Describing Motion Around Us — Practice

Question 1

A boy walks 4 m east, then 3 m west. Find the distance and the displacement.

Accepted Answer

Distance = 4 + 3 = 7 m.

Displacement = 4 − 3 = 1 m towards east.

✦ Answer: distance 7 m; displacement 1 m east.

Question 2

Convert 36 km/h into m/s.

Accepted Answer

Step 1 — multiply by 5/18.

Step 2 — 36 × 5/18 = 10 m/s.

✦ Answer: 10 m/s.

Question 3

A car's velocity rises from 5 m/s to 20 m/s in 3 s. Find its acceleration.

Accepted Answer

Step 1 — a = (v − u)/t = (20 − 5)/3.

Step 2 — a = 15/3 = 5 m/s².

✦ Answer: 5 m/s².

Question 4

A body starts from rest and accelerates at 4 m/s² for 6 s. Find the distance covered.

Accepted Answer

Step 1 — s = ut + ½at², u = 0.

Step 2 — s = 0 + ½ × 4 × 6² = ½ × 4 × 36.

Step 3 — s = 72 m.

✦ Answer: 72 m.

Question 5

A stone is dropped from rest. What is its velocity after 2 s? (g = 9.8 m/s²)

Accepted Answer

Step 1 — v = u + at, u = 0, a = g = 9.8.

Step 2 — v = 0 + 9.8 × 2 = 19.6 m/s.

✦ Answer: 19.6 m/s (downward).

Question 6

On a velocity–time graph, a body moves at constant 10 m/s for 5 s. What does the area under the graph represent and what is its value?

Accepted Answer

Step 1 — Area under a v–t graph = distance travelled.

Step 2 — Area = 10 m/s × 5 s = 50 m.

✦ Answer: distance = 50 m.

Question 7

A car moving at 20 m/s brakes and stops in 50 m. Find its (uniform) retardation.

Accepted Answer

Step 1 — v² = u² + 2as, with v = 0, u = 20, s = 50.

Step 2 — 0 = 20² + 2a(50) → 0 = 400 + 100a.

Step 3 — a = −400/100 = −4 m/s².

✦ Answer: retardation = 4 m/s².

Question 8

A stone tied to a string is whirled in a circle of radius 0.5 m, completing one round in 1 s. Find its speed, and state why the motion is accelerated.

Accepted Answer

Step 1 — v = 2πr/T = 2 × 3.14 × 0.5 / 1.

Step 2 — v = 3.14 m/s.

Step 3 — Although the speed is constant, the direction (hence velocity) changes continuously, so the motion is accelerated.

✦ Answer: 3.14 m/s; accelerated because direction/velocity changes.

Question 9

A ball is thrown straight up at 19.6 m/s. Find (a) the time to reach the top and (b) the maximum height. (g = 9.8 m/s²)

Accepted Answer

Step 1 — At the top v = 0; a = −9.8 m/s², u = 19.6.

Step 2 — v = u + at → 0 = 19.6 − 9.8t → t = 2 s.

Step 3 — v² = u² + 2as → 0 = 19.6² − 2(9.8)s → s = 384.16/19.6 = 19.6 m.

✦ Answer: (a) 2 s, (b) 19.6 m.

Practice — Describing Motion Around Us

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