Question

A force =6t2+4t,is acting on a particle of mass 3 kg then what will be velocity of particle at t=3sec.if at t=0,particle is at rest

Answer 1

Answer: 24 m/s.

Explanation:

F= ma = m (dv/dt) = 6t^2 + 4t { dv/dt= a}

Now, Integrate: ==> m v = 2t^3 + 2t^2.

Since, At t=0, Body is at rest. So, We don't have to cosider t=0. hence, Put the value of t=3.

So, m v = 54+ 18= 72 ==> v= 72/m= 72/3 = 24 m/s.

Answer 2

Answer:

24 m/s

Explanation:

$\bf{\huge\mathcal{\boxed{\rm{\mathfrak\blue{Hey Mate:}}}}}$

F(t) = 6t² + 4t

F= M.(dv/dt)

Since mass is constant ,

[F(t) . dt ]/m= dv

Integrating from t=0 to t=3

v×m = ∫F(t) . dt

= ∫(6t² + 4t)dt

= ∫6t² dt + ∫4tdt

= 6[t³/3] + 4[t²/2]

v(0-3) × m= 6[9-0] + 4[4.5-0]

= 54 + 18

= 72

v = 72/3

= 24 m/s