A body of mass $0.40 kg$moving initially with a 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 of the time when the force is applied to be $t = 0$, and the position of the body at that time $t$ be $x=0$. Predict its position at$t =-5s, 25 s, 100s$?

Here, $m = 0.4 kg, u = 10 m{s^{–1 }}^{ }$due north 

$F = – 8 N$, (negative sign shows the force directed opposite)

Therefore,

 $$\begin{gathered} a=\frac{F}{m}=\frac{-8}{0.4} = -20 m/s^2 (0 \le t \le 30s) when t = –5s \\ x = ut = 10(–5) = –50 m (as a = 0) \\ When t = 25s, x = ut + ½ at2 = 10 \times 25 + ½ (–20)\left(25\right)2 = – 6000 m \\ Upto t = 30s, motion is under acceleration, i.e \\ x = ut + ½ at2 = 10 \times 30 + ½ (–20)\left(30\right)2 = – 8700 m \\ At t = 30s, v = u + at = 10 – 20 \times 30 = – 590 m/s \\ During t =30 to 100s, x_2= vt = – 590 \times 70 = – 41300m \\ [as the force is removed a = 0] \\ Total dis\tan ce, x_1+x_2= –(8700 + 41300) m = –50 km \end{gathered}$$