A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β and comes to rest. If total time elapsed is t, then maximum velocity acquired by car will be (a) ((α² - β²)t)/(αβ) (b) ((α² + β²)t)/(αβ) (c) ((α + β)t)/(αβ) (d) (αβ t)/(α + β)

Answer: (d). Let the maximum speed be v. Time to speed up t₁ = (v)/(α), time to stop t₂ = (v)/(β). t = (v)/(α) + (v)/(β) = v(α + β)/(αβ) v = (αβ t)/(α + β) So the answer is (d) (αβ t)/(α + β).

A car accelerates from rest at a constant rate α for some time after… | NEET 1994 (AIPMT)

Q44NEET 1994 · AIPMTMedium

Equations of Kinematics

A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ and comes to rest. If total time elapsed is $t$ , then maximum velocity acquired by car will be

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Answer: (d)

Let the maximum speed be $v$ . Time to speed up $t_1 = \frac{v}{\alpha}$ , time to stop $t_2 = \frac{v}{\beta}$ .
$t = \frac{v}{\alpha} + \frac{v}{\beta} = v\frac{\alpha + \beta}{\alpha\beta}$
$v = \frac{\alpha\beta t}{\alpha + \beta}$
So the answer is (d) $\frac{\alpha\beta t}{\alpha + \beta}$ .

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