{"id":4992,"date":"2019-01-19T18:05:00","date_gmt":"2019-01-19T12:35:00","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=4992"},"modified":"2022-11-04T12:06:30","modified_gmt":"2022-11-04T06:36:30","slug":"how-to-find-acceleration-with-velocity-and-distance","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/how-to-find-acceleration-with-velocity-and-distance\/","title":{"rendered":"How to find acceleration with velocity and distance"},"content":{"rendered":"\n

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.<\/p>\n\n\n\n

\"accelerating<\/figure><\/div>\n\n\n\n

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<\/a> with velocity<\/a> and distance<\/a><\/p>\n\n\n\n

How to find acceleration with velocity and distance<\/h2>\n\n\n\n

For Uniform acceleration motion in one dimension<\/h3>\n\n\n\n

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}\\)<\/p>\n\n\n\n

Example 1<\/strong>
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<\/strong>
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<\/p>\n\n\n\n

For Non Uniform acceleration motion in one dimension<\/h3>\n\n\n\n

For an object whose velocity varies with position, the instantaneous acceleration can be obtained as
$a = \\frac {dv}{dx}$<\/p>\n\n\n\n

Example 2<\/strong>
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<\/strong>
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<\/sup><\/p>\n\n\n\n

Practice Problems<\/strong><\/p>\n\n\n\n

  1. An object velocity increase from 30 m\/s to 40 m\/s  uniformly while covering the distance 10 m. Find  the acceleration<\/li>
  2. A ball is thrown vertically upwards with the initial velocity 25 m\/s. What height does the ball reach. Take g=10 m\/s2<\/li>
  3.  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<\/li><\/ol>\n\n\n\n

    Hope you like the article on how to find acceleration with velocity and distance. Please do provide the feedback<\/p>\n\n\n\n

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