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

Dec 7, 2016

$1 + \pi$

Explanation:

Velocity is defined as
$v \left(t\right) \equiv \frac{\mathrm{dp} \left(t\right)}{\mathrm{dt}}$

Therefore, in order to find speed we need to differentiate function $p \left(t\right)$ with respect to time. Please remember that $v \mathmr{and} p$ are vector quantities and speed is a scalar.

$\frac{\mathrm{dp} \left(t\right)}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(t - t \sin \left(\frac{\pi}{3} t\right)\right)$
$\implies \frac{\mathrm{dp} \left(t\right)}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} t - \frac{d}{\mathrm{dt}} \left(t \sin \left(\frac{\pi}{3} t\right)\right)$

For the second term will need to use the product rule and chain rule as well. We get

$v \left(t\right) = 1 - \left[t \times \frac{d}{\mathrm{dt}} \sin \left(\frac{\pi}{3} t\right) + \sin \left(\frac{\pi}{3} t\right) \times \frac{d}{\mathrm{dt}} t\right]$
$\implies v \left(t\right) = 1 - \left[t \times \cos \left(\frac{\pi}{3} t\right) \times \frac{\pi}{3} + \sin \left(\frac{\pi}{3} t\right)\right]$
$\implies v \left(t\right) = 1 - \left[\frac{\pi}{3} t \cos \left(\frac{\pi}{3} t\right) + \sin \left(\frac{\pi}{3} t\right)\right]$

Now speed at $t = 3$ is $v \left(3\right)$, therefore we have

$v \left(3\right) = 1 - \left[\frac{\pi}{3} \times 3 \cos \left(\frac{\pi}{3} \times 3\right) + \sin \left(\frac{\pi}{3} \times 3\right)\right]$
$\implies v \left(3\right) = 1 - \left[\pi \cos \left(\pi\right) + \sin \left(\pi\right)\right]$

Inserting values of $\sin \mathmr{and} \cos$ functions
$v \left(3\right) = 1 - \left[\pi \times \left(- 1\right) + 0\right]$
$v \left(3\right) = 1 + \pi$