Question

The velocity of an object with a mass of `3 kg` is given by `v(t)= sin 4 t + cos 3 t`. What is the impulse applied to the object at `t= pi /6`?

Answer 1

`int F*d t=2,598 N*s`

Explanation:

`int F*d t=int m* d v`
`d v=4*cos4 t*d t-3*sin 3 t*d t`
`int F*d t=m(4 int cos 4t d t -3 int sin 3t d t)`
`int F*d t=m(4*1/4sin 4t +3*1/3 cos 3t )`
`int F*d t=m(sin 4t +cos 3t)`
`"for " t=pi/6`
`int F*d t=m (sin 4*pi/6 +cos 3* pi/6 )`
`int F*d t=m(sin (2*pi/3)+ cos ( pi/2))`
`int F*d t=3(0,866+0)`
`int F*d t=3* 0,866`
`int F*d t=2,598 N*s`