## Question

A body moving at 20 m/s decelerates uniformly at 5m/s^{2} till its speed becomes 10 m/s. The distance covered within this period is

### Answer

**The answer to the distance covered within this period is 30 meters.**

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### Here is the explanation to the answer

**Data:** Available values from the question

The initial velocity, u = 20 m/s

Deceleration, a = opposite of acceleration = – 5 ms^{-2}

Final velocity, v = 10 m/s

**Unknown:** The values from we need to find

Distance covered, s = ?

**Formula:** The formula that will help us find the distance covered

From the equations of motion, v^{2} = u^{2} + 2as

after making s subject of the formula, we will have

s = (v^{2} – u^{2}) / 2a

We will apply the above equation to solve the problem

#### Solution

After inserting our data into the formula, we will have

s = (v^{2} – u^{2}) / 2a = (10^{2} – 20^{2}) / 2(-5)

Which will give us

s = (100 – 400) / -10

Thus

s = – 300 / -10 = 30 m

**Therefore, the distance covered s is 30 meters. **