PYQ NEET- Motion In A Straight Line Kinematics L-5

Question: Two cars $\mathrm{P}$ and $\mathrm{Q}$ start from a point at the same time in a straight line and their positions are represented by#

$$ x_P(t)=\left(a t+b t^2\right) \text { and } x_Q(t)=\left(f t-t^2\right) \text {. } $$

At what time do the cars have the same velocity?

A) $\frac{a-f}{1+b}$

B) $\frac{a+f}{2(b-1)}$

C) $\frac{a+f}{2(1+b)}$

D) $\frac{f-a}{2(1+b)}$

Answer: $\frac{f-a}{2(1+b)}$#

Sol:#

For car $\mathrm{P}$, $$ \begin{aligned} & \mathrm{x}{\mathrm{P}}(\mathrm{t})=\left(a t+b t^2\right) \ & \mathrm{v}{\mathrm{P}}(\mathrm{t})=\frac{d x_p(t)}{d t}=a+2 b t \end{aligned} $$

Similarly for car Q, $$ \begin{aligned} & \mathrm{x}{\mathrm{Q}}(\mathrm{t})=\left(f t-t^2\right) \ & \mathrm{v}{\mathrm{Q}}(\mathrm{t})=\frac{d x_Q(t)}{d t}=f-2 t \end{aligned} $$

When they have same velocity then, $v_P(t)=v_Q(t)$ $$ \begin{aligned} & \therefore a+2 b t=f-2 t \ & \Rightarrow 2 t(b+1)=f-a \ & \Rightarrow \mathrm{t}=\frac{f-a}{2(1+b)} \end{aligned} $$