Question

A motorbike initially at rest, picks up a velocity of $72km/h^{-1}$ over a distance of $40m$. Calculate (1)acceleration (2)time in which it picks up above velocity

Answer 1

$\begin{array}{c}u=0\\ v=72km/m=72\times \frac{5}{18}=20m/s\\ v^2=u^2+2as\\ 20\times 20=0+2\times 40\times a\\ a=5m/s^2\\ v=u+at\\ 20-0=at\\ t=4\sec \end{array}$