﻿ 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.   from Physics Laws of Motion Class 11 Manipur Board

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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 = 0, the position of the body at that time to be = 0, and predict its position at t = –5 s, 25 s, 100 s.

Given,

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 =

(i)

At t = –5 s,

Acceleration, a' = 0

Initial velocity, u = 10 m/s

Using the relation,

s = ut + (1/2) a' t

= 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)

= 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"t

= 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" t

= -590 × 70

= -41300 m

∴ Total distance, s" = s1 + s2

= -8700 -41300

= -50000 m

= -50 km.
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