# 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 ?

Feb 15, 2016

$F \left(\frac{\pi}{2}\right) = 40 , 98 N . s$

#### Explanation:

$F \left(t\right) = \frac{d P}{d t}$
$P = m \cdot v$
$F \left(t\right) = \frac{d \left(m \cdot v\right)}{d t}$
$F \left(t\right) = m \cdot \frac{d v}{d t}$
$F \left(t\right) = m \cdot \frac{d \left(\sin 5 t + \cos 3 t\right)}{d t}$
$F \left(t\right) = m \cdot \left(5 \cdot \cos 5 t - 3 \cdot \sin 3 t\right)$
F(pi/2)=6*(5*cos 5pi/2-3*sin3pi/2))
$F \left(\frac{\pi}{2}\right) = 6 \cdot \left(5 \cdot 0 , 766 - 3 \cdot \left(- 1\right)\right)$
$F \left(\frac{\pi}{2}\right) = 6 \cdot \left(6 , 83\right)$
$F \left(\frac{\pi}{2}\right) = 40 , 98 N . s$