The given problem provides the velocity function $v(t) = -10t + 15t^2$ and requires us to find $s(7)$, the position function $s(t)$, given that $s(3) = 72$.

Steps:

1. Find the position function $s(t)$: The velocity function $v(t)$ is the derivative of the position function $s(t)$. Thus, to find $s(t)$, we need to integrate $v(t)$.

   $v(t) = \frac{ds}{dt} = -10t + 15t^2$

   To find $s(t)$, integrate $v(t)$:

   $s(t) = \int (-10t + 15t^2) dt$

   $s(t) = -5t^2 + 5t^3 + C$

   where $C$ is the constant of integration.

2. Use the initial condition $s(3) = 72$ to find $C$: Plug $t = 3$ into $s(t)$ and set it equal to 72:

   $s(3) = -5(3^2) + 5(3^3) + C = 72$

   $-5(9) + 5(27) + C = 72$

   $-45 + 135 + C = 72$

   $90 + C = 72$

   $C = 72 - 90 = -18$

   Thus, the position function is:

   $s(t) = -5t^2 + 5t^3 - 18$

3. Find $s(7)$: Now, substitute $t = 7$ into $s(t)$:

   $s(7) = -5(7^2) + 5(7^3) - 18$

   $s(7) = -5(49) + 5(343) - 18$

   $s(7) = -245 + 1715 - 18$

   $s(7) = 1450 - 18 = 1432$

Thus, $s(7) = 1432$.

Would you like any further clarification or details on these steps?

Here are 5 related questions for further practice:

1. How do you find the velocity function if you're given the position function $s(t)$?
2. What is the physical meaning of the constant $C$ when integrating velocity to find position?
3. If $v(t)$ is a more complex function, does the process of finding $s(t)$ change?
4. What does it mean if the position function $s(t)$ is linear?
5. How can you find acceleration from the velocity function?

Tip: Always check for initial conditions when integrating, as they are crucial in determining the constants.