A car that accelerates uniformly passes at point A with a speed of 30 kph. It reaches point B, 21 km from A, after 18 minutes. Find the acceleration of the car.

$s = v_o t + \frac{1}{2}a t^2$

$21 = 30 \left( \frac{18}{60} \right) + \frac{1}{2}a\left( \frac{18}{60} \right)^2$

$\frac{9}{200}a = 12$

$a = \frac{800}{3} ~ \text{km/hr}^2$
 

$a = \dfrac{800}{3} ~ \dfrac{\text{km}}{\text{hr}^2} \times \dfrac{1000 ~ \text{m}}{1 ~ \text{km}} \times \left( \dfrac{1 ~ \text{hr}}{3600 ~ \text{sec}} \right)^2$

$a = \frac{5}{243} ~ \text{m/sec}^2 = 0.02057 ~ \text{m/sec}^2$