Question

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A. body of mass 0.4 kg moving with a constant speed of 10 M per second to the north is subjected to a constant force of 8 Newton directed towards the south for 30 seconds take the instant the force is applied to be t = 0 the position of the particle at that time to be x is equal to zero and predict its position at t= - 5s, 25s and 100s❓❓

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Answer 1

Here We Have

m = 0.4 Kg

F = 8 N

By Using F = ma

a = 8/0.4

a = 20 m/s²

And We have u = 10 m/s

Distance Travelled in -5 seconds

Here '-' Indicates that The Direction of The Motion and the Direction Of Acceleration is Opposite..

So Using 3rd Equation Of Motion

S = ut - 1/2 *a*t²

S   = 10*5 - 1/2 * 20 * 25

S = 50-250

S = -200 m

Distance Travelled in 25 Seconds

S1 = 10*25 - 1/2 *20*25²

S1 = 250 - 6500

S1 = -6000 m

So Distance Travelled in 30 Seconds

S2 = 10*30 - 1/2*20*900

S2 = 300- 9000

S2 = -8700 m

Now The Velocity is Same upto 30 Seconds but After 30 seconds it's velocity will be

v = u - at

v = 10 - 20*30

v = 10 - 600

v = -590 m/s

Now Distance Travelled Between 30 to 100 seconds

t = 70 seconds

Velocity = S3/ Time

S3= -590*70

S3 = -41300 m

So The Distance Travelled in 100 seconds

S4 = S2 + S3

S4 = -8700 m - 41300

S4 = -50000 m

Answer 2

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$\bold\green{Your\:Answer:-}$

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

____________________________❤

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

____________________________❤

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

____________________________❤

Total distance= -8700-41300 = -50000 m

I hope, this will help you

Thank you___❤

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