# The position of an object moving along a line is given by p(t) = 2t - 3cos(( pi )/2t) + 2 . What is the speed of the object at t = 3 ?

May 7, 2018

$\frac{- 3 \pi}{2} + 2$

#### Explanation:

The position function of the object is given by:

$p \left(t\right) = 2 t - 3 \cos \left(\frac{\pi}{2} t\right) + 2$

Since the velocity is displacement over time, it means that it the rate of changing the position over time, or the derivative of the function.

Then we got,

$v \left(t\right) = p ' \left(t\right)$

Let's find the hardest part, which I think is:

$\frac{d}{\mathrm{dt}} \left(3 \cos \left(\frac{\pi}{2} t\right)\right)$

We get:

$= 3 \frac{d}{\mathrm{dt}} \left(\cos \left(\frac{\pi}{2} t\right)\right)$

$= 3 \frac{d}{\mathrm{dt}} \left(\cos \left(\frac{\pi t}{2}\right)\right)$

Let $u = \frac{\pi t}{2} , \therefore \frac{\mathrm{du}}{\mathrm{dt}} = \frac{\pi}{2}$.

Then, $y = \cos u , \therefore \frac{\mathrm{dy}}{\mathrm{du}} = - \sin u$.

Combine to get:

$= - \sin u \cdot \frac{\pi}{2}$

$= - \sin \left(\frac{\pi t}{2}\right) \cdot \frac{\pi}{2}$

$= - \frac{\pi}{2} \sin \left(\frac{\pi t}{2}\right)$

So,
$\frac{d}{\mathrm{dt}} \left(3 \cos \left(\frac{\pi}{2} t\right)\right) = \frac{- 3 \pi}{2} \sin \left(\frac{\pi t}{2}\right)$

And now, we get:

$p ' \left(t\right) = 2 - \left[\frac{- 3 \pi}{2} \sin \left(\frac{\pi t}{2}\right)\right] + 0$

$= \frac{3 \pi}{2} \sin \left(\frac{\pi t}{2}\right) + 2$

So, at $t = 3$, we get:

$= \frac{3 \pi}{2} \sin \left(\frac{3 \pi}{2}\right) + 2$

$= \frac{- 3 \pi}{2} + 2$