# An object's two dimensional velocity is given by v(t) = ( 2t^2 , t^3 - 4t ). What is the object's rate and direction of acceleration at t=7 ?

Jun 9, 2016

$| \vec{a} \left(7\right) | = 145.7 u n i t s$, rounded to one decimal place
Angle with $x$-axis $\theta = {78.9}^{\circ}$, rounded to one decimal place

#### Explanation:

Given two dimensional velocity
$v \left(t\right) = \left(2 {t}^{2} , {t}^{3} - 4 t\right)$
Let the two dimensions be $x \mathmr{and} y$. Writing explicitly
$\vec{v} \left(t\right) = 2 {t}^{2} \hat{i} + \left({t}^{3} - 4 t\right) \hat{j}$

To find acceleration $a \left(t\right)$, differentiating with time
$\vec{v} ' \left(t\right) = \vec{a} \left(t\right) = \frac{d}{\mathrm{dt}} \left(2 {t}^{2} \hat{i} + \left({t}^{3} - 4 t\right) \hat{j}\right)$
or $\vec{a} \left(t\right) = 4 t \hat{i} + \left(3 {t}^{2} - 4\right) \hat{j}$
We need to find acceleration at $t = 7$
or $\vec{a} \left(7\right) = 4 \times 7 \hat{i} + \left(3 \times {7}^{2} - 4\right) \hat{j}$
$\implies \vec{a} \left(7\right) = 28 \hat{i} + 143 \hat{j}$
Now $| \vec{a} \left(7\right) | = \sqrt{{28}^{2} + {143}^{2}}$
$| \vec{a} \left(7\right) | = 145.7 u n i t s$, rounded to one decimal place
If $\theta$ is the angle acceleration vector makes with $x$-axis
$\theta = {\tan}^{-} 1 \left(\frac{143}{28}\right)$
$\theta = {78.9}^{\circ}$, rounded to one decimal place