An object's velocity is given by v(t) = (t^2 +t +1 , sin4t ). What is the object's rate and direction of acceleration at t=6 ?

Jul 17, 2017

$a = 11.1$ ${\text{LT}}^{-} 2$

$\theta = {8.77}^{\text{o}}$

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

${v}_{x} \left(t\right) = {t}^{2} + t + 1$

${v}_{y} \left(t\right) = \sin \left(4 t\right)$

Finding the derivatives:

${a}_{x} \left(t\right) = \frac{d}{\mathrm{dx}} \left[{t}^{2} + t + 1\right] = 2 t + 1$

${a}_{y} \left(t\right) = \frac{d}{\mathrm{dx}} \left[\sin \left(4 t\right)\right] = 4 \cos \left(4 t\right)$ (I'll assume this is in radians)

Plugging in $t = 6$ (no units), we have

${a}_{x} = 2 \left(6\right) - 1 = 11$ ${\text{LT}}^{-} 2$

${a}_{y} = 4 \cos \left(4 \left(6\right)\right) = 1.70$ ${\text{LT}}^{-} 2$

(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(11^2 + (1.70)^2) = color(red)(11.1 color(red)("LT"^-2

And the direction is

theta = arctan((a_y)/(a_x)) = arctan((1.70)/11) = color(blue)(8.77^"o"

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