1. a+f/2(b-1) 2. a+f/2(1+b) 3. f-a/2(1+b) 4. a-f/1+b

Answer: 3. f-a/2(1+b) Velocity is the derivative of position; equate the two velocities. v_P=a+2bt ; v_Q=f-2t . a+2bt=f-2t⇒ 2t(b+1)=f-a . t=f-a/2(1+b) Option (3).

Topic: Frame of reference, motion in a straight line, speed and velocity· Kinematics

Two cars P and Q start from a point at the same time in a straight line and their positions are represented by x_P(t)=at+bt^2 and x_Q(t)=ft-t^2. At what time do the cars have the same velocity?

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

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

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

4. $\frac{a-f}{1+b}$

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Answer: 3. $\frac{f-a}{2(1+b)}$

Velocity is the derivative of position; equate the two velocities.

$v_P=a+2bt$; $v_Q=f-2t$.
$a+2bt=f-2t\Rightarrow 2t(b+1)=f-a$.

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

Option (3).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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Two cars P and Q start from a point at the same time in a straight line and their positions are represented by x_P(t)=at+bt^2 and x_Q(t)=ft-t^2. At what time do the cars have the same velocity?

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