# An object's two dimensional velocity is given by v(t) = ( t-2t^3 , 4t^2-3t). What is the object's rate and direction of acceleration at t=3 ?

Dec 21, 2016

The acceleration is the first derivative of the velocity, $a \left(t\right) = \left(1 - 6 {t}^{2} , 8 t - 3\right)$

which, evaluated at t=3 in standard form is

$a = 59.8 \frac{m}{s} ^ 2$ at a direction of 158.4°.

#### Explanation:

Taking the first derivative of the velocity, we get

$a \left(t\right) = \left(1 - 6 {t}^{2} , 8 t - 3\right)$

Evaluating this at $t = 3$, we get a(3) = (-53, 21)

The magnitude of this acceleration is found by Pythagorus' theorem because the two components above are perpendicular:

$a = \sqrt{{\left(- 53\right)}^{2} + {21}^{2}} = 59.8 \frac{m}{s} ^ 2$

The direction is found from theta = tan^(-1) (21/-53) = -21.6°

But this angle cannot be right, because it would describe an acceleration vector pointing into quadrant 4 - with positive $x$-component and negative $y$-component.

So, you must be careful in how you interpret this last result. If you did the ${\tan}^{- 1}$ operation on a calculator (instead of using a rectangular to polar coordinator conversion function), you may not have the correct angle. This has happened here. When we did the division $\left(\frac{21}{-} 53\right)$, the calculator evaluates this as -0.39622.., but tan^(-1) treats the negative result as though the numerator caused it to be negative, when in fact it was the denominator.

A diagram will sort out the problem. With a negative $x$-component and positive $y$-component, the direction must point into the second quadrant - we are off by 180°! The correct angle is 158.4°

Always include a diagram when solving two-dimensional vector problems. It will help you avoid errors in calculation.