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/s2
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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