A body of mass 0.40 kg moving initially with a constant speed of `10 m s^(-1)` to the north is subject 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, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.

Explanation

`a = - 20 ms^(-2) 0lt= t lt= 30 s `
`t= -5s : x = ut = - 10 xx 5 = -50 m`
`t = 25 s : x = ut + (1//2) at^2 = (10 xx 25 - 10 xx 625 )m = - 6km`
t = 100 s : first consider motion upto 30 s
`x_1 = 10 xx 30 - 10 xx 900 = - 8700m`
At `t = 30s, v = 10 - 20 xx 30 = - 590 ms^(-1)`
for motion from 30s to 100s : `x_2 = -590 xx 70 = -41300 m `
` x = x_1 + x_2 = - 50 km`