Question

A bicycle is moring with uniform velcoity of 20 m/s is behind bus by 96m which is starting from rest with a constant acceleration of$2/m/se{c}^2$Find the time taken by the bicycle to over take the bus

Answer 1

With respect to Bus.

Let after ${}^{\prime }{t}^{\prime }$ sec bicycle will meet bus.

$\begin{array}{c}\Rightarrow 96=20t-\frac{1}{2}\times 2\times t^2\\ \Rightarrow 96=20t-t^2\\ \Rightarrow t^2-20t++96=0\\ \Rightarrow \left(t-12\right)\left(t-8\right)=0\\ t=12\sec t=8\sec \end{array}$

$\therefore$ Hence first the bicycle will catch the bus at $8\sec$ & then bus will catch the bicycle at $12\sec$.