# An object's two dimensional velocity is given by v(t) = ( t^2 - 2t , cospit - t ). What is the object's rate and direction of acceleration at t=5 ?

Jul 16, 2017

$a = 8.06$ ${\text{LT}}^{-} 2$

$\theta = - {7.13}^{\text{o}}$

#### Explanation:

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

To do this, we need too differentiate the velocity component equations, which are given as

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

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

Finding the derivatives:

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

${a}_{y} \left(t\right) = \frac{d}{\mathrm{dx}} \left[\cos \left(\pi t\right) - t\right] = - \pi \sin \left(\pi t\right) - 1$ (I'll assume this is in radians)

Plugging in $t = 5$ (no units), we have

${a}_{x} = 2 \left(5\right) - 2 = 8$ ${\text{LT}}^{-} 2$

${a}_{y} = - \pi \sin \left(\pi \left(5\right)\right) - 1 = - 1$ ${\text{LT}}^{-} 2$

(The ${\text{LT}}^{-} 2$ term is the dimensional form of the units for acceleration (${\text{distance"xx"time}}^{-} 2$). I used this term here since no units were given.)

The magnitude of the acceleration is thus

a = sqrt((a_x)^2 + (a_y)^2) = sqrt(8^2 + (-1)^2) = color(red)(8.06 color(red)("LT"^-2

And the direction is

theta = arctan((a_y)/(a_x)) = arctan((-1)/8) = color(blue)(-7.13^"o"

Always make sure your arctangent calculation is in the right direction; it could be ${180}^{\text{o}}$ off!