Question

Two cars P and Q start from a point at the same time in a straight line and their positions are represented by $x_p\left(t\right)=at+b{t}^2$ and $x_Q\left(t\right)=ft-t^2$ At what time do the cars have the same velocity?

• A. $\frac{a+f}{2(1+b)}$
• B. $\frac{f-a}{2(1+b)}$
• C. $\frac{a-f}{1+b}$
• D. $\frac{a+f}{2(b-1)}$

Answer 1

The correct option is B $\frac{f-a}{2(1+b)}$

$\begin{array}{c}{X}_p=at+b{t}^2\\ \frac{dx}{dt}=a+2bt\\ x_Q=ft-t^2\\ \frac{dx}{dt}=f-2t\end{array}$

If velocities are equal

$\begin{array}{c}\frac{d{x}_P}{dt}=\frac{d{x}_Q}{dt}\\ a+2bt=f-2t\\ Onsolving\left(a+2bt=f-2t\right)\\ 2bt+2t=f-0\\ 2t\left(b+1\right)=f-a\\ t=\frac{f-a}{2(b+1)}\end{array}$