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Laws of Motion

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Physics Part I

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Physics

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Class 10 Class 12
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 = straight F over straight m space equals space fraction numerator negative 8 over denominator 0.40 end fraction space equals space minus 20 space m divided by s squared

(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 + open parentheses 1 half close parentheses a" t

    = -590 × 70

    = -41300 m 

∴ Total distance, s" = s1 + s2 

                          = -8700 -41300

                          = -50000 m

                          = -50 km.
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