An othleter runs a sidtance of ` 1500 m` as follows , (i) He starts from rest and accelerates himself uniformly with acceleration ` 2 ms^(-2)` till he covers a distance or ` 900 m`. (ii) He then runs the remaining distance of `600 m` with a uniform speed developed. Calculate the time taken by the athlete to cover the two parts to the distance central point of the total length of track.

Explanation

Refer to Fig. 2 (b) . 37`
.
Taking movtion of athelete from ` o` to ` A` , we have
` u=0, a=2 ms^(-2), S= 900, t=t_1 , v=v_1` (say)
` v^2 =u^2 + 2 aS` or ` ` v_1^2 =0 + 2 xx 2 xx 900`
or ` v_1 = 60 ms^(-1)`
time ` t_1= (v-u)/a = ( 60-0)/2 = 30 s`
Taking motion of athelete from ` o` to `C`, we have
`u=0, a=2 ms^(-2), t=t_3, S=750 m`
As, `S=ut + 1/2 at^2` so, ` 750=0 + 1/2 xx 2 xx t_3^2`
or ` t-3 = sqrt 750 = 27.4 s`.