Question

An object's velocity is given by `v(t) = (t^2 +t +1 , sin4t )`. What is the object's rate and direction of acceleration at `t=6`?

Answer 1

`a = 11.1` `"LT"^-2`

`theta = 8.77^"o"`

Explanation:

We're asked to find the the magnitude (rate) and direction of an object's acceleration at a given time, given its velocity equation.

To do this, we need too differentiate the velocity component equations, which are given as

`v_x(t) = t^2 + t + 1`

`v_y(t) = sin(4t)`

Finding the derivatives:

`a_x(t) = d/(dx) [t^2 + t + 1] = 2t + 1`

`a_y(t) = d/(dx)[sin(4t)] = 4cos(4t)` (I'll assume this is in radians)

Plugging in `t = 6` (no units), we have

`a_x = 2(6)-1 = 11` `"LT"^-2`

`a_y = 4cos(4(6)) = 1.70` `"LT"^-2`

(The `"LT"^-2` term is the dimensional form of the units for acceleration (`"distance"xx"time"^-2`). I used this term here since no units were given.)

The magnitude of the acceleration is thus

`a = sqrt((a_x)^2 + (a_y)^2) = sqrt(11^2 + (1.70)^2) = color(red)(11.1` `color(red)("LT"^-2`

And the direction is

`theta = arctan((a_y)/(a_x)) = arctan((1.70)/11) = color(blue)(8.77^"o"`

Always make sure your arctangent calculation is in the right direction; it could be `180^"o"` off!