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\(ms^{-1}\)
설명
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