Question

A hundred metre sprinter increases her speed from rest uniformly at the rate 1m/s up to three quarters of the total run and covers the last quarter with the uniform speed . How much time she take to cover the first half and second half of the run .

Answer 1

d = 100 m
$v^2=u^2+2ab$
$\Rightarrow 1=0+2\times a\times \frac{3}{4}d$
$\Rightarrow 1=2\times a\times \frac{3}{4}\times 100$
$\therefore a=\frac{1}{150}m/s^2$
To reach 50 m, $50=\frac{1}{2}\times \frac{1}{150}\times t^2\Rightarrow t=\sqrt{15000}s$
To reach 75 m, $75-\frac{1}{2}\times \frac{1}{150}\times t_1^2\Rightarrow t_1=\sqrt{75\times 300}s$
So, time to cover from 50 m to 75 m $=t_1=t$
Remaining time $=\frac{25m}{1m/s}=s$