# 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 ?

May 13, 2017

The rate of acceleration is $= 145.72 m {s}^{-} 1$ and the direction is $= 169.3$º anticlockwise from the x-axis.

#### Explanation:

The acceleration is the derivative of the velocity.

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

$a \left(t\right) = v ' \left(t\right) = \left(4 t , - 3 {t}^{2} + 4\right)$

Therefore,

$a \left(7\right) = \left(4 \cdot 7 , - 3 \cdot 49 + 4\right) = \left(28 , - 143\right)$

The rate of acceleration is

$| | a \left(4\right) | | = \sqrt{{28}^{2} + {\left(- 143\right)}^{2}}$

$= \sqrt{21233}$

$= 145.72 m {s}^{-} 2$

The direction is

$\theta = \arctan \left(- \frac{28}{143}\right)$

$\theta$ lies in the 2nd quadrant

theta=180-10.7=169.3º