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Question :- A particle of mass 2 kg is subjected to force that varies with time as shown above. Particle is initially at rest.

What is the velocity of particle after 10 s ?

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Force F(t) during the period t=0 to t=2 s is

F(t) = ( 5 t )

Acceleration a(t) = force/mass = (5t)/2 = 2.5 t

acceleration a(t) = dv/dt = 2.5 t

velecity as a function of time from t=0 to t=2 s is determined by integrating above expression

dv = 2.5 t dt

v(2) - v(0) = 1.25 × 4 = 5 m/s

hence velocity at t =2 is  v(2) = 5 m/s

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Force F(t) during the period t= 2 s to t = 4 s is

F(t) = 10 N

Acceleration a(t) = force/mass = 10/2 m/s2  = 5 m/s2

acceleration a(t) = dv/dt = 5 m/s2

velocity as a function of time from t = 2 s to t = 4 s is determined from equation of motion " v(t) = v(2) + ( a ×  Δt ) "

where time interval Δt = (6-4) s  = 2 s

v(t) = 5 + ( 5 × Δt )

velocity at t = 4 s is   , v(4) = 5 + ( 5 × 2) m/s = 15 m/s

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Force F(t) during the period t= 4 s to t = 6 s is

F(t) = 10 + (t-4)5 = 5t - 10

Acceleration a(t) = force/mass = 2.5 t - 5

acceleration a(t) = dv/dt = 2.5 t - 5

velocity as a function of time from t = 4 s to t = 6 s is determined by integrating above expression

dv = 2.5 t dt - 5 dt

v(6)-v(4) = 1.25 × (36-16) - 5 × (6-4) = 15 m/s

velocity at t = 6 s is   , v(6) =  ( 15 + 15 ) m/s = 30 m/s

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Force F(t) during the period t= 6 s to t = 10 s is

F(t) = 20 - (t-6)5 = 50 - ( 5 t )

Acceleration a(t) = force/mass = 25 - (2.5 t )

acceleration a(t) = dv/dt = 25 - (2.5 t )

velocity as a function of time from t = 6 s to t = 10 s is determined by integrating above expression

dv = 25dt - (2.5 t dt )

v(10)-v(6) = 25(4) - 2.5(64) = 20 m/s

velocity at t = 10 s is   , v(10) =  ( 30+20 ) m/s = 50 m/s

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