Question

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

Answer 1

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

`theta = 14.0^"o"`

Explanation:

We're asked to find the magnitude and direction of the acceleration of an object at `t = 1`, given the velocity component equations.

Acceleration is the first derivative of velocity, so we differentiate the velocity component equations:

`a_x(t) = d/(dt) [3t^2 - 2t] = ul(6t - 2`

`a_y(t) = d/(dt) [t] = ul(1`

Substituting in `t = 1`:

`a_x = 6(1) - 2 = 4`

`a_y = 1`

The magnitude of the acceleration is given by

`a = sqrt((a_x)^2 + (a_y)^2) = sqrt(4^2 + 1^2) = color(red)(ulbar(|stackrel(" ")(" "4.12color(white)(l)"LT"^-2" ")|)`

The `"LT"^-2` is the dimensional form of the units for acceleration. I used it here because no units were given.

The direction is given by

`theta = arctan((a_y)/(a_x)) = arctan(1/4) = color(red)(ulbar(|stackrel(" ")(" "14.0^"o"" ")|)`

measured anticlockwise from the positive `x`-axis (which is the normal measurement standard for planar angles).