Suppose that a ball is rolling dowri a ramp. The distance traveled by the ball is given by the function \( s(t)=6 t^{2} \), where \( t \) is the time, in seconds, after the ball is released, and \( s(t) \) is measured in feet. Find the balls average velocity in each of the following time intervals. a. \( t_{1}=2 \) to \( t_{2}=3 \) \( \frac{\Delta s}{\Delta t}=\square \mathrm{t} / \mathrm{sec} \)

To find the average velocity of the ball over the interval from \( t_1 = 2 \) seconds to \( t_2 = 3 \) seconds, we first need to calculate the distance traveled by the ball at these two times using the given function \( s(t) = 6t^2 \). 1. Calculate \( s(2) \): \[ s(2) = 6(2^2) = 6(4) = 24 \text{ feet} \] 2. Calculate \( s(3) \): \[ s(3) = 6(3^2) = 6(9) = 54 \text{ feet} \] 3. Now, find the change in distance (\( \Delta s \)) and the change in time (\( \Delta t \)): \[ \Delta s = s(3) - s(2) = 54 - 24 = 30 \text{ feet} \] \[ \Delta t = t_2 - t_1 = 3 - 2 = 1 \text{ second} \] 4. Finally, we calculate the average velocity: \[ \frac{\Delta s}{\Delta t} = \frac{30 \text{ feet}}{1 \text{ second}} = 30 \text{ ft/sec} \] Therefore, the average velocity of the ball from \( t = 2 \) to \( t = 3 \) seconds is \( \boxed{30} \text{ ft/sec} \).