Two cars A and B start from a point at the same time in a straight line and their positions are represented…
Two cars A and B start from a point at the same time in a straight line and their positions are represented by $\mathrm{R}_{\mathrm{A}}(\mathrm{t})=\mathrm{at}+\mathrm{bt}^2$ and $\mathrm{R}_{\mathrm{B}}(\mathrm{t})=x \mathrm{t}-\mathrm{t}^2$. At what time do the cars have same velocity?
$\frac{x-a}{2(b+1)}$
$\frac{x+a}{2(b-1)}$
$\frac{x-a}{(b+1)}$
$\frac{x+a}{(b-1)}$
Solution
$\therefore \quad$ Velocity of car A and B:
$\begin{aligned}
\mathrm{V}_{\mathrm{A}} & =\frac{\mathrm{d}\left(\mathrm{R}_{\mathrm{A}}\right)}{\mathrm{dt}} \\
& =\mathrm{a}+2 \mathrm{bt} \\
\mathrm{V}_{\mathrm{B}} & =\frac{\mathrm{d}\left(\mathrm{R}_{\mathrm{B}}\right)}{\mathrm{dt}} \\
& =\mathrm{x}-2 \mathrm{t}
\end{aligned}$
$\therefore \quad$ So, time at which cars have same velocity is
$\begin{array}{ll}
& V_A=V_B \\
& a+2 b t=x-2 t \\
\therefore \quad & t=\frac{x-a}{2(b+1)}
\end{array}$