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?

Answer:
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. $$\begin{array}{l}{\left(28\right)}^2={\left(4\right)}^2+2a(64)\\ 128a=784-16\\ a=\frac{784-16}{128}=6m{s}^{-2}\end{array}$$

Reference

$$\begin{array}{l}v=u+at\\ s=ut+\frac{1}{2}a{t}^2\\ v^2=u^2+2as\\ s=\frac{1}{2}(u+v)t\end{array}$$

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