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

Nov 18, 2016

$| \overline{a} | \approx 12.04 , \theta \approx 6.2 \text{ radians}$

Explanation:

$\frac{{\mathrm{dv}}_{x}}{\mathrm{dt}} = 2 t + 2$

$\frac{{\mathrm{dv}}_{y}}{\mathrm{dt}} = - 2 \pi \sin \left(\pi t\right) - 1$

Evaluate both at t = 5:

$\frac{{\mathrm{dv}}_{x}}{\mathrm{dt}} {|}_{t = 5} = 12$

$\frac{{\mathrm{dv}}_{y}}{\mathrm{dt}} {|}_{t = 5} = - 1$

The acceleration vector is:

$\overline{a} = 12 \hat{i} - \hat{j}$

The magnitude of the acceleration is:

$| \overline{a} | = \sqrt{{12}^{2} + {\left(- 1\right)}^{2}} = \sqrt{145} \approx 12.04$

The direction is:

$\theta = 2 \pi + {\tan}^{-} 1 \left(- \frac{1}{12}\right)$

Note: we add $2 \pi$, because the angle is in the 4th quadrant.

$\theta = 2 \pi + {\tan}^{-} 1 \left(- \frac{1}{12}\right)$

$\theta \approx 6.2 \text{ radians}$