An object's two dimensional velocity is given by v(t) = ( 3t^2 - 5t , -t^3 +4t ). What is the object's rate and direction of acceleration at t=9 ?

Feb 11, 2017

$\dot{v} \left(9\right) = \left(49 , - 239\right) m {s}^{- 2}$

in direction $\text{ } {\tan}^{- 1} \left(- \frac{239}{49}\right)$

Explanation:

acceleration is given by the first derivative of its velocity

so $v \left(t\right) = \left(3 {t}^{2} - 5 t , - {t}^{3} + 4 t\right)$

$a \left(t\right) = \frac{d}{\mathrm{dt}} \left(v \left(t\right)\right) = \dot{v} \left(t\right)$

$\dot{v} \left(t\right) = \left(6 t - 5 , - 3 {t}^{2} + 4\right)$

$\dot{v} \left(9\right) = \left(6 \times 9 - 5 , - 3 \times {9}^{2} + 4\right)$

$\dot{v} \left(9\right) = \left(49 , - 239\right) m {s}^{- 2}$

direction given by$\text{ } {\tan}^{- 1} \left(\frac{y}{x}\right)$

$\text{ } {\tan}^{- 1} \left(- \frac{239}{49}\right)$