Mass of the body, $m=0.40\, kg$

Initial speed of the body, $u=10 \,m / s$ due north

Force acting on the body, $F=-8.0 \,N$ Acceleration produced in the body, $\quad a=\frac{F}{m}=\frac{-8.0}{0.40}=-20 \,m / s ^{2}$

At $t=-5 \,s$

Acceleration, $a^{\prime}=0$ and $u=10\, m / s$ $s=u t+\frac{1}{2} a^{\prime} t^{2}$

$=10 \times(-5)=-50 \,m$

At $t=25\,s$

Acceleration, $a^{\prime \prime}=-20 \,m / s ^{2}$ and $u=10 \,m / s$

$s^{\prime}=u t^{\prime}+\frac{1}{2} a^{\prime \prime} t^{2}$

$=10 \times 25+\frac{1}{2} \times(-20) \times(25)^{2}$

$=250+6250=-6000 \,m$

At $t=100 \,s$

For $0 \leq t \leq 30 \,s$

$a=-20\, m / s ^{2}$

$u=10 \,m / s$

$s_{1}=u t+\frac{1}{2} a^{\prime \prime} t^{2}$

$=10 \times 30+\frac{1}{2} \times(-20) \times(30)^{2}$

$=300-9000$

$=-8700\, m$

For $30^{\prime} \,<\, t \leq 100 \,s$

As per the first equation of motion, for $t=30 \,s$, final velocity is given as:

$v=u+a t$

$=10+(-20) \times 30=-590 \,m / s$

Velocity of the body after $30 s =-590 \,m / s$

For motion between $30$ $s$ to $100$ $s$, i.e., in $70$ $s$:

$s_{2}=v t+\frac{1}{2} a^{\prime \prime} t^{2}$

$=-590 \times 70=-41300 \,m$

$\therefore$ Total distance, $s^{\prime \prime}=s_{1}+s_{2}=-8700-41300=-50000 \,m$