Solution:

(a) frictional force exerted by ground will be in forward direction.

$f f=m a$

$\mu N=m a$

$\mu m g=m a$

$a=\mu g=0.9 \times 10=9 \mathrm{~m} / \mathrm{s}^{2}$

$u=0 ; a=9 \frac{m}{s^{2}} ; s=50 \mathrm{~m}$

$s=u t+\frac{1}{2} a t^{2}$

$50=0+\frac{1}{2}(9)(t)^{2}$

$t=\frac{10}{3}$ sec

(b) While stopping, frictional force will act in opposite direction of motion.

$0-f f=m a$

$-\mu N=m a$

$-\mu m g=m a$

$a=-\mu g=-0.9 \times 10=-9 \mathrm{~m} / \mathrm{s}^{2}$

After covering $50 \mathrm{~m}$, the velocity of athelete is

$v=u+a t$

$=0+9 \times \frac{10}{3}$

$v=30 \mathrm{~m} / \mathrm{s}$

Now,

$u=30 \frac{m}{s} ; v=0 ; a=-9 m / s^{2}$

$v=u+a t$

$t=3.33 \mathrm{sec}$