# 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=12 ?

Jul 31, 2017

$a = 462$ ${\text{LT}}^{-} 2$

$\theta = - {20.8}^{\text{o}}$

#### Explanation:

We're asked to find the magnitude and direction of the acceleration of an object at a certain time, given the velocity equations as functions of time.

To do this, we need to find the acceleration of the object as a suction of $t$, by differentiating the velocity component equations.

We're given:

${v}_{x} \left(t\right) = {t}^{3}$

${v}_{y} \left(t\right) = t - {t}^{2} \sin \left(\frac{\pi}{8}\right) t$

So

a_x(t) = d/(dt) [t^3] = ul(3t^2

a_y(t) = d/(dt) [t-t^2sin(pi/8)t] = ul(1-3t^2sin(pi/8)

Now, we plug in $t = 12$:

a_x = 3(12)^2 = ul(432

a_y = 1-3(12)^2sin(pi/8) = ul(-164

The magnitude of the acceleration is thus

$a = \sqrt{{\left({a}_{x}\right)}^{2} + {\left({a}_{y}\right)}^{2}} = \sqrt{{\left(432\right)}^{2} + {\left(- 164\right)}^{2}}$

= color(red)(ul(462color(white)(l)"LT"^-2

(the ${\text{LT}}^{-} 2$ term is the dimensional form of the units for acceleration; I used it here since no units were given)

The direction is

theta = arctan((a_y)/(a_x)) = arctan((-164)/(432)) = color(blue)(ul(-20.8^"o"

Always be sure to check your arctangent calculation, as it could be ${180}^{\text{o}}$ off!