Question

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`?

Answer 1

The acceleration is the first derivative of the velocity, `a(t) = (1-6t^2, 8t - 3)`

which, evaluated at t=3 in standard form is

`a=59.8 m/s^2` at a direction of 158.4°.

Explanation:

Taking the first derivative of the velocity, we get

`a(t) = (1-6t^2, 8t - 3)`

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((-53)^2+21^2) = 59.8 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 `(21/-53)`, 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.