Suppose that a ball is rolling down a ramp. The distance traveled by the ball is given by the function \( s(t)=2 t^{2} \), where \( t \) is the time, in seconds, after the ball is released, and \( s(t) \) is measured in feet. Find the ball's average velocity in each of the following time intervals. a. \( t_{1}=2 \) to \( \mathrm{t}_{2}=3 \) \( \frac{\Delta \mathrm{~s}}{\Delta t}=10 \mathrm{ft} / \mathrm{sec} \) b. \( \mathrm{t}_{1}=2 \) to \( \mathrm{t}_{2}=2.5 \) \( \frac{\Delta s}{\Delta t}=\square \mathrm{ft} / \mathrm{sec} \)

To find the average velocity of the ball over the given time intervals, we will use the formula for average velocity \( \frac{\Delta s}{\Delta t} \), where \( \Delta s = s(t_2) - s(t_1) \) and \( \Delta t = t_2 - t_1 \). a. For the interval \( t_1=2 \) to \( t_2=3 \): First, calculate \( s(2) \) and \( s(3) \): \[ s(2) = 2(2^2) = 8 \text{ ft} \] \[ s(3) = 2(3^2) = 18 \text{ ft} \] Now, compute \( \Delta s \) and \( \Delta t \): \[ \Delta s = 18 - 8 = 10 \text{ ft} \] \[ \Delta t = 3 - 2 = 1 \text{ sec} \] Thus, the average velocity is: \[ \frac{\Delta s}{\Delta t} = \frac{10 \text{ ft}}{1 \text{ sec}} = 10 \text{ ft/sec} \] b. For the interval \( t_1=2 \) to \( t_2=2.5 \): Calculate \( s(2) \) again (which we already found as 8 ft) and now compute \( s(2.5) \): \[ s(2.5) = 2(2.5^2) = 2(6.25) = 12.5 \text{ ft} \] Now, compute \( \Delta s \) and \( \Delta t \): \[ \Delta s = 12.5 - 8 = 4.5 \text{ ft} \] \[ \Delta t = 2.5 - 2 = 0.5 \text{ sec} \] Thus, the average velocity is: \[ \frac{\Delta s}{\Delta t} = \frac{4.5 \text{ ft}}{0.5 \text{ sec}} = 9 \text{ ft/sec} \] So, your final answers are: a. \( \frac{\Delta s}{\Delta t} = 10 \) ft/sec b. \( \frac{\Delta s}{\Delta t} = 9 \) ft/sec