# A car starts from rest and accelerates

## Question

A car starts from rest and accelerates uniformly for 10 sec to a velocity of 8 m/s. It then runs at a constant velocity and is finally brought to rest in 64 meters with a constant retardation. The total distance covered by the car is 584 meters. Find the value of acceleration, retardation and total time taken.

The answer to the acceleration, retardation and total time taken is 0.8 m/s2, 0.5 m/s2, and 86 seconds respectively.

### Explanation

We will break down the question and solve it individually in terms of acceleration, retardation, and total time taken.

Data: Revealed information from the question

The car starts from rest. Therefore, the initial velocity, u = 0

Time it takes the car to accelerate, t = 10 s

velocity, v = 8 m/s

The car was brought to rest after covering a distance, s1 = 64 m

Total distance covered, sT = 584 m

To find acceleration, retardation, and the total time taken. We will apply the following methods:

1. Unknown
2. Formula
3. Solution

#### Acceleration

Unknown: Unrevealed information from the question

Acceleration, a = ?

Formula: The equation that will help us solve the problem

We will apply one of the equations of motion that says v = u + at

and by making a subject of the formula, we will obtain

a = (v – u) / t

##### Solution

We will apply the above expression to solve the problem

a = (v – u) / t = (8 – 0) / 10 = 0.8 m/s2

#### Retardation

Unknown: Unrevealed information from the question

Retardation = negative acceleration = a = ?

Formula: The equation that will help us solve the problem

We will start by applying v2 = u + 2as to find the distance

We will make s subject of the formula from the above question to get

s2 = v2 / 2a [Remember that the car starts from rest, which implies that u is zero (0)]

Hence, we will apply the following formulae to find the retardation:

First formula, s2 = v2 / 2a

Second Formula, sT = s1 + s2 + s3

Third formula, distance s3 = speed (v) x time (t)

Fourth formula, retardation a = – (u2 / 2s)

##### Solution

We will insert our data and the value for acceleration into the above formulae to find the retardation

Applying the first formula

s2 = v2 / 2a = 82 / 2 x 0.8 = 64 / 0.16 = 40 m

By applying the second formula

sT = s1 + s2 + s3

Which implies

584 = 64 + 40 + s3

Hence,

s3 = 584 – 64 – 40 = 480 m

We will now apply the third formula

s3 = speed (v) x time (t)

Thus

480 = 8 x t

Which will give us

t = 480 / 8 = 60 s

Applying the fourth formula

a = – (u2 / 2s) = – (82 / 2 x 64) = – (64 / 128) = – 0.5 m/s2

Therefore, retardation is 0.5 meters per second square.

#### Total time taken

Since we know that

Acceleration (retardation) = change in velocity (u – v) / time (t)

We will make t subject of the formula from

a = (u – v) / t

Hence, apply

t = (u – v) / a

We will now insert our data into the above expression

t = (u – v) / a = (0 – 8) / – 0.5 = 16 s

Thus, to find the the total time taken. We will add the time it takes to accelerate (10s), t = 60 s, and t = 16s.

Hence,

ttotal = 10 + 60 + 16 = 86 s

Therefore, the total time taken is 86 seconds