# The position of an object moving along a line is given by p(t) = t - tsin(( pi )/4t) . What is the speed of the object at t = 5 ?

Aug 26, 2016

$| v \left(5\right) | = 1 + \frac{1}{\sqrt{2}} + \frac{5 \pi}{4 \sqrt{2}}$.

#### Explanation:

The velocity of the particle at any time $t$ is given by the derivative with respect to time of the position. In other words, $v \left(t\right) = p ' \left(t\right)$.

$v \left(t\right) = p ' \left(t\right) = 1 - \sin \left(\frac{\pi}{4} t\right) - t \frac{\pi}{4} \cos \left(\frac{\pi}{4} t\right)$. (Note: to do this you must use the product rule).

$v \left(5\right) = 1 - \sin \left(\frac{5 \pi}{4}\right) - \frac{5 \pi}{4} \cos \left(\frac{5 \pi}{4}\right) = 1 + \frac{1}{\sqrt{2}} + \frac{5 \pi}{4 \sqrt{2}}$.

Speed is the absolute value of velocity, but the velocity here is positive so it's not an issue.