question_answer

A particle is moving in a straight line with constant velocity 3 m/s. At t = 0, a force starts acting on the particle in a direction perpendicular to the direction of its initial motion which causes an acceleration of 1 m/s2. Determine the magnitude of particle's velocity at t = 3 s.

A)  3 m/s                          

B)  6 m/s

C)  \[3\sqrt{2}\,\text{m/s}\]                     

D)  \[2\sqrt{3}\,\text{m/s}\]

Correct Answer: C

Solution :

Along initial direction, \[{{v}_{x}}=3\text{m/s}\text{.}\] Along perpendicular direction \[{{v}_{y}}=0+1\times 3=3\,\,\text{m/s}\]             \[\therefore \]    \[|v|=\sqrt{v_{x}^{2}+v_{y}^{2}}=3\sqrt{2}\,\text{m/s}\]