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

Feb 7, 2016

$\int F \cdot d t = 2 , 598 N \cdot s$

Explanation:

$\int F \cdot d t = \int m \cdot d v$
$d v = 4 \cdot \cos 4 t \cdot d t - 3 \cdot \sin 3 t \cdot d t$
$\int F \cdot d t = m \left(4 \int \cos 4 t d t - 3 \int \sin 3 t d t\right)$
$\int F \cdot d t = m \left(4 \cdot \frac{1}{4} \sin 4 t + 3 \cdot \frac{1}{3} \cos 3 t\right)$
$\int F \cdot d t = m \left(\sin 4 t + \cos 3 t\right)$
$\text{for } t = \frac{\pi}{6}$
$\int F \cdot d t = m \left(\sin 4 \cdot \frac{\pi}{6} + \cos 3 \cdot \frac{\pi}{6}\right)$
$\int F \cdot d t = m \left(\sin \left(2 \cdot \frac{\pi}{3}\right) + \cos \left(\frac{\pi}{2}\right)\right)$
$\int F \cdot d t = 3 \left(0 , 866 + 0\right)$
$\int F \cdot d t = 3 \cdot 0 , 866$
$\int F \cdot d t = 2 , 598 N \cdot s$