Question

The velocity of an object with a mass of `3 kg` is given by `v(t)= sin 6 t + cos 9 t`. What is the impulse applied to the object at `t= ( 7 pi)/ 12`?

Answer 1

`F*t=0,7642 N.s`

Explanation:

`int F.d t=int m*d v`
`F*t=m int d v`
`F*t=m*int (sin6t+cos9t)d t`
`F*t=m*[-1/6*cos6t+1/9*sin 9t]+C`
`F(0)=1 ; C=1`
`F*t=m*[-1/6*cos6t+1/9*sin 9t]+1`
`t=(7pi)/12`
`F*t=3[-1/6*cos42pi/12+1/9*sin63pi/12]+1`
`F*t=3[-1/6*0-1/9*0,707]+1`
`F*t=3(-0,0786)+1`
`F*t=-0,2358+1`
`F*t=0,7642 N.s`