# 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=4 ?

May 30, 2016

$a \left(4\right) = 33.84 \text{ } \frac{m}{s} ^ 2$

#### Explanation:

${a}_{x} \left(t\right) = \frac{d}{d t} \left(3 {t}^{2} - 5 t\right) = 6 t - 5$

${a}_{x} \left(4\right) = 6 \cdot 4 - 5 = 24 - 5 = 19$

${a}_{y} \left(t\right) = \frac{d}{d t} \left(- {t}^{3} + 4 t\right) = - 3 {t}^{2} + 4$

${a}_{y} \left(4\right) = - 2 \cdot {4}^{2} + 4 = - 32 + 4 = - 28$

a(4)=sqrt(a_x(4)^2+a_y(4)^2

$a \left(4\right) = \sqrt{{19}^{2} + {\left(- 28\right)}^{2}}$

$a \left(4\right) = \sqrt{361 + 784}$

$a \left(4\right) = \sqrt{1145}$

$a \left(4\right) = 33.84 \text{ } \frac{m}{s} ^ 2$