A body of mass 0 .40 kg moving intially with a constant speed of `10 m//s` to the north is subjected to a constant force of 8.0 N directed towards the south for 30 s Take the instant the force is applied to be t = 0 , and the position of the particle at that time to be x = 0 , predict its position at `t = -5 s , 25 s , 100 s` ?

Explanation

Here , ` m = 0.4 kg u = 10 m//s "due" N,F = - 8.0 N`
`a = (F)/(m) = (-8.0)/(0.40) = - 20 ms^(-2)` for `0 le t le 30 s `
(i) At `t = - 5 s , x = ut = 10 xx (-5) = - 50 m `
(ii) At `t = 25 s , x = ut + (1)/(2) at^(2) = 10 xx 25 (1)/(2) (-20) (25)^(2) = - 600 m`
(iii) At `t = 100 s`, The problem is divided into two parts Upto 30 s there is force/acc.
`:.` from `x_(1) = ut + (1)/(2) at^(2) = 10 xx 30 (1)/(2) (-20) (30)^2) = - 8700 m`
At `t = 30 s,upsilon = u + "at" = 10 - 20 xx 30 = - 590 m//s, :.` for motion from 30 s to 100 s
` x_(2) = vt = - 590 xx 70 = - 41300 m. :. x = x_(1) + x_(2) = - 8700 - 41300 = - 50000 m = - 50 km`.