Question

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

Answer 1

`int F d t=-1,414212 " N.s"`

Explanation:

`J=int F.d t " 'impulse'"`
`M=int m.d v " 'momentum'"`
`int F. d t=int m. d v`
`v(t)=sin5t + cos6t`
`d v= (5 .cos5 t-6.sin6t)d t`
`int F.d t=m int (5. cos5t- 6. sin6t) d t`
`int F d t=2 (5 int cos5t d t- 6 int sin6t d t)`
`int F d t=2(5.1/5 .sin5t+ 6.1/6 cos 6t)`
`int F d t=2(sin 5t + cos 6t)`
`"for t="pi/4`
`int F d t=2(sin 5pi/4 + cos6pi/4)`
`int F d t=2(-0,707106+0)`
`int F d t=-1,414212 " N.s"`