WAEC - Physics (2017 - No. 3)

A body accelerates uniformly from rest at 2\(ms^{-2}\). Calculate its velocity when it has travelled a distance of 9m.

3.0\(ms^{-1}\)

4.5\(ms^{-1}\)

6.0\(ms^{-1}\)

18.0\(ms^{-1}\)

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

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

\(v^{2} = 0^{2} + 2\times 2\times 9\)

\(v^{2} = 36; v= 6.0ms^{-1}\)

Note; u = Initial velocity

v = final velocity

a = acceleration

s = distance covered