Question

Two buses $P$ and $Q$ start from a point at the same time and move in a straight line and their positions are represented by $X_P\left(t\right)=\alpha t+\beta t^2$ and $X_Q\left(t\right)=ft-t^2$. At what time, both the buses have same velocity?

Answer 1

$X_P=\alpha t+\beta t^2$

$\therefore V_P=\frac{d{X}_P}{dt}=\alpha +2\beta t$

$X_Q\left(t\right)=ft-t^2$

$V_Q=\frac{d{X}_Q}{dt}=f-2t$

Let $t$ be the time when velocity becomes equal.

$\because V_P=V_Q$

$\Rightarrow \alpha +2\beta t=f-2t$

$\Rightarrow t=\frac{f-\alpha }{2(1+\beta )}$