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

To find the average velocity of the ball during the time interval from \( t_1 = 2 \) to \( t_2 = 3 \), we first calculate the distance traveled at each time using the function \( s(t) = 2t^2 \). 1. For \( t_1 = 2 \): \[ s(2) = 2(2^2) = 2 \times 4 = 8 \text{ ft} \] 2. For \( t_2 = 3 \): \[ s(3) = 2(3^2) = 2 \times 9 = 18 \text{ ft} \] Now, the change in distance (\( \Delta s \)) and change in time (\( \Delta t \)) can be calculated as follows: - \( \Delta s = s(3) - s(2) = 18 - 8 = 10 \text{ ft} \) - \( \Delta t = t_2 - t_1 = 3 - 2 = 1 \text{ sec} \) Thus, the average velocity \( \frac{\Delta s}{\Delta t} \) is: \[ \frac{\Delta s}{\Delta t} = \frac{10 \text{ ft}}{1 \text{ sec}} = 10 \text{ ft/sec} \] So, the average velocity is \( 10 \text{ ft/sec} \).