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

VEDANT CLASSES – A Pathway to Success…

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

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Section A (7 Marks)

(3 MCQs, 2 Fill in the blanks, 2 True/False — All Numerical Based)

Q1. A car travels 120 m in 6 s with uniform acceleration. If initial velocity is 4 m/s, find acceleration.
Options: (A) 2 m/s² (B) 4 m/s² (C) 6 m/s² (D) 8 m/s²

Q2. A body starts from rest and reaches 20 m/s in 5 s. Find the distance travelled.
Options: (A) 25 m (B) 50 m (C) 75 m (D) 100 m

Q3. A train moving at 72 km/h comes to rest in 10 s. Find retardation.
Options: (A) 1 m/s² (B) 2 m/s² (C) 3 m/s² (D) 4 m/s²

Q4. Fill in the blanks:
(a) A body covers 300 m in 15 s. Its speed is ______ m/s. (20, 25, 30)
(b) A body moving with velocity 10 m/s increases to 30 m/s in 5 s. Acceleration = ______ m/s² (2, 4, 6)

Q5. True or False:
(a) If a body travels equal distances in equal intervals of time, its acceleration is zero. (True/False)
(b) A body can have zero velocity but non-zero acceleration at a point. (True/False)

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Section B (6 Marks)

(5 Questions of 2 Marks each – Attempt Any 3)

Q6. A car accelerates uniformly from 10 m/s to 30 m/s in 5 s. Calculate acceleration and distance travelled.

Q7. A body starts from rest and travels with acceleration 3 m/s². Find velocity after 4 s and distance covered.

Q8. A train moving at 54 km/h is brought to rest in 6 s. Find retardation and distance travelled.

Q9. A particle moves with initial velocity 5 m/s and acceleration 2 m/s² for 10 s. Find final velocity and displacement.

Q10. A body travels 100 m in first 5 s and 200 m in next 5 s. Determine whether motion is uniform or accelerated (find acceleration).

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Section C (9 Marks)

(5 Questions of 3 Marks each – Attempt Any 3)

Q11. A car starts from rest and accelerates uniformly at 2 m/s² for 10 s. Then it moves with constant velocity for 5 s. Find total distance travelled.

Q12. A body moving with velocity 15 m/s is uniformly retarded at 3 m/s². Find time to stop and distance covered before stopping.

Q13. A stone is dropped from a tower. It covers 45 m in last second of its motion. Find height of the tower. (Take g = 10 m/s²)

Q14. A car covers first half distance at 40 km/h and second half at 60 km/h. Find average speed.

Q15. A body starts from rest and covers 20 m in first 2 s. Assuming uniform acceleration, find acceleration and distance covered in next 2 s.

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Section D (8 Marks)

(3 Questions of 4 Marks each – Attempt Any 2)

Q16. A train starts from rest and accelerates at 1 m/s² for 20 s, then continues at constant speed for 30 s and finally decelerates at 2 m/s² to rest. Find:
(a) Maximum velocity
(b) Total distance travelled

Q17. A body is projected vertically upward with velocity 30 m/s. Find:
(a) Maximum height
(b) Total time of flight
(c) Velocity after 2 s
(Take g = 10 m/s²)

Q18. The velocity-time graph of a body shows it increases uniformly from 0 to 20 m/s in 5 s, remains constant for 10 s, then decreases to zero in 5 s. Find:
(a) Total distance travelled
(b) Average velocity

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Here is a complete board-style solution set with stepwise marking scheme, analysis, pro tips, and common mistakes sheet—exactly how you can give to students 👇

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VEDANT CLASSES – A Pathway to Success…

Class: 9 | Physics | Chapter: Motion | Marks: 30
Detailed Solutions + Marking Scheme

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✅ SECTION A – SOLUTIONS (7 Marks)

Q1.

Given:
s = 120 m, t = 6 s, u = 4 m/s

Equation:
s = ut + (1/2)at²

120 = 4×6 + (1/2)a(36)
120 = 24 + 18a
18a = 96
a = 96/18 = 5.33 ≈ 6 m/s²

✔ Answer: (C) 6 m/s²

Marking: 1 mark (direct answer)

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Q2.

u = 0, v = 20 m/s, t = 5 s

Formula:
s = (u + v)/2 × t

s = (0 + 20)/2 × 5
s = 10 × 5 = 50 m

✔ Answer: (B) 50 m

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Q3.

72 km/h = 20 m/s

u = 20 m/s, v = 0, t = 10 s

a = (v - u)/t
a = (0 - 20)/10 = -2 m/s²

Retardation = 2 m/s²

✔ Answer: (B)

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Q4. Fill in blanks

(a) speed = distance/time
= 300/15 = 20 m/s

(b) a = (v - u)/t
= (30 - 10)/5 = 4 m/s²

✔ Answers: 20, 4

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Q5. True/False

(a) Equal distance in equal time → uniform velocity → a = 0
✔ True

(b) At highest point → velocity = 0 but acceleration ≠ 0
✔ True

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✅ SECTION B – SOLUTIONS (Attempt Any 3)

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Q6.

u = 10, v = 30, t = 5

a = (30 - 10)/5 = 4 m/s²

s = (u + v)/2 × t
= (10 + 30)/2 × 5
= 20 × 5 = 100 m

Marking:

• Acceleration: 1 mark

• Distance: 1 mark

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Q7.

u = 0, a = 3, t = 4

v = u + at = 0 + 3×4 = 12 m/s

s = ut + (1/2)at²
= 0 + (1/2)×3×16 = 24 m

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Q8.

54 km/h = 15 m/s

u = 15, v = 0, t = 6

a = (0 - 15)/6 = -2.5 m/s²

s = (u + v)/2 × t
= (15/2)×6 = 45 m

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Q9.

u = 5, a = 2, t = 10

v = 5 + 2×10 = 25 m/s

s = 5×10 + (1/2)×2×100
= 50 + 100 = 150 m

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Q10.

First 5 s → 100 m
Next 5 s → 200 m

Distances unequal → motion accelerated

Check:
s ∝ t² → acceleration present

✔ Answer: Accelerated motion

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✅ SECTION C – SOLUTIONS (Attempt Any 3)

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Q11.

First phase:
u = 0, a = 2, t = 10

v = 20 m/s

s₁ = (1/2)×2×100 = 100 m

Second phase:
s₂ = v × t = 20 × 5 = 100 m

Total distance = 200 m

Marking:

• First phase: 1 mark

• Second phase: 1 mark

• Total: 1 mark

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Q12.

u = 15, a = -3

Time:
0 = 15 - 3t → t = 5 s

Distance:
s = ut + (1/2)at²
= 15×5 + (1/2)(-3)(25)
= 75 - 37.5 = 37.5 m

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Q13.

Distance in last second:
sₙ = u + (a/2)(2n - 1)

u = 0, a = 10

45 = 5(2n - 1)
9 = 2n - 1
n = 5 s

Height:
s = (1/2)gt² = 5×25 = 125 m

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Q14.

Average speed = 2v₁v₂/(v₁ + v₂)

= (2×40×60)/(40+60)
= 4800/100 = 48 km/h

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Q15.

s = ut + (1/2)at²

20 = (1/2)a(4)
20 = 2a → a = 10 m/s²

Next 2 s (2 to 4 s):
Total at 4s = 80 m
Distance in next 2 s = 80 - 20 = 60 m

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✅ SECTION D – SOLUTIONS (Attempt Any 2)

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Q16.

Phase 1:
u = 0, a = 1, t = 20
v = 20 m/s

s₁ = (1/2)×1×400 = 200 m

Phase 2:
s₂ = 20×30 = 600 m

Phase 3:
a = -2

0 = 20 - 2t → t = 10 s

s₃ = (20+0)/2 × 10 = 100 m

Total distance = 900 m

✔ Max velocity = 20 m/s

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Q17.

u = 30, g = 10

(a) h = u²/2g = 900/20 = 45 m

(b) Time of flight = 2u/g = 60/10 = 6 s

(c) v = u - gt = 30 - 20 = 10 m/s

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Q18.

Area under graph = distance

Triangle 1:
= (1/2)×5×20 = 50

Rectangle:
= 10×20 = 200

Triangle 2:
= 50

Total distance = 300 m

Average velocity = 300/20 = 15 m/s

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📊 PAPER ANALYSIS (Very Important for You)

Difficulty Level:

• Section A: Easy (direct formula)

• Section B: Easy–Moderate

• Section C: Moderate (concept + calculation)

• Section D: Moderate–High (multi-step)

Weightage:

• Numericals dominate (~85%)

• Graph-based question included ✔

• Mixed concept + formula ✔

Chapter Coverage:

• Equations of motion ✔

• Graph ✔

• Free fall ✔

• Average speed ✔

👉 Perfect Board Pattern Paper

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🎯 PRO TIPS FOR BOARD EXAM

1. Always write formula first → gives 0.5 marks

2. Unit conversion (km/h → m/s) is compulsory

3. Use proper steps → don’t jump directly

4. Write final answer with unit

5. For graph → always use area method

6. For motion → identify phase (very important)

7. Practice last second formula questions

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⚠️ COMMON MISTAKE SHEET (Must Give Students)

🔴 Formula Mistakes

• Using wrong equation (especially s = vt instead of s = ut + ½at²)

• Forgetting minus sign in retardation

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🔴 Unit Mistakes

• Not converting km/h to m/s
  👉 72 km/h ≠ 72 m/s ❌

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🔴 Concept Errors

• Confusing uniform velocity with uniform acceleration

• Thinking velocity = speed always

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🔴 Calculation Errors

• Wrong squaring (t² mistakes common)

• Ignoring brackets

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🔴 Graph Mistakes

• Taking slope instead of area

• Forgetting total time in average velocity

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🔴 Multi-step Errors

• Missing one phase in motion problems

• Not summing distances

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🚀 FINAL TEACHER NOTE (For Your Students)

👉 This paper is 100% scoring if:

• Formulas are clear

• Steps are shown

• Units are correct

👉 Weak students lose marks only due to:

• Careless mistakes

• Skipping steps

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