‘Master Motion Topic-Wise 25 Easy Questions!” is caefully curated set of 25 questions arranged topic wise as per your NCERT textbook’s chapter 7 Motion

Ready to move beyond theory and truly grasp Motion concepts? While our other posts break down the fundamentals, mastery comes from targeted practice. That’s exactly why we’ve curated this essential set: Master Motion Topic-Wise 25 Effortless Questions!

These 25 precisely crafted, easy questions are your key to building rock-solid confidence. Focused topic-by-topic, they provide the perfect, manageable practice volume to reinforce understanding, identify weak spots, and solidify your skills – effortlessly turning knowledge into problem-solving power.

Before you rush to solve the questions, please understand and master the motion chapter from our below posts:

Motion Chapter Explanantion

Motion Shortes Notes

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Topic 1: Distance and Displacement

Question 1: A person walks 10 meters East and then 5 meters West. What is the total distance covered and the magnitude of their displacement?

Answer 1:

Total Distance Covered: The total path length covered by an object is known as the distance. In this case, the person walks 10 m + 5 m = 15 meters.
Magnitude of Displacement: The shortest distance measured from the initial to the final position of an object is known as the displacement. The person starts at a point, moves 10 m East, and then moves 5 m West from that point. Their final position is 10 m – 5 m = 5 m East of the starting point. So, the magnitude of the displacement is 5 meters.

Question 2: An object moves along a straight path from point O to point A, a distance of 60 km. It then moves back along the same path and reaches point B, which is 25 km from A (towards O). What is the total distance covered by the object and the magnitude of its displacement from O?

Answer 2:

Total Distance Covered: The total path length covered by the object is OA + AB.
- OA = 60 km.
- AB = 25 km.
- Total distance = 60 km + 25 km = 85 km.
Magnitude of Displacement: Displacement is the shortest distance from the initial position to the final position. The object starts at O and ends at B. Point A is 60 km from O. Point B is 25 km back from A.
- The distance of B from O is 60 km (OA) – 25 km (AB) = 35 km. This is the magnitude of the displacement.

Question 3: An object travels from point X to point Y and then returns to point X. If the distance between X and Y is 50 meters, what is the total distance covered and the magnitude of the displacement?

Answer 3:

Total Distance Covered: The object travels 50 m from X to Y and then 50 m from Y back to X.
- Total distance = 50 m + 50 m = 100 meters.
Magnitude of Displacement: The displacement is the shortest distance from the initial to the final position. Since the object starts at X and returns to X, its final position coincides with its initial position.
- Therefore, the displacement is zero meters. This demonstrates that the magnitude of displacement for a course of motion can be zero even if the corresponding distance covered is not zero.

Topic 2: Describing Motion and Reference Points

Question 4: A car’s odometer reads 1500 km at the start of a journey and 1750 km at the end of the journey. How much distance did the car travel during this journey?

Answer 4:

An odometer is a device in automobiles that shows the distance traveled.
To find the distance traveled, we calculate the difference between the final reading and the initial reading of the odometer.
Distance traveled = Final reading – Initial reading = 1750 km – 1500 km = 250 km.

Topic 3: Speed and Velocity

Question 5: A runner covers a distance of 400 meters in 80 seconds. What is the runner’s average speed?

Answer 5:

Speed is defined as the distance travelled by the object in unit time.
Average speed is obtained by dividing the total distance travelled by the total time taken.
Formula: Average speed (v) = Total distance (s) / Total time (t).
Given: s = 400 m, t = 80 s.
Average speed = 400 m / 80 s = 5 m/s.

Question 6: An object travels 20 m in 5 s and then another 30 m in 5 s. What is the average speed of the object?

Answer 6:

Total distance travelled by the object = 20 m + 30 m = 50 m.
Total time taken = 5 s + 5 s = 10 s.
Average speed = Total distance travelled / Total time taken.
Average speed = 50 m / 10 s = 5 m/s.

Question 7: A car travels 100 km North in 2 hours and then 50 km South in 1 hour. What is the average speed and average velocity of the car?

Answer 7:

Average Speed:
- Total distance covered = 100 km + 50 km = 150 km.
- Total time taken = 2 hours + 1 hour = 3 hours.
- Average speed = Total distance covered / Total time taken = 150 km / 3 h = 50 km/h.
Average Velocity:
- Displacement: The car travels 100 km North and then 50 km South. The net displacement is 100 km (North) – 50 km (South) = 50 km North (from the starting point).
- Average velocity = Displacement / Total time taken.
- Average velocity = 50 km (North) / 3 h = 16.67 km/h North.
- Note: Velocity requires both magnitude and direction.

Question 8: A bus travels a distance of 200 km in 4 hours. What is the speed of the bus?

Answer 8:

Speed (v) = Distance (s) / Time (t).
Given: s = 200 km, t = 4 h.
Speed = 200 km / 4 h = 50 km/h.

Question 9: Convert a speed of 36 km/h to m/s.

Answer 9:

To convert km/h to m/s, we use the conversion factors: 1 km = 1000 m and 1 hour = 3600 seconds.
36 km/h = 36 × (1000 m / 3600 s) = 36 × (5/18) m/s.
Speed = 10 m/s.

Question 10: Convert a speed of 15 m/s to km/h.

Answer 10:

To convert m/s to km/h, we use the conversion factors: 1 m = 1/1000 km and 1 s = 1/3600 h.
15 m/s = 15 × ( (1/1000) km / (1/3600) h ) = 15 × (3600/1000) km/h = 15 × (18/5) km/h.
Speed = 54 km/h.

Topic 4: Acceleration

Question 11: A car’s velocity changes from 10 m/s to 20 m/s in 5 seconds. Calculate the acceleration of the car.

Answer 11:

Acceleration is a measure of the change in the velocity of an object per unit time.
Formula: Acceleration (a) = (Final velocity (v) – Initial velocity (u)) / Time taken (t).
Given: u = 10 m/s, v = 20 m/s, t = 5 s.
a = (20 m/s – 10 m/s) / 5 s = 10 m/s / 5 s = 2 m/s².
The SI unit of acceleration is m s⁻².

Question 12: A bicycle is moving at 8 m/s and then applies brakes, slowing down to 2 m/s in 3 seconds. Calculate the acceleration of the bicycle.

Answer 12:

Given: Initial velocity (u) = 8 m/s, Final velocity (v) = 2 m/s, Time (t) = 3 s.
Acceleration (a) = (v – u) / t.
a = (2 m/s – 8 m/s) / 3 s = -6 m/s / 3 s = -2 m/s².
The negative sign indicates that the acceleration is in the opposite direction to the velocity, meaning the bicycle is decelerating.

Question 13: A bus starting from rest attains a speed of 10 m/s in 20 seconds. What is the uniform acceleration of the bus?

Answer 13:

“Starting from rest” means the initial velocity (u) = 0 m/s.
Given: u = 0 m/s, v = 10 m/s, t = 20 s.
Acceleration (a) = (v – u) / t.
a = (10 m/s – 0 m/s) / 20 s = 10 m/s / 20 s = 0.5 m/s².
The motion is with uniform acceleration because its velocity increases by equal amounts in equal intervals of time.

Topic 5: Equations of Motion

Question 14: A car starts from rest and accelerates uniformly at 2 m/s² for 10 seconds. What is its final velocity?

Answer 14:

This uses the velocity-time relation: v = u + at.
Given: u = 0 m/s (starts from rest), a = 2 m/s², t = 10 s.
v = 0 + (2 m/s² × 10 s) = 20 m/s.

Question 15: A train is moving at 10 m/s and accelerates uniformly at 0.5 m/s². How long will it take to reach a velocity of 20 m/s?

Answer 15:

This uses the velocity-time relation: v = u + at.
Given: u = 10 m/s, v = 20 m/s, a = 0.5 m/s².
Rearranging the formula for t: t = (v – u) / a.
t = (20 m/s – 10 m/s) / 0.5 m/s² = 10 m/s / 0.5 m/s² = 20 seconds.

Question 16: A car travelling at 15 m/s applies brakes and comes to a stop in 5 seconds. What is the acceleration produced by the brakes?

Answer 16:

This uses the velocity-time relation: v = u + at.
Given: u = 15 m/s, v = 0 m/s (comes to a stop), t = 5 s.
Rearranging the formula for a: a = (v – u) / t.
a = (0 m/s – 15 m/s) / 5 s = -15 m/s / 5 s = -3 m/s².
The negative sign indicates deceleration.

Question 17: A motorboat starts from rest and accelerates uniformly at 2 m/s² for 5 seconds. How far does the boat travel during this time?

Answer 17:

This uses the position-time relation: s = ut + ½ at².
Given: u = 0 m/s (starts from rest), a = 2 m/s², t = 5 s.
s = (0 m/s × 5 s) + ½ × (2 m/s² × (5 s)²).
s = 0 + ½ × 2 × 25 m = 25 meters.

Question 18: A car accelerates uniformly from 5 m/s to 15 m/s in 5 seconds. What is the distance covered by the car in that time?

Answer 18:

This uses the position-time relation: s = ut + ½ at². First, we need to find the acceleration.
- Acceleration (a) = (v – u) / t = (15 m/s – 5 m/s) / 5 s = 10 m/s / 5 s = 2 m/s².
Now, calculate the distance:
- s = (5 m/s × 5 s) + ½ × (2 m/s² × (5 s)²).
- s = 25 m + ½ × 2 × 25 m = 25 m + 25 m = 50 meters.

Question 19: A ball rolling down an inclined plane has a uniform acceleration of 4 cm/s². If it starts from rest, what distance will it cover when its velocity reaches 12 cm/s?

Answer 19:

This uses the position-velocity relation: 2as = v² – u².
Given: u = 0 cm/s (starts from rest), a = 4 cm/s², v = 12 cm/s.
2 × 4 cm/s² × s = (12 cm/s)² – (0 cm/s)².
8s = 144.
s = 144 / 8 = 18 cm.

Question 20: An object moving with an initial velocity of 2 m/s has a uniform acceleration of 3 m/s². What will be its velocity after it has covered a distance of 8 meters?

Answer 20:

This uses the position-velocity relation: 2as = v² – u².
Given: u = 2 m/s, a = 3 m/s², s = 8 m.
2 × 3 m/s² × 8 m = v² – (2 m/s)².
48 = v² – 4.
v² = 48 + 4 = 52.
v = √52 ≈ 7.21 m/s.

Question 21: A train starting from rest attains a velocity of 30 km/h in 10 minutes. Assuming uniform acceleration, find (i) the acceleration in m/s² and (ii) the distance travelled by the train in meters.

Answer 21:

First, convert units:
- u = 0 m/s (from rest).
- v = 30 km/h = 30 × (1000/3600) m/s = 30 × (5/18) m/s = 8.33 m/s (approx).
- t = 10 minutes = 10 × 60 s = 600 s.
(i) Acceleration (a):
- a = (v – u) / t.
- a = (8.33 m/s – 0 m/s) / 600 s ≈ 0.0139 m/s².
(ii) Distance travelled (s):
- Using s = ut + ½ at².
- s = (0 × 600) + ½ × (0.0139 m/s² × (600 s)²).
- s = ½ × 0.0139 × 360000 = 2502 meters (approx).
- Alternatively, using 2as = v² – u²:
  - 2 × 0.0139 × s = (8.33)² – 0²
  - 0.0278s = 69.39
  - s = 69.39 / 0.0278 = 2496 meters (slight difference due to rounding of velocity, but close).

Question 22: A car accelerates uniformly from 18 km/h to 36 km/h in 5 seconds. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.

Answer 22:

First, convert velocities to m/s:
- u = 18 km/h = 18 × (5/18) m/s = 5 m/s.
- v = 36 km/h = 36 × (5/18) m/s = 10 m/s.
- t = 5 s.
(i) Acceleration (a):
- a = (v – u) / t = (10 m/s – 5 m/s) / 5 s = 5 m/s / 5 s = 1 m/s².
(ii) Distance covered (s):
- Using s = ut + ½ at².
- s = (5 m/s × 5 s) + ½ × (1 m/s² × (5 s)²).
- s = 25 m + ½ × 1 × 25 m = 25 m + 12.5 m = 37.5 meters.

Question 23: A bus decreases its speed from 72 km/h to 36 km/h in 5 seconds. Find the acceleration of the bus and the distance it travels during this time.

Answer 23:

First, convert velocities to m/s:
- u = 72 km/h = 72 × (5/18) m/s = 20 m/s.
- v = 36 km/h = 36 × (5/18) m/s = 10 m/s.
- t = 5 s.
Acceleration (a):
- a = (v – u) / t = (10 m/s – 20 m/s) / 5 s = -10 m/s / 5 s = -2 m/s².
Distance travelled (s):
- Using s = ut + ½ at².
- s = (20 m/s × 5 s) + ½ × (-2 m/s² × (5 s)²)
- s = 100 m + ½ × (-2) × 25 m = 100 m – 25 m = 75 meters.

Topic 6: Uniform Circular Motion

Question 24: An object moves in a circular path of radius 7 meters. If it completes one full round in 11 seconds, what is its speed? (Use π = 22/7)

Answer 24:

The distance covered in one full round of a circular path is its circumference, given by 2πr.
Speed (v) = Distance / Time.
Given: r = 7 m, t = 11 s, π = 22/7.
Distance = 2 × (22/7) × 7 m = 44 m.
Speed (v) = 44 m / 11 s = 4 m/s.
This is an example of uniform circular motion, where an object moves in a circular path with uniform speed.

Question 25: An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve once around the Earth.

Answer 25:

The distance covered in one revolution is the circumference of the orbit: 2πr.
Speed (v) = Distance / Time.
Given: r = 42250 km, t = 24 hours.
Using π ≈ 3.14 (a common approximation, although 22/7 is also used).
Distance = 2 × 3.14 × 42250 km = 265460 km.
Speed (v) = 265460 km / 24 hours = 11060.83 km/h (approx).
To convert to m/s (as often used for satellites):
- 11060.83 km/h × (1000 m / 1 km) × (1 h / 3600 s) = 3072.45 m/s (approx).

Conclusion : Motion Topic-Wise 25 Easy Questions

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