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

We're asked to find the magnitude (rate) and direction of the object's acceleration at `t = 1` `"s"`, given the object's velocity equation.

We need to find the object's acceleration as a function of time, doing so via differentiation of the velocity equation:

`veca = d/(dt) vecv`

`vecv = (t-2)hati + (2cos(pi/2t)-3t)hatj` (given, component form)

`v_x = t-2`

`a_x = 1` (derivative of `t-2`)

`v_y = 2cos(pi/2t)-3t`

`a_y = -pisin(pi/2t)-3` (derivative)

Now that we have the components of the acceleration as a function of `t`, let's plug in `1` for `t` to find the acceleration components at `t = 1` `"s"`:

`a_x = color(red)(1` `color(red)("m/s"^2`

`a_y = -pisin(pi/2(1))-3 = color(green)(-6.14` `color(green)("m/s"^2`

The magnitude of the acceleration at this time is

`a = sqrt((a_x)^2 + (a_y)^2) = sqrt((color(red)(1)color(white)(l)color(red)("m/s"^2))^2 + (color(green)(-6.14)color(white)(l)color(green)("m/s"^2))^2)`

`= color(blue)(6.22` `color(blue)("m/s"^2`

The direction of the acceleration is

`theta = arctan((a_y)/(a_x)) = arctan((color(green)(-6.14)color(white)(l)color(green)("m/s"^2))/(color(red)(1)color(white)(l)color(red)("m/s"^2))) = -1.41"rad" = color(purple)(-80.8^"o"`