A car accelerates uniformly from $18 k m h^{- 1}$ to $36 k m h^{- 1}$ in 5s.
Calculate (i) the acceleration and
(ii) the distance covered by the car at that time.

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Solution

$A c c e l e r a t i o n = \frac{C h a n g e i n V e l o c i t y}{T i m e}$
Change in Velocity = $36 - 18 = 18 k m / h = 5 m / s$
Time= $5 S e c o n d s$
Acceleration = $\frac{5}{5} = 1 m / s^{2}$
Equation of motion can be given as:
$s = u t + (\frac{1}{2}) a t^{2}$
$u = 18 k m / h = 5 m / s$
$t = 5 s$
$a = 1 m / s^{2}$
$s = (5 \times 5) + (\frac{1}{2} \times 1 \times 5 \times 5)$
$s = 25 + 12.5$
$s = 37.5 m$

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