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

Mar 29, 2017

2nd quandrant (negative x and positive y)

Explanation:

When you put 3 (for t), you will get v(t) = (-5, 9). The distance (if speed is expressed as m/s and time is expressed as s) 30.89 meters.

Note that the units of time and velocity were not provided in the original question.

Mar 29, 2017

The rate of acceleration is $= 6.18 m {s}^{-} 2$ in the direction of $= 104$º

Explanation:

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

The acceleration is the derivative of the velocity

$a \left(t\right) = v ' \left(t\right) = \frac{\mathrm{dv}}{\mathrm{dt}}$

$= \left(\frac{1}{2} \cdot \frac{1}{\sqrt{t - 2}} - 2 , 2 t\right)$

Therefore,

$a \left(3\right) = \left(\frac{1}{2} - 2 , 6\right) = \left(- \frac{3}{2} , 6\right)$

The rate of acceleration is

$= \sqrt{{\left(- \frac{3}{2}\right)}^{2} + {6}^{2}}$

$= \sqrt{\frac{9}{4} + 36}$

$= \frac{\sqrt{153}}{2}$

$= 6.18 m {s}^{-} 2$

$\tan \theta = \frac{6}{- \frac{3}{2}} = - 4$

The angle is in the 2nd quadrant and is

$\theta = \arctan \left(- 4\right) = 104$º