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, v_max, is given by: v_max = 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 v_max, and the final velocity is 0.
Using the equation of motion for constant deceleration: 0 = v_max − βt₂

Substituting v_max 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/β

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

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 v_max: v_max = αt₁ = 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