Question

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, the maximum velocity acquired by the car is given by

A (α²+β²/αβ)t
B (α²−β²/αβ)t
C (α+β/αβ)t
D (αβ/α+β)t

Answer 1

The maximum velocity acquired by the car is given by (D) (αβ/α + β)t in terms of the acceleration rates α and β and the total time elapsed t.

Step-by-step explanation:

The maximum velocity acquired by the car can be found using the equations of motion. Let's break down the problem into two parts - acceleration and deceleration.

1. Acceleration: The car starts from rest and accelerates at a constant rate α.

The time taken to reach the maximum velocity during acceleration can be found using the equation v = u + at, where u is the initial velocity and v is the final velocity.

Here, u = 0, v = maximum velocity, and t is the time taken.

Rearranging the equation, we get t = v/a.

2. Deceleration: After reaching the maximum velocity, the car decelerates at a constant rate β to come to rest.

The time taken to decelerate can be found using the same equation as above, but with u = maximum velocity and v = 0.

Rearranging the equation, we get t = -v/a (since deceleration is negative acceleration).

Adding the time taken for acceleration and deceleration, we get the total time elapsed: t = ta + td = v/α - v/β. Rearranging the equation, we find v = (αβ/α + β)t.

So, the correct answer is (D) (αβ/α + β)t.