Question

The velocity of an object with a mass of `6 kg` is given by `v(t)= sin 5 t + cos 3 t`. What is the impulse applied to the object at `t= ( pi)/ 2`?

Answer 1

`F(pi/2)=40,98 N.s`

Explanation:

`F(t)= (d P)/(d t)`
`P=m*v`
`F(t)=(d(m*v))/(d t)`
`F(t)=m*(d v)/(d t)`
`F(t)=m*(d(sin5t+cos3t))/(d t)`
`F(t)=m*(5*cos 5t-3*sin3t)`
`F(pi/2)=6*(5*cos 5pi/2-3*sin3pi/2))`
`F(pi/2)=6*(5*0,766-3*(-1))`
`F(pi/2)=6*(6,83)`
`F(pi/2)=40,98 N.s`