Question

An object's two dimensional velocity is given by `v(t) = ( t^3, t-t^2sin(pi/8)t)`. What is the object's rate and direction of acceleration at `t=4`?

Answer 1

`a = 48.0` `"m/s"^2`

`theta = 1.19^"o"`

Explanation:

We're asked to find the object's acceleration (magnitude and direction) given its velocity equation.

We need to find the object's acceleration as a function of time, by differentiating the velocity component equations:

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

`a_y = d/(dt) [t - t^2 sin((pi/8)t)] = 1 - 2t^2cos((pi/8)t)`

Note: I changed the `y`-velocity equation SLIGHLTY by moving the final `t` inside the sine function, because (1) otherwise it could have said `t - t^3sin(pi/8)` and (2) this is how I've always seen these equations.

How that we know the acceleration components as functions of time, let's plug in `t = 4` `"s"` to find the components at that time:

`a_x (4) = 3(4)^2 = 48` `"m/s"^2`

`a_y = 1 - 2(4)^2cos((pi/8)(4)) = 1` `"m/s"^2` (cosine of `pi/2` is `0`)

The magnitude of the acceleration is thus

`a = sqrt((48color(white)(l)"m/s"^2)^2 + (1color(white)(l)"m/s"^2)^2) = color(red)(48.0` `color(red)("m/s"^2`

And the direction is

`theta = arctan((1cancel("m/s"^2))/(48cancel("m/s"^2))) = color(blue)(1.19^"o"`