# An object's two dimensional velocity is given by v(t) = ( t^3, t-t^2sin(pi/8)t). What is the object's rate and direction of acceleration at t=4 ?

Jul 10, 2017

$a = 48.0$ ${\text{m/s}}^{2}$

$\theta = {1.19}^{\text{o}}$

#### Explanation:

We're asked to find the object's acceleration (magnitude and direction) given its velocity equation.

We need to find the object's acceleration as a function of time, by differentiating the velocity component equations:

${a}_{x} = \frac{d}{\mathrm{dt}} \left[{t}^{3}\right] = 3 {t}^{2}$

${a}_{y} = \frac{d}{\mathrm{dt}} \left[t - {t}^{2} \sin \left(\left(\frac{\pi}{8}\right) t\right)\right] = 1 - 2 {t}^{2} \cos \left(\left(\frac{\pi}{8}\right) t\right)$

Note: I changed the $y$-velocity equation SLIGHLTY by moving the final $t$ inside the sine function, because (1) otherwise it could have said $t - {t}^{3} \sin \left(\frac{\pi}{8}\right)$ and (2) this is how I've always seen these equations.

How that we know the acceleration components as functions of time, let's plug in $t = 4$ $\text{s}$ to find the components at that time:

${a}_{x} \left(4\right) = 3 {\left(4\right)}^{2} = 48$ ${\text{m/s}}^{2}$

${a}_{y} = 1 - 2 {\left(4\right)}^{2} \cos \left(\left(\frac{\pi}{8}\right) \left(4\right)\right) = 1$ ${\text{m/s}}^{2}$ (cosine of $\frac{\pi}{2}$ is $0$)

The magnitude of the acceleration is thus

a = sqrt((48color(white)(l)"m/s"^2)^2 + (1color(white)(l)"m/s"^2)^2) = color(red)(48.0 color(red)("m/s"^2

And the direction is

theta = arctan((1cancel("m/s"^2))/(48cancel("m/s"^2))) = color(blue)(1.19^"o"