A ` ` 100 m` sprinter in creases her speed from rest uniformly at the rate of ` 1. 5 ms^(-2)` up to three quarters of the total run and covers the last quarter with uniform speed. How much time does she take to cover the first half and the second half of the run ?

Explanation

In Fig. 2 (b) . 38,
total distance ` AD= 100 m`.
Three quarter distance `=AV=75 m` and half ot the tatal distance ` = AB=50 m`.
.
Therefore , distance
` CB=AD-AC=100 - 75 = 25 m`
Taking motion of sprinter from (A) to (B), we have,
` u= 0, a= 1.5 ms^(-2) , t=?, S=50 m`
As, ` S= ut + 1/2at^2`,
:. 50 = 0 xx + 1/2 1.5 xx t^1`
Taking motion of sperinter from (A) to ` C, u=0,`
` a= 1.5 ms^(-2) , t=? S= 75 m , v=?` ltbRgt As ` S= ut + 1/2 at^2` ,
:. 75=0 xx t + 1/2 xx 1.5 xxt^2`
or ` t=( 75 xx 2//1.5)^(1//2) = 10 s`
Also , `v=u+ at=0 + 1.5xx 10 =15 ms^(-1)`
Therefore, time taken from ` B` to `C=10-8.165`
`= 1.825 s`
Taking motion of sprinter from ` C` to ` D` ,
` u= 15 ms^(-1) , t=?`
S= 25 m, a=0` ltbRgt As, ` S=ut + 1/2 at^2` ,
:. `25 =15 xx t+ 1/2 (0) t^2 = 15 t`
or ` t= 25// 15 = 1. 667 s`
:. time from ` B` to ` D= 1.835 + 1. 667`
`= 3.502 s`.