# An object's two dimensional velocity is given by v(t) = ( 3t^2 - 2t , 1- 3t ). What is the object's rate and direction of acceleration at t=5 ?

Jun 24, 2017

The acceleration is $28.2$ $m {s}^{-} 2$ at $- {84}^{o}$: that is, at an angle in the third quadrant, downward and to the right.

#### Explanation:

The acceleration is the rate of change of the velocity, and is found as the first derivative of the velocity. We can treat the $x$ and $y$ directions independently.

${v}_{x} = 3 {t}^{2} - 2 t$

$\frac{{\mathrm{dv}}_{x}}{\mathrm{dt}} = 6 t - 2$

${v}_{y} = 1 - 3 t$

$\frac{{\mathrm{dv}}_{y}}{\mathrm{dt}} = - 3$

We can recombine these dimensions:

$a \left(t\right) = \left(6 t - 2 , - 3\right)$

Interestingly, the acceleration in the $y$ direction is constant, while the acceleration in the $x$ direction changes with time.

At $t = 5$ $s$, we substitute in 5, and get:

$a \left(t\right) = \left(6 \left(5\right) - 2 , - 3\right) = \left(28 , - 3\right)$

We can use Pythagoras' theorem to find the magnitude and the tangent to find the direction of the resultant acceleration vector.

$r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{28}^{2} + {\left(- 3\right)}^{2}} = \sqrt{793} = 28.2$ $m {s}^{-} 2$

$\tan \setminus \theta = \frac{28}{- 3} \to \setminus \theta = - {84}^{o}$