A body accelerates uniformly from rest at \(2ms^{-2}\). Calculate the velocity after traveling 9m
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option C
Explanation:
\(V^2= U^2+2as\) =\(0+ 2\times 2ms^{-2}\times 9 =36ms^{-1}\) thus
V=\(\sqrt{36m^2s^{-2}}\) = \(6ms^{-1}\)
\(V^2= U^2+2as\) =\(0+ 2\times 2ms^{-2}\times 9 =36ms^{-1}\) thus
V=\(\sqrt{36m^2s^{-2}}\) = \(6ms^{-1}\)