The driver of a car moving with a speed of $10m{s}^{-1}$ sees a red light ahead, applies brakes and stops after covering 10 m distance. If the same driver would have stopped the car after covering 30 m distance. Within what distance the car can be stopped if travelling with a velocity of  $15m{s}^{-1}$ ? Assume the same reaction time and the same deceleration in each case.

If $t_0$ is the reaction time ,then the distance covered during decelerated motion is $10-10t_0$ 

Now, in the first case, 

${10}^2=2a(10-10t_0)\mathrm{......................}\left(i\right)$

Similarly, in the second case,

${20}^2=2a(30-20t_0)\mathrm{......................}\left(ii\right)$

Again, in the third case, ${15}^2=2a(x-5t_0)\mathrm{......................}\left(iii\right)$

Dividing Eq.(ii) by Eq.(i),

$\frac{{20}^2}{{10}^2}=\frac{30-20t_0}{10-10t_0}$

or $40-40t_0=30-20t_0$

or $20t_0=10$

or $t_0=\frac{1}{2}s$

Dividing Eq (iii) by Eq (i), we get

$\frac{225}{100}=\frac{x-15t_0}{10-10t_0}$

or $\frac{9}{4}=\frac{x-15\times \frac{1}{2}}{10-10\times \frac{1}{2}}$

$45=4x-30$

or $4x=75$

or $x=\frac{75}{4}m=18.75m$