A car, moving with a speed of 50kmph, can be stopped by applying brakes after travelling at least a distance of 6 m. If the same car is moving at a speed of 100kmph, the minimum distance covered by the car before being stopped is     

Given that a car moving with

Initial speed ${\text{u}}_{\text{1}}=50\text{kmph}=50\times \frac{5}{18}=\frac{125}{9}\text{m/s}$

Brought to rest ${\text{v}}_{\text{1}}=0$

Let acceleration caused by brakes $=-a$

Distance travelled before coming to rest ${\text{s}}_{\text{1}}=6\text{m}$

We know ${\text{v}}^{\text{2}}-{\text{u}}^{\text{2}}=2a\text{s}$

$0-{\text{u}}_1^2=-2a{\text{s}}_{\text{1}}$

$\therefore {\text{u}}_1^2=2a{\text{s}}_{\text{1}}\to \left(1\right)$

It same car with same brakes acceleration $=-a$

Has initial velocity ${\text{u}}_{\text{2}}=100\text{kmph}=100\times \frac{5}{18}=\frac{250}{9}\text{m/s}$

Brought to rest i.e. ${\text{v}}_{\text{2}}=0$

Let the distance travelled ${\text{s}}_{\text{2}}=?$

We can have

${\text{v}}^{\text{2}}-{\text{u}}^{\text{2}}=2a\text{s}$

$0-u_2^2=-2a\cdot {\text{s}}_{\text{2}}$

 $\therefore u_2^2=2a{\text{s}}_{\text{2}}\to \left(2\right)$

From (1) & (2) we get

$\frac{u_1^2}{u_2^2}=\frac{2a{\text{s}}_{\text{1}}}{2a{\text{s}}_{\text{2}}}$

$\Rightarrow {\left(\frac{{\text{u}}_{\text{1}}}{{\text{u}}_{\text{2}}}\right)}^2=\frac{{\text{s}}_{\text{1}}}{{\text{s}}_{\text{2}}}$

${\text{u}}_{\text{1}}=\frac{125}{9},{\text{u}}_{\text{2}}=\frac{250}{9},{\text{s}}_{\text{1}}=6\text{m},{\text{s}}_{\text{2}}=?$

${\left(\frac{\frac{125}{9}}{\frac{250}{9}}\right)}^2=\frac{6}{{\text{s}}_{\text{2}}}$

$\Rightarrow \frac{1}{\text{4}}=\frac{6}{{\text{s}}_{\text{2}}}$

$\therefore {\text{s}}_{\text{2}}=24\text{m}$