An object's two dimensional velocity is given by v(t) = ( t-2t^3 , 4t^2-t). What is the object's rate and direction of acceleration at t=1 ?

Dec 18, 2017

The rate of acceleration is $= 8.6 m {s}^{-} 2$ in the direction ${144.5}^{\circ}$ anticlockwise from the x-axis

Explanation:

The acceleration is the derivative of the velocity

$a \left(t\right) = \frac{\mathrm{dv}}{\mathrm{dt}}$

Here,

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

$a \left(t\right) = v ' \left(t\right) = \left(1 - 6 {t}^{2} , 8 t - 1\right)$

Therefore,

When $t = 1$

$a \left(1\right) = \left(1 - 6 , 8 - 1\right) = \left(- 5 , 7\right)$

The rate of acceleration is

$| | a \left(1\right) | | = \sqrt{{\left(- 5\right)}^{2} + {\left(7\right)}^{2}} = \sqrt{25 + 49} = \sqrt{74} = 8.6 m {s}^{-} 2$

The direction of acceleration is

$\theta = \arctan \left(- \frac{5}{7}\right) = {144.5}^{\circ}$ anticlockwise from the x-axis