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 #?

1 Answer
Feb 15, 2016

Answer:

#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#