Question

Two cars P and Q starts from a point at the same time in a straight line and their positions are represents 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 $\frac{f-a}{2(1+b)}$

$\text{Step 1:}$ $\text{Calculate velocity of car P and Q}$

$V_P=\frac{d\left[x_P\right(t\left)\right]}{dt}$ $=\frac{d}{dt}(at+b{t}^2)$ $=a+2bt$ $....\left(1\right)$

$V_Q=\frac{d\left[x_Q\right(t\left)\right]}{dt}=\frac{d}{dt}(ft-t^2)$$=f-2t$ $....\left(2\right)$

$\text{Step 2:}$ $\text{Time at which both have the same velocity}$

Let $t_0$ be the time when $V_P$ and $V_Q$ are same

Equate equation $\left(1\right)$ and $\left(2\right)$

$V_P{|}_{t=t_0}=V_Q{|}_{t=t_0}$

$\Rightarrow a+2b{t}_0=f-2t_0$

$\Rightarrow$ $t_0=\frac{f-a}{2(b+1)}$

Hence, Option(A) is correct.