In this article learn how to find acceleration with velocity and distance. We will look at the cases for motion with uniform acceleration and non-uniform acceleration in one dimension.

We know that acceleration is defined as a change in velocity per unit time. So given velocities and time, we can calculate acceleration. Now here we will be checking how to find acceleration with velocity and distance

## How to find acceleration with velocity and distance

### For Uniform acceleration motion in one dimension

In uniformly acceleration motion, acceleration can be found with velocity and distance using the formula

$v^2= u^2 + 2as$

Where v is the final velocity

\(u\) is the initial velocity

\(a\) is the acceleration

\(s\) is the distance traveled

Rearranging the formula we get

\(a=\frac{v^2-u^2}{2s}\)

**Example 1**

An object slows downs from 50 m/s to rest in a distance of 100m. what is the acceleration of the object assuming constant?**Solution**

We know that

$v^2= u^2 + 2as$

or $a = \frac {v^2 -u^2}{2s} = -50/200 =- .5 m/s^2 $

the negative sign is present as it is deceleration or retardation

### For Non Uniform acceleration motion in one dimension

For an object whose velocity varies with position, the instantaneous acceleration can be obtained as

$a = \frac {dv}{dx}$

**Example 2**

An object moves in a straight line such as $v= x+ x^2 + 5$ where \(v\) is the velocity and \(x\) is the position Find the acceleration equation and acceleration at \(x=0\).**Solution**

Given

$v= x+ x^2 + 5$

Now

$a = v \frac {dv}{dx}$

$a = (x+ x^2 + 5) \frac {d}{dx} (x+ x^2 + 5) $

$a=(x+ x^2 + 5)(2x+ 1) = 2x^3+ 3x^2 +11x + 5$

At \(x=0\)

a= 5 m/s^{2}

**Practice Problems**

- An object velocity increase from 30 m/s to 40 m/s uniformly while covering the distance 10 m. Find the acceleration
- A ball is thrown vertically upwards with the initial velocity 25 m/s. What height does the ball reach. Take g=10 m/s2
- A charged particle in a uniform electric field is accelerated in a straight line from rest to 1000 km/s in a distance 1 m.What is the acceleration

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