An object's two dimensional velocity is given by v(t) = ( t-2t^3 , 4t^2-t). What is the object's rate and direction of acceleration at t=3 ?

Jan 29, 2018

The rate of acceleration (unknown units) is $57.78 \setminus \left(2 \mathrm{dp}\right)$

The direction is at a bearing ${203}^{o}$

Explanation:

The velocity function is:

$\boldsymbol{v} \left(t\right) = \left(t - 2 {t}^{3} , 4 {t}^{2} - t\right)$

Differentiating wrt $t$ we get the acceleration function:

$\boldsymbol{a} \left(t\right) = \left(1 - 6 {t}^{2} , 8 t - 1\right)$

When $t = 3$ we get

$\boldsymbol{a} \left(t\right) = \left(1 - 6 \cdot 9 , 8 \cdot 3 - 1\right) = \left(- 53 , 23\right)$

We can represent this by a vector:

$\boldsymbol{a} \left(3\right) = - 53 \boldsymbol{\hat{i}} + 23 \boldsymbol{\hat{j}}$

So we gain the rate using Pythagoras:

$| \boldsymbol{a} \left(3\right) | = | - 53 \boldsymbol{\hat{i}} + 23 \boldsymbol{\hat{j}} |$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{- {53}^{2} + {23}^{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{2809 + 529}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{3338}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 57.78 \left(2 \setminus \mathrm{dp}\right)$

And for the direction using Trigonometry:

$\tan \alpha = \frac{23}{53} \implies \alpha = {23}^{o}$ (nearest degree)

So the direction is at a bearing ${203}^{o}$