#### 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.

#### Solution 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 a't^2`

= 10 × (–5) = –50 m

(ii) __At ____t____ = 25 s__

Acceleration, *a*'' = –20 m/s^{2} and *u* = 10 m/s

s' = ut' + 1/2 a"t^{2}

`= 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/s^{2}

u = 10 m/s

S_{1} = ut + 1/2 a^{"}t^{2}

`= 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:

s_{2} = vt + 1/2 a"t^{2}

= –590 × 70 = –41300 m

∴Total distance, S" = s_{1} + s_{2} = -8700 - 41300 = -50000 m

#### Solution 2

Here m = 0.40 kg, u = `10 ms^(-1)`, F = -8 N(retarding force)

As `a = F/m = - 8/0.4 = - 20 ms^(-2)`

Also `S = ut + 1/2 at^2`

1) Position at t = - 5s

`S = 10(-5) + 1/2 xx 0 xx (-5)^2 = -50 m`

2) Position at t = 25 s

`S_1 = 10 xx 25 + 1/2 xx (-20)xx(25)^2 = -6000 m = - 6 km`

3) Postion at t = 100 s

`S_2 = 10xx30 + 1/2 xx (-20) xx (30)^2 = -8700 m`

At t = 30 s, v = u + at

`v = 10 - 20 xx 30 = -590 ms^(-1)`

Now for motion from 30 s to 100s

`S_3 = -590 xx 70 + 1/2 (0) xx (70)^2 = -41300 m`

`"Total distance" = S_2 + S_3 = -8700 - 41300 = -50000 m = - 50 km`