Given,

Mass of the body, m = 0.40 kg

Initial speed of the body, u = 10 m/s

Force acting on the body, F = -8 N

The force is negative because it acts in the opposite direction i.e. south to the motion of the body i.e. north.

Time for which the force acts on the body, t =30 s

According to the second law of motion, the acceleration of the body is:

The third equation of motion is given as,

…..(i)

Where,

‘s’ is the position of the body,

‘u’ is the initial speed of the body,

‘a’ is the acceleration of the body and

‘t’ is the time.

Given,

1. At t=-5 s

The acceleration of the body is zero since the force was applied starting t=0

u= 10 m/s

using equation (i) we get,

s=10 m/s × (-5 sec)

⇒ s= -50 m

2. At t= 25s

Acceleration, a’ = -20 m/s² and

u = 10 m/s

using equation (i) we get,

⇒ s = 250 m- 6250 m = -6000 m

3. At t=100 s

We divide this time period in two parts with displacements s’ and s”,

The first part is time 0<t≤30s,

Given,

Acceleration of the body, a = -20 m/s²

u = 10 m/s

The displacement travelled by the body using equation (i) during this time t’ is,

⇒ s’= 300 m – 9000 m = -8700 m

The displacement travelled by the body till t=30 s , s’=-8700 m

The second part is 30s<t≤100s,

Given,

Acceleration of the body=0

The final velocity for the first part is the initial velocity for the second part,

From the first equation of motion, The final velocity after t=30 s is,

v= u + at

⇒ v=10 m/s+ (-20) m/s²× 30 s = -590 m/s

Thus, velocity of the body after 30 sec= -590 m/s

For, motion between 30 s to 100 s, i.e. t = 70 s, we use equation (i):

s”=v t + 0

⇒ s”= -590 m/s × 70 s = -41300 m

Thus, s = s’+ s”= -8700 m – 41300 m

⇒ s = -50000 m

The position of the body at t=-5 s, 25 s, 100 s is -50 m, -6000 m, -50000 m respectively.