The velocity of an object with a mass of 3 kg3kg is given by v(t)= sin 4 t + cos 3 t v(t)=sin4t+cos3t. What is the impulse applied to the object at t= pi /6 t=π6?

1 Answer
ali ergin
Feb 7, 2016

int F*d t=2,598 N*s∫F⋅dt=2,598N⋅s

Explanation:

int F*d t=int m* d v∫F⋅dt=∫m⋅dv
d v=4*cos4 t*d t-3*sin 3 t*d tdv=4⋅cos4t⋅dt−3⋅sin3t⋅dt
int F*d t=m(4 int cos 4t d t -3 int sin 3t d t)∫F⋅dt=m(4∫cos4tdt−3∫sin3tdt)
int F*d t=m(4*1/4sin 4t +3*1/3 cos 3t )∫F⋅dt=m(4⋅14sin4t+3⋅13cos3t)
int F*d t=m(sin 4t +cos 3t)∫F⋅dt=m(sin4t+cos3t)
"for " t=pi/6for t=π6
int F*d t=m (sin 4*pi/6 +cos 3* pi/6 )∫F⋅dt=m(sin4⋅π6+cos3⋅π6)
int F*d t=m(sin (2*pi/3)+ cos ( pi/2))∫F⋅dt=m(sin(2⋅π3)+cos(π2))
int F*d t=3(0,866+0)∫F⋅dt=3(0,866+0)
int F*d t=3* 0,866∫F⋅dt=3⋅0,866
int F*d t=2,598 N*s∫F⋅dt=2,598N⋅s

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