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 #?

1 Answer
Feb 18, 2016

Answer:

#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#