Question

What is the instantaneous velocity of an object moving in accordance to `f(t)= (sin(2t-pi/2),cost/t )` at `t=(3pi)/8`?

Answer 1

`v((3pi)/8) = 1.77` `"LT"^-1`

`phi = -37.9^"o"`

Explanation:

We're asked to find the instantaneous velocity of an object, given its position equation.

In component equations, the position is

`x(t) = sin(2t-pi/2)`

`y(t) = (cost)/t`

The velocity can be found by differentiating the position equations:

`v_x(t) = d/(dt) [sin(2t-pi/2)] = 2sin(2t)`

`v_y(t) = d/(dt) [(cost)/t] = -(tsint + cost)/(t^2)`

To find the velocity components at `t = (3pi)/8`, we'll plug it in:

`v_x((3pi)/8) = 2sin(2((3pi)/8)) = 2sin((3pi)/4) = sqrt2` `"LT"^-1`

`v_y((3pi)/8) = -(((3pi)/8)sin((3pi)/8) + cos((3pi)/8))/(((3pi)/8)^2) = -1.06` `"LT"^-1`

The term "`"LT"^-1`" is there to remind us velocity is measured in units of distance divided by time, which in dimensional form is `"LT"^-1`. I wrote this because no units for either were given.

The magnitude of the instantaneous velocity is thus

`v((3pi)/8) = sqrt((sqrt2)^2 + (-1.06)^2) = color(red)(1.77` `color(red)("LT"^-1`

And the direction is

`phi = arctan((-1.06)/(sqrt2)) = color(blue)(-37.9^"o"`

with respect to the positive `x`-axis.