# The velocity of an object with a mass of 6 kg is given by v(t)= sin 2 t + cos 4 t . What is the impulse applied to the object at t= (5pi)/12 ?

Oct 11, 2016

#### Explanation:

Impulse is $\vec{J} = {\int}_{a}^{b} \vec{F} \mathrm{dt}$

$= {\int}_{{t}_{1}}^{{t}_{2}} \frac{d \vec{p}}{\mathrm{dt}} \mathrm{dt}$

$= \vec{p} \left({t}_{2}\right) - \vec{p} \left({t}_{1}\right)$

So we need a time period for there to be an impulse within the definition provided, and the Impulse is the change of momentum over that time period.

We can calculate the momentum of the particle at $t = \frac{5 \pi}{12}$ as

$v = 6 \left(\sin \frac{10 \pi}{12} + \cos \frac{20 \pi}{12}\right) = 6 \setminus k g \setminus m \setminus {s}^{- 1}$

But that is the instantaneous momentum.

We can try

$\setminus \vec{J} = {\lim}_{\Delta t = 0} \vec{p} \left(t + \Delta t\right) - \vec{p} \left(t\right)$

$= 6 {\lim}_{\Delta t = 0} \sin 2 \left(t + \Delta t\right) + \cos 4 \left(t + \Delta t\right) - \sin 2 t - \cos 4 t$

$= 6 {\lim}_{\Delta t = 0} \sin 2 t \cos 2 \Delta t + \cos 2 t \sin 2 \Delta t + \cos 4 t \cos 4 \Delta t - \sin 4 t \sin 4 \Delta t - \sin 2 t - \cos 4 t = 0$

No luck :-(

The next port of call might be the Dirac delta function but I'm not sure where that might lead as it's been a while.