Question

A body of mass 0.40 kg moving initially with a constant speed of 10 ms-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 = 0.4kgforce = 8Nacceleration = 8/0.4 = 20 m/s²
Assume direction from south to north as positive,at the instant when the force acts,u = 10 m/sa = -20 m/s²x₀ = 0x = x₀ + ut + 0.5at²
a)before t=0s, force was not actingso 0 = x + ut⇒ 0 = x + 10×(5)⇒ x = -50mSo particle was 50m south of x=0 at t=-5s.
b)at t = 25s,x = 10×25 - 0.5×20×25² = -6000 mSo particle is at 6000m south of x=0 at t=25s.
c)at t = 30s,x = 10×30 - 0.5×20×30² = -8700 mSo particle is at 8700m south of x=0 at t=30s.

Answer 2

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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

At t = –5 s

Acceleration, a' = 0 and u = 10 m/s

S = ut + 1/2 a' t^2 = 10× (-5)= -50 m

At t = 25 s

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

S' = ut' + 1/2 a'' t^2 = (10×25)+ 1/2 (-20)× (25)^2

= 250+ 6250= - 6000 m

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At t = 100 s For 0≤ t ≤ 30

a = –20 m/s2

u = 10 m/s

S' = ut' + 1/2 a'' t^2 = (10×30)+ 1/2 (-20)× (30)^2

= 300- 9000 = - 8700 m

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For 30≤ t ≤ 100

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' = ut' + 1/2 a'' t^2

= –590 × 70 = –41300 m

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Total distance= -8700-41300 = -50000 m

I hope, this will help you

Thank you___❤

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