An object's two dimensional velocity is given by v(t) = ( sqrt(t^2-1)-2t , t^2-5). What is the object's rate and direction of acceleration at t=2 ?

May 6, 2017

$| \vec{a} | \approx 4.09 {\text{ms}}^{-} 2$
$\theta \approx 1.779 \text{rad or } {101.9}^{\circ}$

Explanation:

Here we have velocity along $x$ direction:

${v}_{x} \left(t\right) = \sqrt{{t}^{2} - 1} - 2 t$

$\therefore$ acceleration along $x$ direction:

${a}_{x} \left(t\right) = \frac{{\mathrm{dv}}_{x}}{\mathrm{dt}} = \frac{t}{\sqrt{{t}^{2} - 1}} - 2$

At $t = 2 s$:

a_x(2"s")=(2/ sqrt3-2)"ms"^-2

Similarly

${a}_{y} \left(t\right) = 2 t$

At $t = 2 s$:

a_y(2"s")=4"ms"^-2

The acceleration vector at $t = 2$, is $\vec{a} = 2 \left(\frac{\sqrt{3}}{3} - 1\right) \hat{i} + 4 \hat{j}$

The acceleration is the magnitude of this vector:

$| \vec{a} | = \sqrt{{\left(2 \left(\frac{\sqrt{3}}{3} - 1\right)\right)}^{2} + {4}^{2}}$

$| \vec{a} | \approx 4.09 {\text{ms}}^{-} 2$

The direction is:

$\theta = {\tan}^{-} 1 \left(\frac{4}{2 \left(\frac{\sqrt{3}}{3} - 1\right)}\right) + \pi$

Please notice that we have added $\pi$ because the vector points to the 2nd quadrant.

$\theta \approx 1.779 \text{rad or } {101.9}^{\circ}$