Question

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

Answer 1

The rate of acceleration (unknown units) is `57.78 \ (2dp)`

The direction is at a bearing `203^o`

Explanation:

The velocity function is:

`bbv(t) = ( t-2t^3 , 4t^2-t)`

Differentiating wrt `t` we get the acceleration function:

`bba(t) = ( 1-6t^2 , 8t-1)`

When `t=3` we get

`bba(t) = ( 1-6*9 , 8*3-1) = (-53,23)`

We can represent this by a vector:

`bba(3)= -53bbhati + 23bbhatj`

So we gain the rate using Pythagoras:

`|bba(3) |= |-53bbhati + 23bbhatj |`
`\ \ \ \ \ \ \ \ \ = sqrt(-53^2+23^2)`
`\ \ \ \ \ \ \ \ \ = sqrt(2809+529)`
`\ \ \ \ \ \ \ \ \ = sqrt(3338)`
`\ \ \ \ \ \ \ \ \ = 57.78 (2 \ dp)`

And for the direction using Trigonometry:

`tan alpha = 23/53 => alpha = 23^o` (nearest degree)

So the direction is at a bearing `203^o`