Two cars A and B start from a point at the same time in a straight line and their positions are represented by RA(t) = at +bt² and RB(t) = xt - t². At what time do the cars have same velocity? (a) (x - a)/(2(b + 1)) (b) (x + a)/(2(b - 1)) (c) (x - a)/((b + 1)) (d) (x + a)/((b - 1))

Answer: (a). vA = a + 2bt, vB = x - 2t. a + 2bt = x - 2t ⇒ 2t(b + 1) = x - a So the answer is (a) t = (x - a)/(2(b + 1)).

Two cars A and B start from a point at the same time in a straight… | MHT CET 2023 (11 May, Shift 2)

Q16MHT CET 2023 · 11 MAY · SHIFT 2Easy

Velocity, Acceleration & Variable Motion

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 same velocity?

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Answer: (a)

$v_A = a + 2bt$ , $v_B = x - 2t$ .
$a + 2bt = x - 2t \Rightarrow 2t(b + 1) = x - a$
So the answer is (a) $t = \frac{x - a}{2(b + 1)}$ .

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