Question

Two cars P and Q starts 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. $\frac{f-a}{2(1+b)}$
• B. $\frac{a-f}{1+b}$
• C. $\frac{a+f}{2(b-1)}$
• D.
  $\frac{a+f}{2(1+b)}$

Answer 1

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

Position of car P,
$x_p\left(t\right)=at+b{t}^2$
Thus, velocity of car P
$v_p=\frac{d\left[x_p\right(t\left)\right]}{dt}=a+2bt$

Position of car Q
$x_Q\left(t\right)=ft-t^2$
Thus, velocity of car Q
$v_Q=\frac{d\left[x_Q\right(t\left)\right]}{dt}=f-2t$

According to the question,
$v_p{|}_{t=t_o}=$ $v_Q{|}_{t=t_o}$

$\therefore a+2b{t}_o=f-2t_o$
$⟹t_o=\frac{f-a}{2(1+b)}$