An object's two dimensional velocity is given by v(t) = ( cost , -t^3 +4t ). What is the object's rate and direction of acceleration at t=1 ?

Apr 13, 2018

Answer:

The rate of acceleration is $= 1.31 m {s}^{-} 1$ in the direction $= {130}^{\circ}$ anticlockwise from the x-axis

Explanation:

The acceleration is the derivative of the velocity

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

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

When $t = 1$

$a \left(1\right) = \left(- \sin 1 , - 3 + 4\right) = < - 0.84 , 1 >$

The rate of acceleration of the object is

$| | a \left(1\right) | | = | | < - 0.84 , 1 > | | = \sqrt{{\left(- 0.84\right)}^{2} + {1}^{2}} = \sqrt{1.71} = 1.31 m {s}^{-} 2$

The direction is

$\theta = \arctan \left(- \frac{1}{0.84}\right) = \arctan \left(- 1.19\right) = {130}^{\circ}$ anticlockwise from the x-axis