MHT CET · Physics · Motion In One Dimension
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 \(\frac{\mathrm{a}+\mathrm{F}}{2(1+\mathrm{b})}\)
- D \(\frac{\mathrm{F}-\mathrm{a}}{2(1+\mathrm{b})}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{F}-\mathrm{a}}{2(1+\mathrm{b})}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text { Velocity } v=\frac{d x}{d t} \\
\therefore \quad & v_P=\frac{d x_P}{d t}=a+2 b t \\
& v_Q=\frac{d x_Q}{d t}=F-2 t \\
& a v_P=v_Q ...(given)\\
& a+2 b t=F-2 t \\
\therefore \quad & (2+2 b) t=F-a \Rightarrow t=\frac{F-a}{2(1+b)}
\end{aligned}\)
& \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}\)
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