m/s² for the next 6 seconds.
    5. Calculate the distance covered in the second part using s = ut + ½at²: (6 × 6) + ½ × 1 × 6² = 36 + 18 = 54 m.
    6. Add the distances from both parts to find the total displacement: 720 m + 54 m = 774 m.

    Answer: 774 m

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15. Exercises Q15

    Cars A and B start from rest with constant acceleration: A reaches 5 m s⁻¹ in 5 s, B reaches 3 m s⁻¹ in 10 s. Plot both velocity-time graphs and find each car's displacement over its interval.

    1. Car A starts from rest, so initial velocity u = 0 m/s.
    2. Car A reaches final velocity v = 5 m/s in time t = 5 s.
    3. The displacement of Car A is the area of the triangle under its velocity-time graph.
    4. Displacement of A = 1/2 × base × height = 1/2 × 5 s × 5 m/s.
    5. Displacement of A = 12.5 m.
    6. Car B starts from rest, so initial velocity u = 0 m/s.
    7. Car B reaches final velocity v = 3 m/s in time t = 10 s.
    8. The displacement of Car B is the area of the triangle under its velocity-time graph.
    9. Displacement of B = 1/2 × base × height = 1/2 × 10 s × 3 m/s.
    10. Displacement of B = 15 m.

    Answer: Displacement of Car A = 12.5 m; Displacement of Car B = 15 m

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16. Exercises Q16

    The minute hand of a clock is 7 cm long. From 6 PM to 7:30 PM, find for its tip (i) the distance travelled (ii) the displacement (iii) the average speed (iv) the average velocity.

    1. The time interval from 6:00 PM to 7:30 PM is 90 minutes, which equals 5400 seconds.
    2. In 90 minutes, the minute hand completes 1.5 full revolutions around the clock face.
    3. The path of the tip is a circle with radius r = 7 cm, so the circumference is 2 × π × 7 = 14π cm.
    4. (i) Distance travelled is 1.5 times the circumference, so 1.5 × 14π = 21π cm.
    5. Using π ≈ 22/7, the distance is 21 × 22/7 = 66 cm.
    6. (ii) At 6:00 PM the tip is at the 12 position, and at 7:30 PM it is at the 6 position.
    7. These two positions are diametrically opposite on the clock face.
    8. The displacement is the straight-line distance between these points, which is the diameter 2 × 7 = 14 cm.
    9. (iii) Average speed is total distance divided by total time, so 66 cm / 5400 s.
    10. Simplifying the fraction gives 11/900 cm s⁻¹.
    11. (iv) Average velocity is total displacement divided by total time, so 14 cm / 5400 s.
    12. Simplifying the fraction gives 7/2700 cm s⁻¹.

    Answer: (i) 66 cm; (ii) 14 cm; (iii) 11/900 cm s⁻¹; (iv) 7/2700 cm s⁻¹

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