A car acceleration uniformaly from $20 k m / h$ to $35 k m / h$ in $5 s e c$ calculate:
(i) the acceleration
(ii) the distance curved by the car in that time.

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Solution

Here, $u = 20 (\frac{5}{18}) m / s$ and $v = 35 (\frac{5}{18}) m / s$

So, $a = \frac{v - u}{t} = \frac{35 (\frac{5}{18}) - 20 (\frac{5}{18})}{5}$

$\Rightarrow a = \frac{5}{6} m / s^{2}$

Using third equation of motion $s = \frac{v^{2} - u^{2}}{2 a}$ $= \frac{35^{2} (\frac{5}{18})^{2} - 20^{2} (\frac{5}{18})^{2}}{2 \times \frac{5}{18}} = \frac{4125}{36} = 114.58 m$

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