A car accelerates uniformly from $18 km/h$ to $36 km/h$ in $5\ second$. Calculate the distance covered by the car in that time.

Here given, the initial velocity of the car $u=18\ km/h=18\times\frac{5}{18}=5\ m/s$

The final velocity of the car $v=36\ km/h=36\times\frac{5}{18}\ m/s=10\ m/s$

Time $t=5\ second$

Therefore, acceleration $a=\frac{v-u}{t}$

$=\frac{10-5}{5}$

$=\frac{5}{5}$

$=1\ m/s^2$

Let $s$ be the distance covered by the car. On using the equation of motion, $s=ut+\frac{1}{2}at^2$

$s=5\times5+\frac{1}{2}\times1\times5^2$

Or $s=25+12.5$

Or $s=37.5/ m$

Therefore, the distance covered by the car is $37.5\ m$.