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

Apr 26, 2017

The acceleration at $t = 2$ is $a \left(t\right) = \left(- \frac{1}{4} , 4\right)$, whose absolute value is $\frac{\sqrt{257}}{4}$ and the direction is ${93.576}^{\circ}$ w.r.t. $x$-axis.

#### Explanation:

Acceleration is the first derivative of velocity, so if we differentiate the velocity function, we get a function for the acceleration:

$\frac{\mathrm{dv}}{\mathrm{dt}} = \left(- \frac{1}{t} ^ 2 , 2 t\right)$

Inserting $t = 2$ gives us the acceleration at $t = 2 s$, namely

$a \left(t\right) = \left(- \frac{1}{4} , 4\right)$, whose absolute value is $\sqrt{{\left(- \frac{1}{4}\right)}^{2} + {4}^{2}} = \sqrt{\frac{257}{16}} = \frac{\sqrt{257}}{4}$

The direction is given by $\alpha = {\tan}^{- 1} \left(\frac{4}{- \frac{1}{4}}\right) = {\tan}^{- 1} \left(- 16\right) = {93.576}^{\circ}$ w.r.t. $x$-axis.