$A$ 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 $t=-5 \;s, 25 \;s, 100 \;s$.

(N/A) Mass of the body,$m = 0.40 \;kg$.
Initial velocity,$u = 10 \;m s^{-1}$ (North is positive).
Force,$F = -8.0 \;N$ (South is negative).
Acceleration,$a = \frac{F}{m} = \frac{-8.0}{0.40} = -20 \;m s^{-2}$.
At $t = -5 \;s$:
The force has not been applied yet,so $a = 0$.
$x = u t = 10 \times (-5) = -50 \;m$.
At $t = 25 \;s$:
The force is applied for the entire duration.
$x = u t + \frac{1}{2} a t^2 = 10 \times 25 + \frac{1}{2} \times (-20) \times (25)^2 = 250 - 6250 = -6000 \;m$.
At $t = 100 \;s$:
For $0 \leq t \leq 30 \;s$,$x_1 = 10 \times 30 + \frac{1}{2} \times (-20) \times (30)^2 = 300 - 9000 = -8700 \;m$.
Velocity at $t = 30 \;s$ is $v = u + at = 10 + (-20) \times 30 = -590 \;m s^{-1}$.
For $30 < t \leq 100 \;s$,the force is zero,so $a = 0$.
$x_2 = v \times \Delta t = -590 \times (100 - 30) = -590 \times 70 = -41300 \;m$.
Total position $x = x_1 + x_2 = -8700 - 41300 = -50000 \;m$.