A car, initially at rest, is given a uniform acceleration of a 0.5 $m / s^{2}$ for 20 s. The car then moves with uniform velocity for 20 s. Brakes are then applied and the car is brought to rest in10 s. Find the total distance covered by the car.

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Solution

Velocity attained at end of 10 s, $v = u + a t$
$= 0 + 0.5 \times 20 \Rightarrow v = 10 m / s$
Distance travelled in first 20 s, $s_{1} = u t + \frac{1}{2} a t^{2} = \frac{1}{2} \times 0.5 \times 20^{2} = 100 m$
Distance travelled in next 20 s, $s_{2} = v t = 10 \times 20 = 200 m$
Velocity at end of 50 s = 0
So, acceleration in last 10 s, $a^{'} = \frac{v - u}{t} = \frac{0 - 10}{10} = - 1 m / s^{2}$
Distance travelled in last 10 s, $s_{3} = u t + \frac{1}{2} a t^{2} = 10 \times 10 + \frac{1}{2} \times (- 1) \times 10^{2}$
$= 100 - 50 = 50 m$
Total distance covered = $100 + 200 + 50 = 350 m$

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A car, initially at rest, is given a uniform acceleration of a 0.5 $m / s^{2}$ for 20 s. The car then moves with uniform velocity for 20 s. Brakes are then applied and the car is brought to rest in10 s. Find the total distance covered by the car.