Motion – Numerical Problems with Stepwise Solutions
CBSE Class 9 — Physics
Chapter 8: Motion — 20 Numerical Problems with Stepwise Solutions (NCERT-aligned)
Content Bank — Key formulas (useful for problems)
- v = u + at — final velocity after time t (m/s).
- s = ut + ½at² — displacement in time t (m).
- v² = u² + 2as — relation without time (m²/s²).
- Average speed = total distance / total time.
- Area under v–t graph = displacement; slope of s–t = velocity; slope of v–t = acceleration.
- Free fall: take a = g ≈ 9.8 m/s² (or g = 10 m/s² for approximate answers).
- Unit conversions: 1 km/h = 5/18 m/s; 1 m/s = 3.6 km/h.
Note: These formulas are the basis for all NCERT Class 9 Motion numericals. Show units and sign conventions while solving.
Topic 1 — Basic Kinematics (Q1–Q5)
Q1.
A car starts from rest and accelerates uniformly at 3 m/s² for 6 s. Find (a) final velocity, (b) distance covered. Show steps and units.
Step 1 — Given: u = 0 m/s, a = 3 m/s², t = 6 s.
Step 2 — Final velocity (v = u + at):
v = 0 + (3)(6) = 18 m/s. [m/s]
v = 0 + (3)(6) = 18 m/s. [m/s]
Step 3 — Distance (s = ut + ½at²):
s = 0 + 0.5×3×6² = 1.5×36 = 54 m. [m]
s = 0 + 0.5×3×6² = 1.5×36 = 54 m. [m]
Answer: v = 18 m/s; s = 54 m.
Q2.
A scooter moves at constant speed 18 m/s for 20 s. Calculate distance travelled and average velocity (assume straight-line motion).
Step 1 — Given: v = 18 m/s (constant), t = 20 s.
Step 2 — Distance: s = v × t = 18 × 20 = 360 m. [m]
Step 3 — Average velocity: For constant velocity, average velocity = instantaneous velocity = 18 m/s. [m/s]
Answer: Distance = 360 m; Average velocity = 18 m/s.
Q3.
A runner covers 100 m in 12 s and then 200 m in 20 s. Find (a) total distance, (b) total time, (c) average speed.
Step 1 — Given: d1 = 100 m (t1 = 12 s), d2 = 200 m (t2 = 20 s).
Step 2 — Total distance: D = 100 + 200 = 300 m.
Step 3 — Total time: T = 12 + 20 = 32 s.
Step 4 — Average speed: v_avg = D / T = 300 / 32 = 9.375 m/s ≈ 9.38 m/s. [m/s]
Answer: Distance = 300 m; Time = 32 s; Average speed ≈ 9.38 m/s.
Q4.
A car moves 60 km east, then 40 km west. Find total distance and displacement. Express displacement with direction.
Step 1 — Total distance: 60 + 40 = 100 km.
Step 2 — Displacement: Net = 60 east − 40 west = 20 km east. (Because east considered positive.)
Answer: Distance = 100 km; Displacement = 20 km east.
Q5.
If a body moves with constant velocity 5 m/s for 2 minutes, what is the distance covered? Convert your answer to kilometers.
Step 1 — Convert time: 2 minutes = 120 s.
Step 2 — Distance: s = v × t = 5 × 120 = 600 m.
Step 3 — Convert to km: 600 m = 0.6 km.
Answer: 600 m = 0.6 km.
Topic 2 — Uniform Acceleration & Equations (Q6–Q11)
Q6.
A vehicle accelerates uniformly from 15 m/s to 25 m/s in 5 s. Find acceleration and distance covered in this time.
Step 1 — Given: u = 15 m/s, v = 25 m/s, t = 5 s.
Step 2 — Acceleration (a = (v − u)/t):
a = (25 − 15) / 5 = 10 / 5 = 2 m/s².
a = (25 − 15) / 5 = 10 / 5 = 2 m/s².
Step 3 — Distance (s = ut + ½at²):
s = 15×5 + 0.5×2×25 = 75 + 25 = 100 m.
s = 15×5 + 0.5×2×25 = 75 + 25 = 100 m.
Answer: a = 2 m/s²; s = 100 m.
Q7.
From rest, a particle accelerates at 4 m/s². How long will it take to reach 20 m/s and what distance will it cover in that time?
Step 1 — Given: u = 0, a = 4 m/s², v = 20 m/s.
Step 2 — Time (v = u + at): t = (v − u)/a = 20 / 4 = 5 s.
Step 3 — Distance (s = ut + ½at²): s = 0 + 0.5×4×5² = 2×25 = 50 m.
Answer: t = 5 s; s = 50 m.
Q8.
A car moving at 20 m/s applies brakes and stops in 8 s with uniform deceleration. Find deceleration and stopping distance.
Step 1 — Given: u = 20 m/s, v = 0, t = 8 s.
Step 2 — Deceleration (a = (v − u)/t): a = (0 − 20)/8 = −2.5 m/s² (negative indicates deceleration).
Step 3 — Stopping distance (s = ut + ½at²): s = 20×8 + 0.5×(−2.5)×64 = 160 − 80 = 80 m.
Answer: a = −2.5 m/s²; stopping distance = 80 m.
Q9.
A train increases speed uniformly from 36 km/h to 72 km/h in 20 s. Compute acceleration in m/s². (Show unit conversions.)
Step 1 — Convert speeds to m/s: 36 km/h = 36÷3.6 = 10 m/s; 72 km/h = 72÷3.6 = 20 m/s.
Step 2 — Given: u = 10 m/s, v = 20 m/s, t = 20 s.
Step 3 — Acceleration: a = (v − u)/t = (20 − 10)/20 = 10/20 = 0.5 m/s².
Answer: a = 0.5 m/s².
Q10.
A sprinter accelerates uniformly from rest to 8 m/s in 4 s. Using v² = u² + 2as, find the distance covered during acceleration (verify using s = ut + ½at²).
Method A — Using v² = u² + 2as:
u = 0, v = 8 → 8² = 0 + 2a s → 64 = 2a s → s = 32 / a (need a).
Find a from v = u + at: a = (v − u)/t = 8/4 = 2 m/s².
Now s = 32 / 2 = 16 m.
Method B — s = ut + ½at²: s = 0 + 0.5×2×4² = 1×16 = 16 m (matches).
Answer: Distance = 16 m.
Topic 3 — Graphical Interpretation & Average Values (Q11–Q14)
Q11.
From t = 0 to 4 s an object has velocity increasing linearly from 0 to 12 m/s. Find displacement during this interval using area under v–t graph.
Step 1 — v–t graph is a triangle with base 4 s and height 12 m/s.
Step 2 — Area of triangle (displacement): Area = ½ × base × height = 0.5 × 4 × 12 = 24 m.
Answer: Displacement = 24 m.
Q12.
An s–t graph is a straight line with slope 3 m/s. What is the displacement between t = 2 s and t = 7 s?
Step 1 — Slope = velocity = 3 m/s constant.
Step 2 — Δt = 7 − 2 = 5 s. Displacement = v × Δt = 3 × 5 = 15 m.
Answer: Displacement = 15 m.
Q13.
A v–t graph shows constant 5 m/s from 0–3 s, then 0 m/s from 3–5 s. Find total displacement and average speed for 0–5 s.
Step 1 — Displacement 0–3 s: area = 5×3 = 15 m. 3–5 s: area = 0×2 = 0 m.
Step 2 — Total displacement = 15 m. Total time = 5 s.
Step 3 — Average speed = total distance / total time = 15 / 5 = 3 m/s.
Answer: Displacement = 15 m; Average speed = 3 m/s.
Q14.
An object has velocity v(t) = 2t (m/s). Find displacement from t = 1 to t = 4 s using integration idea (area under v–t curve).
Step 1 — v(t) = 2t. Displacement = ∫(1→4) 2t dt = [t²]₁⁴ = 4² − 1² = 16 − 1 = 15 m.
Answer: Displacement = 15 m.
Topic 4 — Free Fall & Vertical Motion (Q15–Q17)
Q15.
A stone is dropped from rest from a height of 80 m. Find (a) time to hit the ground, (b) velocity on impact. Use g = 9.8 m/s².
Step 1 — Given: u = 0, s = 80 m, a = g = 9.8 m/s² (downwards).
Step 2 — Time (s = ½ g t²): t² = 2s/g = 160 / 9.8 ≈ 16.3265 → t ≈ 4.04 s.
Step 3 — Velocity (v = u + gt): v = 0 + 9.8×4.04 ≈ 39.6 m/s downward.
Answer: t ≈ 4.04 s; v ≈ 39.6 m/s downward.
Q16.
A ball is thrown vertically upward with speed 20 m/s. Find maximum height (use g = 10 m/s² for simplicity) and time to reach maximum height.
Step 1 — Given: u = 20 m/s, a = −g = −10 m/s².
Step 2 — Time to max (v = u + at, v=0): t = u/g = 20/10 = 2 s.
Step 3 — Max height (H = u²/(2g)): H = 20² / (2×10) = 400 / 20 = 20 m.
Answer: H = 20 m; time = 2 s.
Q17.
A ball is thrown horizontally from the top of a cliff with speed 10 m/s and lands 2 s later. Find (a) horizontal range, (b) height of cliff (take g = 9.8 m/s²).
Step 1 — Horizontal motion independent: x = ux × t = 10 × 2 = 20 m.
Step 2 — Vertical motion (dropped): s = ½ g t² = 0.5×9.8×4 = 19.6 m.
Answer: Range = 20 m; Height ≈ 19.6 m.
Topic 5 — Relative Motion & Trains (Q18–Q19)
Q18.
Two trains 100 m and 150 m long approach each other with speeds 36 km/h and 54 km/h. How long to completely pass each other?
Step 1 — Total length to clear = 100 + 150 = 250 m.
Step 2 — Relative speed = 36 + 54 = 90 km/h. Convert to m/s: 90×5/18 = 25 m/s.
Step 3 — Time = distance / relative speed = 250 / 25 = 10 s.
Answer: 10 seconds.
Q19.
A boat's speed in still water is 12 km/h. River current is 3 km/h. Find time to go 15 km upstream and return (downstream) to starting point.
Step 1 — Upstream speed = 12 − 3 = 9 km/h. Downstream speed = 12 + 3 = 15 km/h.
Step 2 — Time upstream = 15 / 9 ≈ 1.6667 h = 1 h 40 min.
Step 3 — Time downstream = 15 / 15 = 1 h.
Step 4 — Total time ≈ 1.6667 + 1 = 2.6667 h ≈ 2 h 40 min.
Answer: Total ≈ 2 h 40 min (or 2.6667 h).
Topic 6 — Mixed / Challenge Numericals (Q20)
Q20.
A car travels 100 m with uniform acceleration from 10 m/s to 20 m/s. (a) Find acceleration using v² = u² + 2as. (b) Find time taken using v = u + at.
Step 1 — Given: u = 10 m/s; v = 20 m/s; s = 100 m.
Step 2 — Use v² = u² + 2as: 20² = 10² + 2a(100) → 400 = 100 + 200a → 200a = 300 → a = 1.5 m/s².
Step 3 — Time (v = u + at): t = (v − u)/a = (20 − 10)/1.5 = 10 / 1.5 ≈ 6.6667 s ≈ 6.67 s.
Answer: a = 1.5 m/s²; t ≈ 6.67 s.