Question

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

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

Answer 1

The Correct Option is D

To solve this problem, we need to find the maximum velocity acquired by a car that accelerates from rest at a constant rate \( \alpha \), and then decelerates at a constant rate \( \beta \), with the total time elapsed being \( t \).

1. Let \( V \) be the maximum velocity achieved by the car.
2. During the acceleration phase, using the equation of motion: \( V = \alpha \cdot t_1 \), where \( t_1 \) is the time period of acceleration.
3. During the deceleration phase, the car slows down to rest, so: \( 0 = V - \beta \cdot t_2 \), where \( t_2 \) is the time period of deceleration. Hence, \( V = \beta \cdot t_2 \).
4. Since the car comes to rest from its maximum velocity, \( \alpha \cdot t_1 = \beta \cdot t_2 \).
5. The total time \( t \) is the sum of the acceleration and deceleration times: \( t = t_1 + t_2 \).
6. From the equation \( \alpha \cdot t_1 = \beta \cdot t_2 \), we get: \( t_1 = \frac{\beta}{\alpha} \cdot t_2 \).
7. Substitute \( t_1 = \frac{\beta}{\alpha} \cdot t_2 \) in the total time equation: \( t = \frac{\beta}{\alpha} \cdot t_2 + t_2 = t_2\left(\frac{\beta}{\alpha} + 1\right) = t_2\left(\frac{\beta + \alpha}{\alpha}\right).\)
8. Simplifying gives: \( t_2 = \frac{\alpha}{\alpha + \beta} \cdot t.\)
9. Substitute the value of \( t_2 \) into the equation \( V = \beta \cdot t_2 \): \( V = \beta \cdot \frac{\alpha}{\alpha + \beta} \cdot t = \frac{\alpha \beta}{\alpha + \beta} \cdot t.\)

Therefore, the maximum velocity acquired by the car is \( \frac{\alpha \beta}{\alpha + \beta} \cdot t \).

Hence, the correct answer is: \( \left(\frac{\alpha \beta}{\alpha+\beta}\right) t \).