Question

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.

Answer 1

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,  `a = "F"/"m" = (-8.0)/(0.40) = -20 "m/s"^2`

(i) At t = -5s

Acceleration, a' = 0 and u = 10 m/s

`"S" = "ut" + 1/2  "at"^2`

= 10 × (-5) = -50 m

(ii) At t = 25 s

Acceleration, a'' = -20 m/s² and u = 10 m/s

S = ut + `1/2` at²

`= 10 xx 25 + 1/2 xx (-20) xx (25)^2`

= 250 + 6250 = -6000 m

(iii) At t = 100 s

For `0 <= t <= 30`

a = –20 m/s²

u = 10 m/s

S₁ = ut + `1/2` at²

`= 10 xx 30 + 1/2 xx (-20) xx (30)^2`

= 300 - 9000

= -8700 m

For 30' < t <= 100s

As per the first equation of motion, for t = 30 s, 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 

= -590 × 70 = -41300 m

∴ Total distance, S = s₁ + s₂ = -8700 - 41300 = -50000 m