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WAEC Physics 2020 Paper

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Question : 29

Total: 51

A motorcycle starting from rest is uniformly accelerated such that its velocity in 10 s is

72 \mathrm{kmhr}^{-1}

. What is its acceleration?

$108 \mathrm{ms}^{2}$
$86 \mathrm{ms}^{2}$
$4 \mathrm{ms}^{2}$
$2 \mathrm{ms}^{2}$

Solution:

Concept:

Use the first equation of motion for uniform acceleration:

a = \frac{v-u}{t}

.

Explanation:

The motorcycle starts from rest, so initial velocity

u = 0

.

Time taken,

t = 10\ \mathrm{s}

.

Final velocity is

v = 72\ \mathrm{kmh}^{-1}

.

Convert to

\mathrm{ms}^{-1}

:

v = \frac{72 \times 1000}{3600} = 20\ \mathrm{ms}^{-1}

.

Now, acceleration

a = \frac{v-u}{t} = \frac{20-0}{10}

.

So,

a = \frac{20}{10} = 2\ \mathrm{ms}^{-2}

.

Answer:

The acceleration is

2\ \mathrm{ms}^{-2}

, which is option D.

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