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 traveled is :

(D) $\frac{\alpha \beta}{2(\alpha+\beta)} \mathrm{t}^{2}$

(A) $\frac{4 \alpha \beta}{(\alpha+\beta)} t^{2}$

(B) $\frac{\alpha \beta}{4(\alpha+\beta)} t^{2}$

(C) $\frac{2 \alpha \beta}{(\alpha+\beta)} t^{2}$

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Kinematics

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 traveled is :

Answer: (D)

Let the car accelerate for a time

t_{1}

and decelerate for

t_{2}

, with

t_{1} + t_{2} = t

. The peak speed occurs at the changeover, so

v_{m a x} = α t_{1} = β t_{2}

Adding

t_{1} = v_{m a x} / α

and

t_{2} = v_{m a x} / β

v_{m a x} (\frac{1}{α} + \frac{1}{β}) = t \Rightarrow v_{m a x} = \frac{α β t}{α + β}

The velocity-time graph is a triangle of base

t

and height

v_{m a x}

, so the distance is its area:

s = \frac{1}{2} v_{m a x} t = \frac{α β}{2 ( α + β )} t^{2}

The total distance travelled is

\frac{α β}{2 ( α + β )} t^{2}

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Single correct option