# An object's two dimensional velocity is given by v(t) = ( t^2, t-t^2sin(pi/3)t). What is the object's rate and direction of acceleration at t=4 ?

Dec 26, 2016

#### Explanation:

${v}_{x} \left(t\right) = {t}^{2}$

$\frac{{\mathrm{dv}}_{x} \left(t\right)}{\mathrm{dt}} = 2 t$

${v}_{y} \left(t\right) = t - {t}^{2} \sin \left(\frac{\pi t}{3}\right)$

$\frac{{\mathrm{dv}}_{y} \left(t\right)}{\mathrm{dt}} = 1 - 2 t \sin \left(\frac{\pi t}{3}\right) - \frac{\pi {t}^{2}}{3} \cos \left(\frac{\pi t}{3}\right)$

The magnitude of the acceleration is

$a = \sqrt{{\left(2 t\right)}^{2} + {\left(1 - 2 t \sin \left(\frac{\pi t}{3}\right) - \frac{\pi {t}^{2}}{3} \cos \left(\frac{\pi t}{3}\right)\right)}^{2}}$

I used WolframAlpha to evaluate the above at t = 4:

$a \approx 9.43$

The direction is along the unit vector:

$\left(\frac{1}{9.43}\right) \left(64 \hat{i} - 4.99 \hat{j}\right)$