Question

The velocity of an object with a mass of `2 kg` is given by `v(t)= 3 t^2+ 2 t +8`. What is the impulse applied to the object at `t= 4`?

Answer 1

The impulse at `t=4` is `52 kg ms^-1`

Explanation:

Impulse is equal to the rate of change of momentum:

`I = Delta p = Delta (mv)`.

In this instance the mass is constant so `I = mDeltav`.

The instantanteous rate of change of the velocity is simply the slope (gradient) of the velocity-time graph, and can be calculated by differentiating the expression for the velocity:

`v(t)=3t^2+2t+8`
`(dv)/dt = 6t +2`

Evaluated at t = 4, this gives `Delta v=26 ms^-1`

To find the impulse, then, `I = mDeltav = 2*26 = 52 kgms^-1`