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WAEC Physics 2017 Paper

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Question : 3

Total: 50

A body accelerates uniformly from rest at 2

\mathrm{ms}^{-2}

. Calculate its velocity when it has travelled a distance of 9m.

3.0 $\mathrm{ms}^{-1}$
4.5 $\mathrm{ms}^{-1}$
6.0 $\mathrm{ms}^{-1}$
18.0 $\mathrm{ms}^{-1}$

Solution:

Given that $u = 0\,\mathrm{ms}^{-1}, a = 2\,\mathrm{ms}^{-2}, s = 9\,\mathrm{m}$

We can use the equation, $v^{2}=u^{2}+2as$ so that we have,

$v^{2}=0^{2}+2\times 2\times 9$

$v^{2}=36; v=6.0\,\mathrm{ms}^{-1}$

Note; u = Initial velocity

v = final velocity

a = acceleration

s = distance covered

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