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

Jul 15, 2017

$a = 4.12$ ${\text{LT}}^{-} 2$

$\phi = - {14.0}^{\text{o}}$

#### Explanation:

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

In component form, the velocity is

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

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

To find the acceleration as a function of time, we must differentiate the velocity equations:

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

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

At $t = 2$ (ambiguous units here), we have

${a}_{x} \left(2\right) = 2 \left(2\right) = 4$ ${\text{LT}}^{-} 2$

${a}_{y} \left(2\right) = - \pi \sin \left(2 \pi\right) - 1 = - 1$ ${\text{LT}}^{-} 2$

(The term ${\text{LT}}^{-} 2$ is the dimensional form the units for acceleration, $\text{length}$ $\times$ ${\text{time}}^{-} 2$. I used it here simply because no units were given.)

The magnitude of the acceleration is

a = sqrt((a_x)^2 + (a_y)^2) = sqrt(4^2 + (-1)^2) = color(red)(4.12 color(red)("LT"^-2

The direction is

phi = arctan((a_y)/(a_x)) = arctan((-1)/4) = color(blue)(-14.0^"o"

Always check the arctangent calculation to make sure your answer is not ${180}^{\text{o}}$ off!