Two cars P and Q start simultaneously from the same point. Their positions are x_P(t) = at + bt^2 and x_Q(t) = ft - t^2. At what time do they have equal velocities?

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NEET Physics QUESTION #7322

Question 1

Two cars P and Q start simultaneously from the same point. Their positions are $x_P(t) = at + bt^2$ and $x_Q(t) = ft - t^2$. At what time do they have equal velocities?

$\dfrac{a+f}{2(1+b)}$
$\dfrac{f-a}{2(1+b)}$✔️
$\dfrac{a-f}{1+b}$
$\dfrac{a+f}{2(b-1)}$

Correct Answer Explanation

$v_P = a + 2bt$, $\quad v_Q = f - 2t$

Setting equal: $a + 2bt = f - 2t \Rightarrow 2t(b+1) = f-a \Rightarrow t = \mathbf{\dfrac{f-a}{2(1+b)}}$

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