Question

An object's two dimensional velocity is given by `v(t) = ( cost , -t^3 +4t )`. What is the object's rate and direction of acceleration at `t=7`?

Answer 1

`a = 143"m"/("s"^2)`

`theta = 270^"o"`

Explanation:

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

We need to find the object's acceleration as a function of time, which we can do by differentiating the velocity equation.

The derivative of `t^n` is `nt^(n-1)`, and the derivative of `cos(at)` is `-asin(at)`, so let's use these to find the acceleration vs. time equation:

Velocity equation (component form):

`v(t) = (cost)hati + (-t^3 + 4t)hatj`

• `d/(dt) (cost) = color(red)(-sint`

• `d/(dt) (-t^3) = color(darkorange)(-3t^2`

• `d/(dt) 4t = color(green)(4`

Therefore,

`a(t) = (color(red)(-sint))hati + (color(darkorange)(-3t^2) + color(green)(4))hatj`

Now, let's plug in `7` for `t` to find the acceleration at that time:

`a(7) = (-sin7)hati + (-3(7)^2 + 4)hatj`

`= -0.657hati - 143hatj`

The magnitude of the acceleration at this time is

`a = sqrt((a_x)^2 + (a_y)^2) = sqrt((-0.657)^2 + (-143)^2)`

`= color(blue)(143"m"/("s"^2)`

The direction of the acceleration is given by

`theta = arctan((a_y)/(a_x)) = arctan((-143)/(-0.657)) = 4.71"rad" = color(purple)(270^"o"`

Therefore, at `t = 7` `"s"`, the acceleration is `color(blue)(143"m"/("s"^2)`, at a direction of `color(purple)(270^"o"` anticlockwise from the positive `x`-axis.