## 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 t
_{1}. - The velocity at the end of this phase, v
_{max}, is given by: v_{max}= αt_{1}

### Phase 2: Deceleration

- The car then decelerates at a constant rate β for a time t
_{2} and comes to rest. - The initial velocity for this phase is v
_{max}, and the final velocity is 0. - Using the equation of motion for constant deceleration: 0 = v
_{max}− βt_{2}

Substituting v_{max} from the first phase: 0 = αt_{1} − βt_{2}

### Total Time

- The total time elapsed is t: t = t
_{1}+ t_{2}

### Solving for t1t_1t1 and t2t_2t2

From the equation αt_{1} = βt_{2}, we can express t_{2} in terms of t_{1}: t_{2} = αt_{1}/β

We will now Substitute t_{2} into the total time equation: t = t_{1} + αt_{1}/β

Factor out t_{1}: t = t_{1}(1 + α/β)

Solving for t_{1}: t_{1} = (t / (1 + α/β)) = (t / ((β + α)/β)) = (tβ / (α + β))

Now, we use t_{1} to find v_{max}: v_{max} = αt_{1} = α(tβ / (α + β)) = αβt / (α + β) = (αβ/(α+β))t

**Therefore, the maximum velocity acquired by the car is given by: v _{max} = αβt / (α + β) = (αβ/(α+β))t**