Q148 / 150 · MHT CET 2024 (11 May Shift 2) · Physics

Two cars start from a point at the same time in a straight line and their positions are represented by $\mathrm{x}_1(\mathrm{t})=\mathrm{at}+\mathrm{bt}^2$ and $\mathrm{x}_2(\mathrm{t})=\mathrm{Ft}-\mathrm{t}^2$. At what time do the cars have the same velocity?

  • A
    $\frac{a+F}{2(b-1)}$
  • B
    $\frac{\mathrm{a}-\mathrm{F}}{1+\mathrm{b}}$
  • C
    $\frac{\mathrm{a}+\mathrm{F}}{2(1+\mathrm{b})}$
  • Correct answer
    D
    $\frac{\mathrm{F}-\mathrm{a}}{2(1+\mathrm{b})}$

Solution

$\begin{aligned} & \text { Velocity } v=\frac{d x}{d t} \\ \therefore \quad & v_P=\frac{d x_P}{d t}=a+2 b t \\ & v_Q=\frac{d x_Q}{d t}=F-2 t \\ & a v_P=v_Q ...(given)\\ & a+2 b t=F-2 t \\ \therefore \quad & (2+2 b) t=F-a \Rightarrow t=\frac{F-a}{2(1+b)} \end{aligned}$

Hence, the correct answer is option: 'D'.

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