Question

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 :

• A. $\frac{\alpha \beta }{2\left(\alpha +\beta \right)}{t}^2$
• B. $\frac{\alpha \beta }{4\left(\alpha +\beta \right)}{t}^2$
• C. $\frac{4\alpha \beta }{\left(\alpha +\beta \right)}{t}^2$
• D. $\frac{2\alpha \beta }{\left(\alpha +\beta \right)}{t}^2$

Answer 1

The correct option is A

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

Step 1. Given data

Let $t$ be the total time.

Let $t_2$ be the time for the second type of motion.

Let $t_1$ is the time for which the particle accelerates.

Velocity is the product of the acceleration and the time.

Let $v_1$ be the velocity of the first type of motion.

$v_1=\alpha t_1$ [$\alpha$ is acceleration]

Let $v_2$ be the velocity of the second type of motion

$v_2=\beta \left(t-t_1\right)$ [ $\beta$is acceleration]

Step 2. Finding the distance traveled

Both the velocities are equal because the particle comes to rest at last.

$\alpha t_1=\beta \left(t-t_1\right)$

$t_1=\frac{\beta t}{\alpha +\beta }$

Distance is equal to the product of the speed and time.

Distance$=\frac{1}{2}\left(t_1+t_2\right)\times \alpha t_1$

$=\frac{1}{2}t\times \alpha \times \frac{\beta t}{\alpha +\beta }$

Distance$=\frac{\alpha \beta }{2\left(\alpha +\beta \right)}{t}^2$

Hence, the correct option is $\left(A\right)$