WAEC - Physics (1994 - No. 34)
A body accelerates uniformly from rest at \(2ms^{-2}\). Calculate the velocity after traveling 9m
36\(ms^{-1}\)
18\(ms^{-1}\)
6\(ms^{-1}\)
4.5\(ms^{-1}\)
4.24\(ms^{-1}\)
Explanation
\(V^2= U^2+2as\) =\(0+ 2\times 2ms^{-2}\times 9 =36ms^{-1}\) thus
V=\(\sqrt{36m^2s^{-2}}\) = \(6ms^{-1}\)
V=\(\sqrt{36m^2s^{-2}}\) = \(6ms^{-1}\)
Comments (0)
