An object's two dimensional velocity is given by v(t) = (t^2 -t +1 , t^3- 3t ). What is the object's rate and direction of acceleration at t=5 ?

May 11, 2016

$a \left(5\right) = 72.56$
$\text{angle=} {69.44}^{o}$

Explanation:

${a}_{x} \left(t\right) = \frac{d}{d t} \left({t}^{2} - t + 1\right) = 2 t - 1$

$\text{for t="5 " ; } {a}_{x} \left(5\right) = 2.5 - 1 = 10 - 1 = 9$

${a}_{x} \left(5\right) = 9$

${a}_{y} \left(t\right) = \frac{d}{d t} \left({t}^{3} - 3 t\right) = 3 {t}^{2} - 3$

$\text{for t=5 ;} {a}_{y} \left(5\right) = 3 \cdot {5}^{2} - 3 = 75 - 3$

${a}_{y} \left(5\right) = 72$

$a \left(5\right) = \sqrt{{72}^{2} + {9}^{2}} \text{ } a \left(5\right) = \sqrt{5265}$

$a \left(5\right) = 72.56$