Suppose that a ball is rolling down a ramp The distanco traveled by tho ball is given by the function \( s(t)=G t^{2} \), 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 \( _{2}=3 \) \( \frac{\Delta s}{\Delta t}=30 \mathrm{ft} / \mathrm{sec} \) b \( t_{1}=2 \) to \( _{2}=25 \) \( \frac{\Delta s}{\Delta t}=\square \mathrm{tt} / \mathrm{sec} \)

To find the average velocity of the ball rolling down the ramp in the given time intervals, we can use the formula for average velocity: \[ \frac{\Delta s}{\Delta t} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \] For part (a) from \( t_1 = 2 \) to \( t_2 = 3 \): First, calculate \( s(2) \) and \( s(3) \): \[ s(2) = G \cdot (2^2) = 4G \] \[ s(3) = G \cdot (3^2) = 9G \] Now, substituting into the average velocity formula: \[ \frac{\Delta s}{\Delta t} = \frac{9G - 4G}{3 - 2} = \frac{5G}{1} = 5G \] Now, for part (b) from \( t_1 = 2 \) to \( t_2 = 25 \): Calculate \( s(2) \) as before: \[ s(2) = 4G \] Calculate \( s(25) \): \[ s(25) = G \cdot (25^2) = 625G \] Now, substituting this into the average velocity formula: \[ \frac{\Delta s}{\Delta t} = \frac{625G - 4G}{25 - 2} = \frac{621G}{23} \approx 27G \] So, the answers are: a) Average velocity = \( 5G \text{ ft/sec} \) b) Average velocity = \( \frac{621G}{23} \text{ ft/sec} \)