A particle starts from rest at time `t=0` and moves on a straight line with acceleration a (ms^-2) as plotted in Fig. 2 (b) .17. Find the time at which the speed og the particle is maximum. Also calculate the displacement of theparticle from starting point after `4 s`.
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Explanation

Taking motion from `0 to 2 seconds` , we have
`u=0, a=- `10 m//s^(2), t=2 s, v=?
`v=u + at =0 + (-10) xx 2= =- 20 ms^(-1)`
Taking motion from `2 to 4 swconsa ` we have ltBrgt `u==- 20 ms^(-1) , a=10 m//s^(2) , t=2 s, v=?
Theregore, the speed is maximum at time
`t=2 seconds`
Taking motion from `0t0 2swons, it `S_(1) is the distance coverd, then`
`S_91) =(v^(2)-u^(2))/(2a) =((-20)^(2) -(0)^(2))/(2(-10)) =-20m`
Taking motion from `2 to 4` seconds it `S_(2) is the distance covered then
`S_(2) =(v^(2)-u^(2))/(2a) =(-(-20)^(2))/(2 xx 10) =- 20 m
:. Total displacement `=S_(1) +S_(2)`
`=- 20 +(-20)`
`=- 20 + (-20)`
=-40 m` lt.