Question

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

Answer 1

`a = 6.52` `"m/s"^2`

`phi = 259^"o"`

Explanation:

We're asked to find the object's acceleration at time `t = 1` `"s"` from a given velocity equation.

To do this, we need to find the object's acceleration components as a function of `t`, via differentiation of the velocity equations:

Finding the derivative of velocity component equations, we have

`v_x = e^t - 2t^2`

`v_y = t - te^2`

`a_x = d/(dt) [e^t - 2t^2] = e^t - 4t`

`a_y = d/(dt) [t - te^2] = 1 - e^2`

At time `t = 1` `"s"`, the acceleration components are thus

`a_x = e^((1)) - 4(1) = -1.28` `"m/s"^2`

`a_y = 1 - e^2 = -6.39` `"m/s"^2`

The magnitude (rate) of the acceleration is

`a = sqrt((-1.28color(white)(l)"m/s"^2)^2 + (-6.39color(white)(l)"m/s"^2)^2) = color(red)(6.52` `color(red)("m/s"^2`

The direction of the acceleration is

`phi = arctan((-6.39color(white)(l)"m/s"^2)/(-1.28color(white)(l)"m/s"^2)) = 78.7^"o"` `+ 180^"o" = color(blue)(259^"o"`

measured anticlockwise from the positive `x`-axis.

The `180^"o"` was added to fix the calculator angle error.