A car accelerates uniformly from ${18 \mathrm{kmh}}^{-1}$ to ${36 \mathrm{km}h}^{-1}$ within a distance of $15 \mathrm{m}.\mathrm{ }$The uniform acceleration of the car is

A car accelerates uniformly from ${18 \mathrm{kmh}}^{-1}$ to ${36 \mathrm{km}h}^{-1}$ within a distance of $15 \mathrm{m}.\mathrm{ }$The uniform acceleration of the car is $2.5 \mathrm{m} {\mathrm{s}}^{-2}$.
Given data in the problem,
Initial velocity, $\mathrm{u} =\mathrm{ }18 \mathrm{km} {\mathrm{h}}^{-1}$
Final velocity, $\mathrm{v} =\mathrm{ }36 \mathrm{km} {\mathrm{h}}^{-1}$
Distance covered, $\mathrm{s} =\mathrm{ }15 \mathrm{m}$
Converting ${\mathrm{kmh}}^{-1}$ to ${\mathrm{ms}}^{-1}$
$\Rightarrow 18 {\mathrm{kmh}}^{-1}=\mathrm{ }18\mathrm{ }\times \mathrm{ }\frac{5}{18} {\mathrm{ms}}^{-1}\mathrm{ }$$=\mathrm{ }5 {\mathrm{ms}}^{-1}$
$\Rightarrow 36 \mathrm{km} {\mathrm{h}}^{-1}=\mathrm{ }36\mathrm{ }\times \mathrm{ }\frac{5}{18} \mathrm{m} {\mathrm{s}}^{-1}$$=\mathrm{ }10 {\mathrm{ms}}^{-1}$
By using the third law of motion,
${\mathrm{v}}^2\mathrm{ }-\mathrm{ }{\mathrm{u}}^2=2\mathrm{as}$ 

${\Rightarrow 10}^2 - 5^2=2 \times \mathit{ a}\times 15$ 

$\Rightarrow 100\mathit{ – }25 = 30 \times \mathit{ a}$ 

$\Rightarrow \mathit{a }= 75/30$ 

$\Rightarrow \mathrm{a} =\mathrm{ }2.5 {\mathrm{ms}}^{-1}$ 

So, the acceleration of the car is $2.5 {\mathrm{ms}}^{-1}$.