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 #?

1 Answer
Feb 5, 2016

Answer:

#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"#