Find the average value of the function $f(t)=(t-5)^{2}$ on $[0,12]$ The average value of the function $f(t)=(t-5)^{2}$ on $[0,12]$ is (Simplify your answer.)

Step 1 ：The average value of a function \(f(t)\) on the interval \([a, b]\) is given by the formula: \[\frac{1}{b-a}\int_{a}^{b}f(t)dt\]

Step 2 ：In this case, \(f(t) = (t-5)^2\), \(a = 0\), and \(b = 12\). So we need to compute the integral of \((t-5)^2\) from \(0\) to \(12\) and then divide by \(12 - 0 = 12\).

Step 3 ：The average value of the function is calculated to be 13.

Step 4 ：Final Answer: The average value of the function \(f(t)=(t-5)^{2}\) on \([0,12]\) is \(\boxed{13}\).