Question

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

Answer 1

`a = 4.12` `"LT"^-2`

`phi = -14.0^"o"`

Explanation:

We're asked to find the magnitude and direction of an object's acceleration at a certain time given its velocity equation.

In component form, the velocity is

`v_x(t) = t^2+2`

`v_y(t) = cos(pit)-t`

To find the acceleration as a function of time, we must differentiate the velocity equations:

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

`a_y(t) = d/(dt) [cos(pit)-t] = -pisin(pit) - 1`

At `t = 2` (ambiguous units here), we have

`a_x(2) = 2(2) = 4` `"LT"^-2`

`a_y(2) = -pisin(2pi) - 1 = -1` `"LT"^-2`

(The term `"LT"^-2` is the dimensional form the units for acceleration, `"length"` `xx` `"time"^-2`. I used it here simply because no units were given.)

The magnitude of the acceleration is

`a = sqrt((a_x)^2 + (a_y)^2) = sqrt(4^2 + (-1)^2) = color(red)(4.12` `color(red)("LT"^-2`

The direction is

`phi = arctan((a_y)/(a_x)) = arctan((-1)/4) = color(blue)(-14.0^"o"`

Always check the arctangent calculation to make sure your answer is not `180^"o"` off!