Two cars start from a point at the same time in a straight line and their positions are represented by x₁(t) = at + bt² and x₂(t) = Ft - t². At what time do the cars have the same velocity? (a) (a + F)/(2(b - 1)) (b) (a - F)/(1 + b) (c) (a + F)/(2(1 + b)) (d) (F - a)/(2(1 + b))

Answer: (d). v₁ = (dx₁)/(dt) = a + 2bt, v₂ = (dx₂)/(dt) = F - 2t. a + 2bt = F - 2t ⇒ 2t(1 + b) = F - a So the answer is (d) t = (F - a)/(2(1 + b)).

Two cars start from a point at the same time in a straight line and… | MHT CET 2024 (11 May, Shift 2)

Q6MHT CET 2024 · 11 MAY · SHIFT 2Easy

Velocity, Acceleration & Variable Motion

Two cars start from a point at the same time in a straight line and their positions are represented by $x_1(t) = at + bt^2$ and $x_2(t) = Ft - t^2$ . At what time do the cars have the same velocity?

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Answer: (d)

$v_1 = \frac{dx_1}{dt} = a + 2bt$ , $v_2 = \frac{dx_2}{dt} = F - 2t$ .
$a + 2bt = F - 2t \Rightarrow 2t(1 + b) = F - a$
So the answer is (d) $t = \frac{F - a}{2(1 + b)}$ .

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Velocity, Acceleration & Variable Motion question from MHT CET 2024 (11 May, Shift 2)

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