Motion with Uniform Acceleration - Example 5

Example 5
A car accelerates from 4 ms-1 reaches a velocity of 28 ms-1 after traveling for 64m. What is the deceleration of the car?

It's advisable to list down all the information that we have.

Initial velocity, u = 4 ms-1
Final velocity, v = 28 ms-1
Displacement, s = 64m
Acceleration, a = ?

The time taken, t, is not involved, hence we select the equation $${v^2} = {u^2} + 2as$$ to solve the problem. $( 28 ) 2 = ( 4 ) 2 +2a(64) 128a=784−16 a= 784−16 128 =6m s −2$

Reference

$v=u+at s=ut+ 1 2 a t 2 v 2 = u 2 +2as s= 1 2 (u+v)t$