Motion — Practice

Question 1

A person walks 4 m east, then 3 m north. Find the distance and magnitude of displacement.

Accepted Answer

Step 1 — Distance = 4 + 3 = 7 m.

Step 2 — Displacement: a right-triangle from start to end. Magnitude = √(4² + 3²) = √(16 + 9) = √25 = 5 m.

✦ Answer: Distance = 7 m, displacement (magnitude) = 5 m (toward NE, at 37° N of E).

Question 2

Convert 72 km/h to m/s.

Accepted Answer

Step 1 — Use 1 km/h = 5/18 m/s.

Step 2 — 72 × 5/18 = 360/18 = 20 m/s.

✦ Answer: 20 m/s.

Question 3

A car starts from rest and acquires a velocity of 20 m/s in 4 s. Find its acceleration.

Accepted Answer

Step 1 — Identify givens.
  u = 0 (starts from rest), v = 20 m/s, t = 4 s.

Step 2 — Use a = (v − u)/t.
  a = (20 − 0)/4 = 5 m/s².

✦ Answer: 5 m/s².

Question 4

A car moving at 10 m/s accelerates at 2 m/s² for 5 s. Find the distance covered.

Accepted Answer

Step 1 — Givens: u = 10 m/s, a = 2 m/s², t = 5 s. Unknown s, no v needed → use eqn 2.

Step 2 — s = ut + ½at² = 10(5) + ½(2)(5)² = 50 + 25 = 75 m.

✦ Answer: 75 m.

Question 5

State the SI units of (a) displacement, (b) velocity, (c) acceleration.

Accepted Answer

Step 1 — Recall the basic SI units.
  (a) Displacement: metre (m). Same unit as distance.
  (b) Velocity: metres per second (m/s). 
  (c) Acceleration: metres per second per second (m/s²).

✦ Answer: (a) m, (b) m/s, (c) m/s².

Question 6

A bullet leaves a 50-cm-long rifle barrel with a velocity of 400 m/s. Find its acceleration assuming uniform acceleration in the barrel.

Accepted Answer

Step 1 — Identify givens.
  u = 0 (starts from rest in the barrel), v = 400 m/s, s = 50 cm = 0.5 m.
  No t given → use 3rd equation.

Step 2 — v² = u² + 2as → 400² = 0 + 2 × a × 0.5.

Step 3 — Solve.
  160000 = a → a = 160000 m/s² = 1.6 × 10⁵ m/s².

✦ Answer: a ≈ 1.6 × 10⁵ m/s².

For comparison, gravity is ~10 m/s². The bullet's acceleration is ~ 16,000 times stronger — that's why rifles produce muzzle velocities of hundreds of m/s in fractions of a second.

Question 7

A train travels half its journey at 60 km/h and the other half at 40 km/h. Find the average speed.

Accepted Answer

Step 1 — Let total distance be 2d.
  First half (d) at 60 km/h: time t₁ = d/60.
  Second half (d) at 40 km/h: time t₂ = d/40.

Step 2 — Total time.
  t = t₁ + t₂ = d/60 + d/40 = (4d + 6d)/240 = 10d/240 = d/24.

Step 3 — Average speed.
  v_avg = total distance / total time = 2d / (d/24) = 2d × 24/d = 48 km/h.

✦ Answer: 48 km/h.

Note: NOT (60+40)/2 = 50. Average speed uses harmonic mean when distances are equal:
   v_avg = 2v₁v₂ / (v₁ + v₂) = 2 × 60 × 40 / (60 + 40) = 4800/100 = 48 km/h. ✓

Question 8

A stone is dropped from a height of 45 m. Find (a) the time to reach the ground, (b) the velocity just before hitting the ground. (g = 10 m/s²)

Accepted Answer

Step 1 — Givens.
  u = 0 (dropped, not thrown), a = g = 10 m/s² (downward), s = 45 m (downward; take downward positive).

Step 2 (a) — Find time.
  Use s = ut + ½at² ⇒ 45 = 0 + ½ × 10 × t² ⇒ t² = 9 ⇒ t = 3 s.

Step 3 (b) — Final velocity.
  Use v = u + at = 0 + 10 × 3 = 30 m/s.
  OR v² = u² + 2as = 0 + 2 × 10 × 45 = 900 ⇒ v = 30 m/s. ✓

✦ Answer: (a) 3 s; (b) 30 m/s.

Question 9

A car moving at 36 km/h is brought to rest by applying brakes over a distance of 10 m. Find the deceleration produced.

Accepted Answer

Step 1 — Convert speed.
  36 km/h = 36 × 5/18 = 10 m/s.

Step 2 — Givens.
  u = 10 m/s, v = 0, s = 10 m. Use v² = u² + 2as.

Step 3 — Substitute.
  0² = 10² + 2 × a × 10
  0 = 100 + 20a
  a = −5 m/s².

✦ Answer: Deceleration = 5 m/s² (negative sign indicates motion is decelerating).

Real-world: braking decelerations on dry road are typically 3–7 m/s². Wet road or worn tyres reduce this.

Question 10

A ball is thrown vertically upward with initial velocity 20 m/s. (a) How high does it rise? (b) How long before it hits the ground? (g = 10 m/s²)

Accepted Answer

Step 1 — Pick up as positive. Givens: u = +20, a = −10 m/s² (gravity opposes upward motion).

Step 2 (a) — At max height, v = 0. Use v² = u² + 2as.
  0 = 400 + 2(−10)s
  s = 400/20 = 20 m.

Step 3 (b) — Total time = time to reach max height + time to fall back (symmetric, no air resistance).
  Time up: v = u + at ⇒ 0 = 20 − 10t ⇒ t_up = 2 s.
  By symmetry, time down = 2 s.
  Total = 4 s.

✦ Answer: (a) Maximum height = 20 m. (b) Total time in the air = 4 s.

Question 11

On a velocity-time graph, a body's velocity increases linearly from 0 to 20 m/s in 10 s, stays constant for 5 s, then decreases linearly to 0 in 5 s. Find: (a) acceleration in each phase, (b) total distance covered.

Accepted Answer

Step 1 — Phase 1 (0 to 10 s, velocity 0 → 20 m/s):
  a₁ = (20 − 0)/10 = 2 m/s².

Step 2 — Phase 2 (10 to 15 s, velocity constant at 20 m/s):
  a₂ = 0.

Step 3 — Phase 3 (15 to 20 s, velocity 20 → 0):
  a₃ = (0 − 20)/5 = −4 m/s² (deceleration of 4 m/s²).

Step 4 — Total distance = total area under v-t graph.
  Phase 1: triangle of base 10, height 20 → area = ½ × 10 × 20 = 100 m.
  Phase 2: rectangle of width 5, height 20 → 5 × 20 = 100 m.
  Phase 3: triangle of base 5, height 20 → ½ × 5 × 20 = 50 m.
  Total = 100 + 100 + 50 = 250 m.

✦ Answer: (a) Phase 1: 2 m/s²; Phase 2: 0; Phase 3: −4 m/s² (deceleration 4 m/s²). (b) Total distance = 250 m.

Question 12

Two cars start from the same point at the same time. Car A moves with constant velocity 20 m/s. Car B starts from rest with acceleration 2 m/s². At what time will they meet again?

Accepted Answer

Step 1 — Set up equations for position.
  Car A: x_A = 20t (uniform motion).
  Car B: x_B = 0 × t + ½ × 2 × t² = t² (uniform acceleration from rest).

Step 2 — They meet again when x_A = x_B.
  20t = t² ⇒ t² − 20t = 0 ⇒ t(t − 20) = 0.
  Solutions: t = 0 (initial start) or t = 20 s.

Step 3 — Meaningful answer.
  They meet again at t = 20 s.
  Distance from start: x = 20 × 20 = 400 m (or t² = 400 m ✓).

✦ Answer: They meet again at t = 20 s, at a distance of 400 m from the start. Car B was slower initially but kept accelerating, so it eventually caught up.

Question 13

A stone is dropped from a tower. It covers 25 m in the LAST second of its fall. Find the height of the tower. (g = 10 m/s²)

Accepted Answer

Step 1 — Let total time of fall be n seconds.
  Distance fallen in n seconds: S_n = ½ × g × n² = 5n².
  Distance fallen in (n − 1) seconds: S_{n−1} = 5(n − 1)².

Step 2 — Distance covered in the last second = S_n − S_{n−1}.
  = 5n² − 5(n − 1)²
  = 5n² − 5(n² − 2n + 1)
  = 5(2n − 1)
  = 10n − 5.

Step 3 — Set equal to 25 m.
  10n − 5 = 25 ⇒ 10n = 30 ⇒ n = 3 s.

Step 4 — Total height = S_n = 5n² = 5 × 9 = 45 m.

✦ Answer: Tower height = 45 m. Fall time = 3 s.

Sanity check: Distance in 2 s = 5 × 4 = 20 m. Distance in 3 s = 45 m. Difference = 25 m ✓.

Question 14

The Earth orbits the Sun in a roughly circular path of radius 1.5 × 10⁸ km. It takes 365.25 days. Find the orbital speed of the Earth in km/s.

Accepted Answer

Step 1 — Use v = 2πr/T.
  r = 1.5 × 10⁸ km.
  T = 365.25 days = 365.25 × 24 × 60 × 60 s = 3.156 × 10⁷ s.

Step 2 — Plug in.
  v = 2π × 1.5 × 10⁸ / 3.156 × 10⁷
   = 9.425 × 10⁸ / 3.156 × 10⁷
   ≈ 29.86 km/s.

✦ Answer: Earth's orbital speed ≈ 30 km/s.

That's about 108,000 km/h. We're hurtling through space at this speed without feeling anything, because we're all moving together.

Question 15

A bus accelerates uniformly from rest and covers 36 m in the 4th second. Find (a) the acceleration, (b) the velocity at the end of the 4th second.

Accepted Answer

Step 1 — Use the 'distance covered in nth second' formula derived from equations of motion.
  S_n = u + a(n − 1/2) = u + ½ × a × (2n − 1).
  For u = 0: S_n = ½ × a × (2n − 1).

Step 2 — Substitute.
  For n = 4: S_4 = ½ × a × (2 × 4 − 1) = ½ × a × 7 = 3.5a.

Step 3 — Set S_4 = 36 m.
  3.5a = 36 ⇒ a = 36/3.5 ≈ 10.29 m/s².
  (Or exactly: a = 72/7 m/s².)

Step 4 — Velocity at end of 4 s.
  v = u + at = 0 + 10.29 × 4 ≈ 41.14 m/s.
  (Or exactly: 288/7 m/s.)

✦ Answer: (a) a ≈ 10.29 m/s²; (b) v ≈ 41.14 m/s at the end of the 4th second.

The 'nth second' formula S_n = u + ½a(2n−1) is a clever shortcut — derived from S_n = S(n) − S(n−1) where S(t) = ut + ½at². Worth memorising.

Practice — Motion

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