(A) The velocity of an object is the time derivative of its position function,$V(t) = \frac{dR}{dt}$.For car $A$: $V_{A}(t) = \frac{d}{dt}(at + bt^2)

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(A) The velocity of an object is the time derivative of its position function,$V(t) = \frac{dR}{dt}$.For car $A$: $V_{A}(t) = \frac{d}{dt}(at + bt^2) = a + 2bt$.For car $B$: $V_{B}(t) = \frac{d}{dt}(xt - t^2) = x - 2t$.To find the time $t$ when the cars have the same velocity,we set $V_{A} = V_{B}$:$a + 2bt = x - 2t$Rearranging the terms to solve for $t$:$2bt + 2t = x - a$$t(2b + 2) = x - a$$t = \frac{x - a}{2(b + 1)}$.

Class 11 Physics Motion in Straight Line Instantaneous Velocity and Speed and Velocity-time Graph

Medium

Two cars $A$ and $B$ start from a point at the same time in a straight line and their positions are represented by $R_{A}(t) = at + bt^2$ and $R_{B}(t) = xt - t^2$. At what time do the cars have the same velocity?

MHT CET 2023 All MHT CET

A
$\frac{x-a}{2(b+1)}$
B
$\frac{x+a}{2(b-1)}$
C
$\frac{x-a}{(b+1)}$
D
$\frac{x+a}{(b-1)}$

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Motion in Straight Line

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Instantaneous Velocity and Speed and Velocity-time Graph

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Two cars $A$ and $B$ start from a point at the same time in a straight line and their positions are represented by $R_{A}(t) = at + bt^2$ and $R_{B}(t) = xt - t^2$. At what time do the cars have the same velocity?

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