Class 9 Science Test Paper Chapter 7 Motion Numericals | CBSE Board | Detailed Solutions & Marking Scheme

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VEDANT IGNITE TEST SERIES

Class: 9  Subject: Physics  Chapter: Motion  Board: CBSE
Marks: 30

Section A (7 Marks)

(5 MCQs + 2 Assertion-Reason, 1 mark each — Competency-based Numericals)

Q1. Riya cycles from home to school. She covers 150 m in first 10 s and 250 m in next 10 s. Find her average speed for entire journey.
Options: (A) 15 m/s (B) 20 m/s (C) 25 m/s (D) 30 m/s

Q2. A delivery van starts from rest and reaches 18 m/s in 6 s while moving in a straight road. Find distance covered during this time.
Options: (A) 36 m (B) 54 m (C) 72 m (D) 108 m

Q3. A train moving at 90 km/h applies brakes and stops in 10 s. Find retardation.
Options: (A) 2.5 m/s² (B) 3 m/s² (C) 2 m/s² (D) 4 m/s²

Q4. A ball is thrown vertically upward with velocity 20 m/s. Find time taken to reach maximum height. (g = 10 m/s²)
Options: (A) 1 s (B) 2 s (C) 3 s (D) 4 s

Q5. A car travels half distance at 30 km/h and remaining half at 60 km/h. Find average speed.
Options: (A) 40 km/h (B) 45 km/h (C) 36 km/h (D) 50 km/h

Q6. Assertion-Reason
Assertion (A): A body moving with constant speed in a circular path has acceleration.
Reason (R): Velocity changes due to change in direction.
Options: (A) Both A and R true, R correct explanation
(B) Both A and R true, R not correct explanation
(C) A true, R false
(D) A false, R true

Q7. Assertion-Reason
Assertion (A): If velocity of a body is zero, acceleration must also be zero.
Reason (R): Acceleration depends on change in velocity.
Options: (A) Both A and R true, R correct explanation
(B) Both A and R true, R not correct explanation
(C) A true, R false
(D) A false, R true

Section B (8 Marks)

(4 Questions of 2 Marks each — Attempt all)

Q8. A student runs from one end of a 200 m track to the other in 20 s and returns in 30 s. Calculate:
(a) Average speed
(b) Average velocity

Q9. A bus starts from rest and accelerates uniformly at 2 m/s² for 10 s. Find:
(a) Final velocity
(b) Distance travelled

Q10. A car moving at 72 km/h slows down uniformly and stops in 8 s. Calculate:
(a) Retardation
(b) Distance travelled

Q11. A stone is thrown vertically upward with velocity 30 m/s. Find:
(a) Time to reach maximum height
(b) Maximum height (g = 10 m/s²)

Section C (6 Marks)

(2 Questions of 3 Marks each — Attempt both)

Q12. A cab driver travels 10 km at 40 km/h, then gets stuck in traffic and travels next 10 km at 20 km/h.
Find:
(a) Total time taken
(b) Average speed for entire journey

Q13. A cyclist starts from rest and accelerates at 1.5 m/s² for 8 s, then moves with constant velocity for next 12 s.
Find:
(a) Velocity after 8 s
(b) Total distance travelled

Section D (5 Marks)

(1 Question of 5 Marks — Case-based Numerical)

Q14. A metro train starts from rest at a station and accelerates uniformly at 1 m/s² for 20 s. It then travels at constant speed for 40 s and finally decelerates uniformly at 2 m/s² to rest.

Find:
(a) Maximum velocity attained
(b) Distance travelled during acceleration
(c) Total distance travelled
(d) Total time of journey

Section E (4 Marks)

(Case Study – Competency Based)

Q15. Rahul observes a speed-time graph of a car. The car starts from rest, reaches 20 m/s in 5 s, continues at this speed for 10 s, and then comes to rest in next 5 s.

Answer the following:

(a) What is acceleration during first 5 s? (1 mark)
(b) What is retardation during last 5 s? (1 mark)
(c) Calculate total distance travelled by the car. (2 marks)

Here is a complete board-level solution package (step-by-step + marking scheme + analysis + mistakes + pro tips) exactly in your preferred easy-to-copy format 👇

🟦 VEDANT IGNITE TEST SERIES – SOLUTIONS

Class: 9 | Physics | Chapter: Motion | CBSE | Marks: 30

🟩 SECTION A – SOLUTIONS (7 Marks)

Q1

Total distance = 150 + 250 = 400 m
Total time = 10 + 10 = 20 s

Average speed = Total distance / Total time
= 400 / 20 = 20 m/s

✅ Answer: (B) 20 m/s

Q2

u = 0, v = 18 m/s, t = 6 s

Distance:
s = ut + (1/2)at²

First find a:
a = (v − u)/t = 18/6 = 3 m/s²

Now,
s = 0 + (1/2 × 3 × 6²)
= 1.5 × 36 = 54 m

✅ Answer: (B) 54 m

Q3

u = 90 km/h = 25 m/s
v = 0, t = 10 s

a = (v − u)/t = (0 − 25)/10 = −2.5 m/s²

Retardation = 2.5 m/s²

✅ Answer: (A)

Q4

u = 20 m/s, v = 0

v = u − gt
0 = 20 − 10t
t = 2 s

✅ Answer: (B) 2 s

Q5

Average speed (equal distances):

Formula:
Average speed = (2v₁v₂)/(v₁ + v₂)

= (2 × 30 × 60)/(30 + 60)
= 3600 / 90 = 40 km/h

✅ Answer: (A)

Q6 (Assertion-Reason)

A: True
R: True and correct explanation

✅ Answer: (A)

Q7 (Assertion-Reason)

A: False (velocity zero but acceleration can exist, e.g., top point of motion)
R: True

✅ Answer: (D)

🟩 SECTION B – SOLUTIONS (8 Marks)

Q8 (2 Marks)

Total distance = 200 + 200 = 400 m
Total time = 20 + 30 = 50 s

(a) Average speed
= 400 / 50 = 8 m/s

(b) Average velocity
Displacement = 0
= 0 m/s

🟢 Marking:

  • Speed (1)

  • Velocity (1)

Q9 (2 Marks)

u = 0, a = 2, t = 10

(a) v = u + at
= 0 + 2×10 = 20 m/s

(b) s = ut + (1/2)at²
= 0 + (1/2 × 2 × 100) = 100 m

🟢 Marking:

  • v (1)

  • s (1)

Q10 (2 Marks)

u = 72 km/h = 20 m/s
v = 0, t = 8

(a) a = (0 − 20)/8 = −2.5 m/s²

(b) s = ut + (1/2)at²
= 20×8 + (1/2 × −2.5 × 64)
= 160 − 80 = 80 m

🟢 Marking:

  • a (1)

  • s (1)

Q11 (2 Marks)

u = 30 m/s

(a) t = u/g = 30/10 = 3 s

(b) h = u² / (2g)
= 900 / 20 = 45 m

🟢 Marking:

  • t (1)

  • h (1)

🟩 SECTION C – SOLUTIONS (6 Marks)

Q12 (3 Marks)

Time₁ = 10/40 = 0.25 h
Time₂ = 10/20 = 0.5 h

(a) Total time = 0.75 h

(b) Average speed = Total distance / Total time
= 20 / 0.75 = 26.67 km/h

🟢 Marking:

  • Time₁, Time₂ (1)

  • Total time (1)

  • Avg speed (1)

Q13 (3 Marks)

u = 0, a = 1.5, t₁ = 8

(a) v = u + at
= 1.5×8 = 12 m/s

Distance₁:
s₁ = (1/2)at²
= 0.5×1.5×64 = 48 m

Distance₂:
s₂ = v × t = 12 × 12 = 144 m

Total distance = 48 + 144 = 192 m

🟢 Marking:

  • v (1)

  • s₁ (1)

  • total (1)

🟩 SECTION D – CASE NUMERICAL (5 Marks)

Q14

Phase 1 (Acceleration):
u = 0, a = 1, t = 20

(a) v = at = 20 m/s

(b) s₁ = (1/2)at²
= 0.5×1×400 = 200 m

Phase 2 (Constant speed):
s₂ = vt = 20×40 = 800 m

Phase 3 (Deceleration):
u = 20, v = 0, a = −2

t₃ = (0 − 20)/−2 = 10 s

s₃ = ut + (1/2)at²
= 20×10 + (1/2×−2×100)
= 200 − 100 = 100 m

(c) Total distance = 200 + 800 + 100 = 1100 m

(d) Total time = 20 + 40 + 10 = 70 s

🟢 Marking:

  • v (1)

  • s₁ (1)

  • s₂/s₃ (1)

  • total distance (1)

  • total time (1)

🟩 SECTION E – CASE STUDY (4 Marks)

Q15

(a) a = (20 − 0)/5 = 4 m/s²

(b) retardation = (0 − 20)/5 = −4 m/s²

(c) Distance = Area under graph

Triangle₁ = (1/2 × 5 × 20) = 50
Rectangle = (10 × 20) = 200
Triangle₂ = (1/2 × 5 × 20) = 50

Total = 300 m

🟢 Marking:

  • a (1)

  • retardation (1)

  • area calc (2)

📊 PAPER ANALYSIS (VERY IMPORTANT)

Difficulty Level: ⭐⭐⭐☆☆ (Moderate)

Chapter Weightage:

  • Equations of motion → 60%

  • Graph-based → 15%

  • Concept (AR) → 10%

  • Average speed → 15%

Student Performance Prediction:

  • 25–30 → Strong conceptual clarity

  • 18–24 → Formula based learning

  • <15 → Weak basics in motion

🚀 PRO TIPS (BOARD EXAM)

  1. Always convert units first

    • km/h → m/s (×5/18)

  2. Use correct formula selection

    • v = u + at

    • s = ut + ½at²

    • v² = u² + 2as

  3. Write steps → marks guaranteed
    Even if final answer wrong → 50–70% marks

  4. Graphs = easy scoring

    • Area under graph = distance

    • Slope = acceleration

  5. Equal distance → harmonic mean formula

❌ COMMON MISTAKES SHEET (VERY IMPORTANT)

🔴 1. Unit Conversion Mistake

❌ 72 km/h = 72 m/s
✅ Correct: 72 × 5/18 = 20 m/s

🔴 2. Sign Mistake in Retardation

❌ a = positive
✅ Always negative in deceleration

🔴 3. Using wrong formula

❌ s = vt in accelerated motion
✅ Use s = ut + ½at²

🔴 4. Average Speed vs Velocity

  • Speed → total distance

  • Velocity → displacement

🔴 5. Forgetting 3 phases (Case study)

Students often miss:

  • acceleration

  • constant speed

  • deceleration

🔴 6. Graph Question Mistake

❌ adding speeds
✅ Take area under graph

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