# An object's two dimensional velocity is given by v(t) = ( cost , -t^3 +4t ). What is the object's rate and direction of acceleration at t=7 ?

Jun 17, 2017

a = 143"m"/("s"^2)

$\theta = {270}^{\text{o}}$

#### Explanation:

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

We need to find the object's acceleration as a function of time, which we can do by differentiating the velocity equation.

The derivative of ${t}^{n}$ is $n {t}^{n - 1}$, and the derivative of $\cos \left(a t\right)$ is $- a \sin \left(a t\right)$, so let's use these to find the acceleration vs. time equation:

Velocity equation (component form):

$v \left(t\right) = \left(\cos t\right) \hat{i} + \left(- {t}^{3} + 4 t\right) \hat{j}$

• d/(dt) (cost) = color(red)(-sint

• d/(dt) (-t^3) = color(darkorange)(-3t^2

• d/(dt) 4t = color(green)(4

Therefore,

$a \left(t\right) = \left(\textcolor{red}{- \sin t}\right) \hat{i} + \left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{- 3 {t}^{2}} + \textcolor{g r e e n}{4}\right) \hat{j}$

Now, let's plug in $7$ for $t$ to find the acceleration at that time:

$a \left(7\right) = \left(- \sin 7\right) \hat{i} + \left(- 3 {\left(7\right)}^{2} + 4\right) \hat{j}$

$= - 0.657 \hat{i} - 143 \hat{j}$

The magnitude of the acceleration at this time is

$a = \sqrt{{\left({a}_{x}\right)}^{2} + {\left({a}_{y}\right)}^{2}} = \sqrt{{\left(- 0.657\right)}^{2} + {\left(- 143\right)}^{2}}$

$= \textcolor{b l u e}{143 {\text{m"/("s}}^{2}}$

The direction of the acceleration is given by

$\theta = \arctan \left(\frac{{a}_{y}}{{a}_{x}}\right) = \arctan \left(\frac{- 143}{- 0.657}\right) = 4.71 \text{rad" = color(purple)(270^"o}$

Therefore, at $t = 7$ $\text{s}$, the acceleration is $\textcolor{b l u e}{143 {\text{m"/("s}}^{2}}$, at a direction of color(purple)(270^"o" anticlockwise from the positive $x$-axis.