We know that acceleration is defined as 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

__Uniform acceleration motion in one dimension__

In uniformly acceleration motion , acceleration can be find 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

**Example 1**

A object slows downs from 50 m/s to rest in a distance 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 $

negative sign is present as it is deceleration or retardation

__Non Uniform acceleration motion in one dimension__

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

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

**Example 2**

A 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 throw vertically upwards with the initial velocity 25 m/s. What height does the ball reaches . Take g=10 m/s
^{2} - A charge particle in an 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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