# An object's two dimensional velocity is given by v(t) = ( t-2 , 2cos(pi/2t )- 3t ). What is the object's rate and direction of acceleration at t=1 ?

Jun 26, 2017

$a = 6.22$ ${\text{m/s}}^{2}$

$\theta = - {80.8}^{\text{o}}$

#### Explanation:

We're asked to find the magnitude (rate) and direction of the object's acceleration at $t = 1$ $\text{s}$, given the object's velocity equation.

We need to find the object's acceleration as a function of time, doing so via differentiation of the velocity equation:

$\vec{a} = \frac{d}{\mathrm{dt}} \vec{v}$

$\vec{v} = \left(t - 2\right) \hat{i} + \left(2 \cos \left(\frac{\pi}{2} t\right) - 3 t\right) \hat{j}$ (given, component form)

${v}_{x} = t - 2$

${a}_{x} = 1$ (derivative of $t - 2$)

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

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

Now that we have the components of the acceleration as a function of $t$, let's plug in $1$ for $t$ to find the acceleration components at $t = 1$ $\text{s}$:

a_x = color(red)(1 color(red)("m/s"^2

a_y = -pisin(pi/2(1))-3 = color(green)(-6.14 color(green)("m/s"^2

The magnitude of the acceleration at this time is

$a = \sqrt{{\left({a}_{x}\right)}^{2} + {\left({a}_{y}\right)}^{2}} = \sqrt{{\left(\textcolor{red}{1} \textcolor{w h i t e}{l} \textcolor{red}{{\text{m/s"^2))^2 + (color(green)(-6.14)color(white)(l)color(green)("m/s}}^{2}}\right)}^{2}}$

= color(blue)(6.22 color(blue)("m/s"^2

The direction of the acceleration is

theta = arctan((a_y)/(a_x)) = arctan((color(green)(-6.14)color(white)(l)color(green)("m/s"^2))/(color(red)(1)color(white)(l)color(red)("m/s"^2))) = -1.41"rad" = color(purple)(-80.8^"o"