A point moves along the curve that has the parametric equations:

x=9t^3- 〖6t〗^2 and y= 〖9t〗^3+27
x and y being in yards and t being in minutes. Find the magnitude and direction of the acceleration when t= 5/2.
Round answers to 3 significant figures.

To find the magnitude and direction of acceleration at a given point on a parametric curve, we need to find the second derivatives of x and y with respect to time (t) and evaluate them at the specific value of t.

First, let's find the first derivatives of x and y:
dx/dt = 27t^2 - 12t
dy/dt = 27t^2

Next, let's find the second derivatives by differentiating the first derivatives with respect to t:
d²x/dt² = 54t - 12
d²y/dt² = 54t

Now, substitute t = 5/2 into the second derivatives to find the acceleration at t = 5/2:
d²x/dt² = 54(5/2) - 12 = 135 - 12 = 123
d²y/dt² = 54(5/2) = 135

The magnitude of acceleration (a) can be found using the formula:
|a| = √(d²x/dt²)² + (d²y/dt²)²

Substituting the values we found earlier:
|a| = √(123)² + (135)²
|a| = √(15129) + (18225)
|a| = √(33354)
|a| ≈ 182.602 (rounded to 3 significant figures)

To find the direction of acceleration, we need to find the angle (θ) it makes with the positive x-axis. The formula to find the angle is:
θ = tan^(-1)(d²y/dt² / d²x/dt²)

Substituting the values we found earlier:
θ = tan^(-1)(135 / 123)
θ ≈ 48.895 degrees (rounded to 3 significant figures)

Therefore, at t = 5/2, the magnitude of acceleration is approximately 182.602 yards/minute² and the direction of acceleration is approximately 48.895 degrees from the positive x-axis.