A particle whose mass is 3.0 kg is pushed along the x-axis with a time-varying force given by: F = 6.0 t2 − 2.0 t + 4.0. At t = 0, the particle is moving at a speed of 15 m/s. How fast (speed) is the particle moving at t = 3.0 s? ( F is in newton and t is in second)

Respuesta :

Answer:

[tex]v = 34 m/s[/tex]

Explanation:

We know that the acceleration a = F/m

  • As the force is a function of time, it varies with time,

[tex]a(t) = F(t) /m[/tex]

  • The mass is constant

The velocity at t = 3 seconds is expressed as an integral of acceleartion over time, which is represented as

[tex]V (t=3)- V(t=0) =\int\limits^3_0 {a(t)} \, dt[/tex]

[tex]a(t) = (6.0t^{2} - 2.0t+4)/3.0[/tex]

[tex]V (t=3)- V(t=0) =(1/3)\int\limits^3_0 {(6.0t^{2}- 2.0t +4) } \, dt[/tex]

[tex]V(t=3) - 15 = 19[/tex]

V (t=3) = 19+15 = 34 m/s