# a car accelerates from rest at a constant rate alpha

## Question

A car accelerates from rest at a constant rate alpha (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. ((α+β)/αβ)t
c. ((α2−β2)/αβ)t
d. (αβ/(α+β))t

## Solution

To determine the maximum velocity acquired by the car, we will analyze the two phases of the car’s motion: acceleration and deceleration.

### Phase 1: Acceleration

• The car starts from rest and accelerates at a constant rate α for a time t1​.
• The velocity at the end of this phase, vmax, is given by: vmax = αt1

### Phase 2: Deceleration

• The car then decelerates at a constant rate β for a time t2​ and comes to rest.
• The initial velocity for this phase is vmax​, and the final velocity is 0.
• Using the equation of motion for constant deceleration: 0 = vmax − βt2

Substituting vmax from the first phase: 0 = αt1 − βt2

### Total Time

• The total time elapsed is t: t = t1 + t2

### Solving for t1t_1t1​ and t2t_2t2​

From the equation αt1 = βt2, we can express t2 in terms of t1​: t2 = αt1

We will now Substitute t2 into the total time equation: t = t1 + αt1

Factor out t1​: t = t1(1 + α/β)

Solving for t1​: t1 = (t / (1 + α/β)) = (t / ((β + α)/β)) = (tβ / (α + β))

Now, we use t1​ to find vmax​: vmax = αt1 = α(tβ / (α + β)) = αβt / (α + β) = (αβ/(α+β))t

Therefore, the maximum velocity acquired by the car is given by: vmax = αβt / (α + β) = (αβ/(α+β))t