A particle initially at rest, moves with an acceleration 5 m `s^(-2)` for 5 s. Find the distance travelled in (a) 4 s, (b) 5 s and (c) 5th second.

Explanation

Given initial velocity u =0
acceleration a= 5 `m s^(-2)`
(a) Distance travelled in t =4 s,
`S_(1) =ut +(1)/(2) at^(2) =0xx4 +(1)/(2) xx5 xx(4)^(2)`
= 40 m
(b) Distance travelled in t = 5 s ,
`S_(2) = ut +(1)/(2) at^(2) = 0 xx5 +(1)/(2) xx5 xx(5)^(2)`
= 62.5 m
(c ) Distance travelled in `5^(th)` second ,
= Distance travelled in 5 s - Distance travelled in 4 s
= `S_(2) - S_(1) = (62.5 -40) m = 22.5 m`