Cars with Same Velocity – Rankers Physics
Topic: Kinematics
Subtopic: Relative Motion in One Dimension

Cars with Same Velocity

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?

(2016 - II)

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

Solution:

Concept: Velocity is the time derivative of position. Calculate \( V_P(t) = \frac{dX_P}{dt} = a + 2bt \) and \( V_Q(t) = \frac{dX_Q}{dt} = f - 2t \). Equate \( V_P(t) = V_Q(t) \) to find time \( t \). \( a + 2bt = f - 2t \) ⇒ \( 2t(b+1) = f-a \), so \( t = \frac{f-a}{2(b+1)} \).

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