Question

An object's two dimensional velocity is given by `v(t) = ( t^3, t-t^2sin(pi/8)t)`. What is the object's rate and direction of acceleration at `t=12`?

Answer 1

`a = 462` `"LT"^-2`

`theta = -20.8^"o"`

Explanation:

We're asked to find the magnitude and direction of the acceleration of an object at a certain time, given the velocity equations as functions of time.

To do this, we need to find the acceleration of the object as a suction of `t`, by differentiating the velocity component equations.

We're given:

`v_x(t) = t^3`

`v_y(t) = t-t^2sin(pi/8)t`

So

`a_x(t) = d/(dt) [t^3] = ul(3t^2`

`a_y(t) = d/(dt) [t-t^2sin(pi/8)t] = ul(1-3t^2sin(pi/8)`

Now, we plug in `t = 12`:

`a_x = 3(12)^2 = ul(432`

`a_y = 1-3(12)^2sin(pi/8) = ul(-164`

The magnitude of the acceleration is thus

`a = sqrt((a_x)^2 + (a_y)^2) = sqrt((432)^2 + (-164)^2)`

`= color(red)(ul(462color(white)(l)"LT"^-2`

(the `"LT"^-2` term is the dimensional form of the units for acceleration; I used it here since no units were given)

The direction is

`theta = arctan((a_y)/(a_x)) = arctan((-164)/(432)) = color(blue)(ul(-20.8^"o"`

Always be sure to check your arctangent calculation, as it could be `180^"o"` off!