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=7`?

Answer 1

The x & y components of the velocity vector are given as separate functions of t. You can take the derivative of each function, which will give the components of the acceleration vector.

`x'(t) = e^t - 4t`

`y'(t) = 1 - e^2` <- (remember that e^2 is a constant)

...evaluate these at time t = 7:

`x'(t) = e^7 - 28 = 1097 - 28 = 1069`
`y'(t) = 1 - 7.39 = -6.39`

The rate of acceleration is the magnitude of this vector.
`=sqrt(1069^2 + (-6.39)^2) = sqrt(1142761 + 40.8)`

`= 1069.02` (units unspecified).

The direction is given by the inverse trig functions, since we know

`sin(theta) = -6.39/1069`

`cos(theta) = 1069/1069`

(NOTE that I am assigning the zero degree direction to be along the positive x axis).
...the cosine of the angle is very, very close to 1, so therefore the direction of acceleration is very very close to the zero direction, about 0.536 degrees clockwise from it.