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.

 

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, a =
(i)
At t = –5 s,
Acceleration, a' = 0
Initial velocity, u = 10 m/s 
Using the relation, 
         s = ut + (1/2) a' t² 

           = 10 × (–5) = –50 m
(ii)
At t = 25 s
Acceleration, a'' = –20 m/s² 
Initial velocity, u = 10 m/s
USing the relation, 
            s' = ut' + (1/2) a" t²

               = 10 × 25 + (1/2) × (-20) × (25)² 

               = 250 - 6250
               = -6000 m
(iii)
At t = 100 s, 
For 0 ≤ t ≤ 30 s 
Acceleration, a = -20 ms^-2
Initial velocity, u = 10 m/s
Now, using the equation of motion, we have
               s₁ = ut + (1/2)a"t² 

                   = 10 × 30 + (1/2) × (-20) × (30)²

                   = 300 - 9000 
                   =  -8700 m
For 30 < t ≤ 100 s,
For t= 30 sec, as per the first equation of motion final velocity is given as, 

                   v = u + at 

                     = 10 + (–20) × 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: 
s₂ = vt + a" t² 

    = -590 × 70
    = -41300 m 
∴ Total distance, s" = s₁ + s₂ 
                          = -8700 -41300
                          = -50000 m