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CBSE Class 9  >  Class 9 Test  >  Science New NCERT 2026-27 (New Syllabus)  >  Test: Describing Motion Around Us - 2 - Class 9 MCQ

Describing Motion Around Us - 2 - Free MCQ Practice Test with solutions,


MCQ Practice Test & Solutions: Test: Describing Motion Around Us - 2 (15 Questions)

You can prepare effectively for Class 9 Science Class 9 New NCERT 2026-27 (New Syllabus) with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Describing Motion Around Us - 2". These 15 questions have been designed by the experts with the latest curriculum of Class 9 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 15 minutes
  • - Number of Questions: 15

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Test: Describing Motion Around Us - 2 - Question 1
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A cyclist travels 3 km East, then 4 km North. What is the magnitude of his displacement?

Detailed Solution: Question 1

The path forms a right-angled triangle. Using Pythagorean theorem: Displacement = √(3² + 4²) = √(9 + 16) = √25 = 5 km. Note that total distance = 3 + 4 = 7 km, which is greater than displacement.

Test: Describing Motion Around Us - 2 - Question 2
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A train moves at 72 km/h. What is its speed in m/s?

Detailed Solution: Question 2

To convert km/h to m/s, multiply by 5/18. Speed = 72 × 5/18 = 360/18 = 20 m/s.

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Test: Describing Motion Around Us - 2 - Question 3
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A ball is thrown upward and returns to the same point after 6 seconds. If it travelled 30 m upward before returning, what is the average speed?

Detailed Solution: Question 3

Total distance = 30 + 30 = 60 m (up and back). Total time = 6 s. Average speed = 60/6 = 10 m/s. Average velocity = 0 (displacement is zero since it returns to starting point).

Test: Describing Motion Around Us - 2 - Question 4
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On a position-time graph, object P has a slope of 10 m/s and object Q has a slope of 6 m/s. Which object is moving faster and by how much?

Detailed Solution: Question 4

The slope of a position-time graph gives the velocity. P has slope 10 m/s and Q has slope 6 m/s. Therefore P is faster. Difference = 10 − 6 = 4 m/s. Steeper slope always means higher velocity.

Test: Describing Motion Around Us - 2 - Question 5
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A rocket starting from rest achieves a velocity of 600 m/s in 30 seconds. What is its acceleration?

Detailed Solution: Question 5

u = 0 m/s (starts from rest), v = 600 m/s, t = 30 s. Using a = (v − u)/t = (600 − 0)/30 = 20 m/s².

Test: Describing Motion Around Us - 2 - Question 6
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A scooter moving at 18 m/s applies brakes and stops in 9 seconds. What is the deceleration?

Detailed Solution: Question 6

u = 18 m/s, v = 0 m/s (stops), t = 9 s. a = (v − u)/t = (0 − 18)/9 = −2 m/s². The negative sign indicates deceleration (acceleration opposite to direction of motion).

Test: Describing Motion Around Us - 2 - Question 7
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A merry-go-round completes one full revolution of radius 3.5 m in 22 seconds. What is the average speed of a child sitting on it? (Use π = 22/7)

Detailed Solution: Question 7

Distance for one revolution = 2πR = 2 × (22/7) × 3.5 = 22 m. Average speed = 22/22 = 1 m/s. Average velocity = 0 (displacement is zero after one complete revolution).

Test: Describing Motion Around Us - 2 - Question 8
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Which of the following motions has zero displacement but non-zero distance?

Detailed Solution: Question 8

When a planet completes one full orbit, it returns to its starting position — so displacement = 0. However, the total path (circumference of orbit) is non-zero, so distance ≠ 0. This is similar to circular motion where one full revolution gives zero displacement.

Test: Describing Motion Around Us - 2 - Question 9
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A bus accelerates uniformly from 10 m/s to 25 m/s over a distance of 105 m. What is its acceleration?

Detailed Solution: Question 9

u = 10 m/s, v = 25 m/s, s = 105 m. Using v² = u² + 2as: 25² = 10² + 2a(105) → 625 − 100 = 210a → 525 = 210a → a = 2.5 m/s².

Test: Describing Motion Around Us - 2 - Question 10
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A stone dropped from a height reaches the ground with a velocity of 20 m/s. If acceleration due to gravity is 10 m/s², from what height was it dropped?

Detailed Solution: Question 10

u = 0 (dropped from rest), v = 20 m/s, a = 10 m/s². Using v² = u² + 2as: 400 = 0 + 2 × 10 × s → s = 400/20 = 20 m.

Test: Describing Motion Around Us - 2 - Question 11
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A velocity-time graph shows a straight horizontal line at v = 15 m/s from t = 0 to t = 8 s. What is the displacement and acceleration?

Detailed Solution: Question 11

A horizontal line on a velocity-time graph means constant velocity, so acceleration = 0. Displacement = area under graph = velocity × time = 15 × 8 = 120 m.

Test: Describing Motion Around Us - 2 - Question 12
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Which of the following is an example of motion where speed is constant but the object is still accelerating?

Detailed Solution: Question 12

In uniform circular motion, speed remains constant but the direction of velocity changes continuously, meaning velocity changes. Since acceleration = rate of change of velocity, the object is accelerating even though speed is constant. This is acceleration due to change in direction alone.

Test: Describing Motion Around Us - 2 - Question 13
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A particle starts from rest and travels with constant acceleration. Which graph correctly represents its motion?

Detailed Solution: Question 13

For constant acceleration starting from rest, velocity increases uniformly with time. Since position = ut + ½at², and u = 0, position = ½at², which is a parabola. Hence the position-time graph is a curve, not a straight line.

Test: Describing Motion Around Us - 2 - Question 14
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A motorcycle travels 150 m in 10 seconds starting from rest with uniform acceleration. What is its final velocity at the end of 10 seconds?

Detailed Solution: Question 14

u = 0 (starts from rest), s = 150 m, t = 10 s. Using s = ut + ½at²: 150 = 0 + ½ × a × 100 → a = 3 m/s². Now, v = u + at = 0 + 3 × 10 = 30 m/s.

Test: Describing Motion Around Us - 2 - Question 15
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Two objects X and Y start from the same point. X travels 60 m East and returns 60 m West. Y travels 60 m East only. After their journeys, which statement is correct?

Detailed Solution: Question 15

X travels 60 m East then 60 m West, returning to start — distance = 120 m, displacement = 0. Y travels 60 m East only — distance = displacement = 60 m. This demonstrates that distance ≥ magnitude of displacement, and displacement can be zero even when distance is non-zero.

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About this Class 9 Practice Test

This test contains 15 MCQs on Describing Motion Around Us - 2 with detailed solutions for each question. It is part of the Science Class 9 New NCERT 2026-27 (New Syllabus) course on EduRev, designed to align with the current Class 9 syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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