A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t seconds, the total distance travelled is: (a) (4αβ)/((α+β))t² (b) (2αβ)/((α+β))t² (c) (αβ)/(2(α+β))t² (d) (αβ)/(4(α+β))t²

Answer: (c). αβvtv Maximum speed: (v)/(α) + (v)/(β) = t ⇒ v = (αβ t)/(α + β). Distance = area of the triangular v - t graph = 1/2vt. s = 1/2 · (αβ t)/(α + β) · t = (αβ)/(2(α + β))t² So the answer is (c) (αβ)/(2(α+β))t².

A car accelerates from rest at a constant rate α for some time after… | JEE Main 2021 (17 Mar, Shift 1)

Q26JEE Main 2021 · 17 MAR · SHIFT 1Medium

Equations of Kinematics

A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is t seconds, the total distance travelled is:

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Answer: (c)

Maximum speed: $\frac{v}{\alpha} + \frac{v}{\beta} = t \Rightarrow v = \frac{\alpha\beta t}{\alpha + \beta}$ .
Distance = area of the triangular $v$ - $t$ graph $= \frac{1}{2}vt$ .
$s = \frac{1}{2} \cdot \frac{\alpha\beta t}{\alpha + \beta} \cdot t = \frac{\alpha\beta}{2(\alpha + \beta)}t^2$
So the answer is (c) $\frac{\alpha\beta}{2(\alpha+\beta)}t^2$ .

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