If a car at rest, accelerates uniformly to a speed of $$144\,km/h$$   in $$20\,s,$$ it covers a distance of

Solution :
First of all find acceleration from the given values and then using equation of motion calculate distance travelled.
Given, Initial velocity $$u = 0,$$   time $$t = 20\,s$$
Final velocity $$v = 144\,km/h$$
$$= 40\,m/s$$
From 1st equation of motion,
$$\eqalign{ & v = u + at \cr & \Rightarrow a = \frac{{v - u}}{t} = \frac{{40 - 0}}{{20}} = 2\,m/{s^2} \cr} $$
Now, from 2nd equation of motion, distance covered, $$s = ut + \frac{1}{2}a{t^2} = 0 + \frac{1}{2} \times 2 \times {\left( {20} \right)^2}$$
$$ = 400\,m$$
NOTE
If the car decelerates, the acceleration will be negative and is termed as retardation.