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

u = `10 ms^(-1)`
m= 0.40 kg (North direction)
t=30 s
Here north direction is considered as positive and south direction is considered as negative
Force is applied for t= 30 s

(i) At t= 5 s position
`X_(1) = ut`
`=10 xx (-5)`
(ii) at t= 25 s
`x_(2) ut + (1)/(2) 4r^(2)`
where `a = (F )/( m ) = (-8)/(0.4) = - 20 ms^(-2)`
`=250 - 6250`
`-=6 km`
when force is applied for 30 s distance covered
velocity of object after 30 s
= u + 9t
=10 -600
(iii) Position of object at t= 100s
`:. x_(100) = x_(30) a + x_(70)`
=-8700 41300
`=-5000 m`
`=-50 km`
Thus position at -5 s 25 s and 10 willbe -50 m - 6 km and -50 km