If the athlete finishes the race in 10.4sec, determine: (i) his initial acceleration; (ii) his maximum velocity.

Let t1 be time in which athlete achieves max. Velocity & let t2 be the remaining time

Let v be the maximum velocity achieved bt athlete & a be his acceleration

$So, t1+t2=10.4 \text {____}(i)$

Also for uniformly acc. Motion , we get

$4= (1/2)*a*(t1)^2 \text {____}(ii) \space\space\&\space\space v=a*(t1) \text {_____}(iii)$

For uniform velocity motion

$96= v*(t2) \text {______}(iv)$

Putting eqn(iii) in eqn(iv)

$96=a*(t1)*(t2) \text {____}(v)$

Putting value of a from eqn (ii) & value of t2 from eqn (i) in eqn (v)

$96 = [8/(t1)^2]*(t1)*(10.4-t1) \\ 96(t1) = 83.2 - 8(t1)$

So, we get, $t1=0.8 sec$ & $t2=9.6 sec$

Putting t1 & t2 in eqn (iv) & (v),we get

$a = 12.5 m/s^2$ & $v = 10 m/s$

His initial acceleration is $12.5 m/s^2$ & his maximum velocity is 10 m/s