## 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.

### Answer

**The answer to the acceleration, retardation and total time taken is 0.8 m/s ^{2}, 0.5 m/s^{2}, 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, s_{1} = 64 m

Total distance covered, s_{T} = 584 m

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

- Unknown
- Formula
- 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/s ^{2}**

#### 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 **v ^{2} = u + 2as** to find the distance

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

**s _{2} = v^{2} / 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, **s _{2} = v^{2} / 2a**

Second Formula, **s _{T} = s_{1} + s_{2} + s_{3}**

Third formula, distance **s _{3} = speed (v) x time (t)**

Fourth formula, retardation **a = – (u ^{2} / 2s)**

##### Solution

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

**Applying the first formula**

**s _{2} = v^{2} / 2a** = 8

^{2}/ 2 x 0.8 = 64 / 0.16 = 40 m

**By applying the second formula**

**s _{T} = s_{1} + s_{2} + s_{3}**

Which implies

584 = 64 + 40 + **s _{3}**

Hence,

**s _{3}** = 584 – 64 – 40 = 480 m

**We will now apply the third formula**

**s _{3} = speed (v) x time (t)**

Thus

480 = 8 x t

Which will give us

t = 480 / 8 = 60 s

**Applying the fourth formula**

**a = – (u ^{2} / 2s)** = – (8

^{2}/ 2 x 64) = – (64 / 128) = – 0.5 m/s

^{2}

**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,

**t _{total} = 10 + 60 + 16 = 86 s**

**Therefore, the total time taken is 86 seconds**