Question

A parked motorcycle started a trip towards east with uniform acceleration of 0.62m/s2 travelled for 90 s. What is the final velocity at t = 90 s?

Answer 1

Given data

*The given acceleration is a = 0.62 m/s^2

*The given time is t = 90 s

*The initial velocity of the parked motorcycle is u = 0 m/s

The formula for the final velocity of the motorcycle is given by the equation of motion as

$v=u+at$

Substitute the known values in the above expression as

$\begin{gathered} v=(0)+(0.62)(90) \\ =55.8\text{ m/s} \end{gathered}$

Hence, the final velocity of the parked motorcycle is v = 55.8 m/s