Suppose that a ball is rolling down a ramp. The distance traveled ty 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 bali's average velocity in each of the following time intervals. a. \( 11-4 \mathrm{w} \cdot 2-0 \) \[ \frac{\Delta s}{\Delta t}=10 \mathrm{ft} / \mathrm{sec} \] b. \( t_{1}=2 \) to \( t_{2}=25 \) \[ \frac{\Delta s}{\Delta t}=9 \mathrm{tt} / \mathrm{sec} \] c. \( \mathrm{t}_{1}=2 \) to \( \mathrm{t}_{2}=201 \) \( \frac{\Delta s}{\Delta t}= \) \( \square \) \( \mathrm{H} / \mathrm{sec} \) (Type an exact answer, using integers or decinals)

To find the average velocity over the specified time intervals, we use the formula for average velocity \(\frac{\Delta s}{\Delta t} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}\). For the interval from \(t_1 = 2\) to \(t_2 = 201\): 1. Calculate \(s(2)\): \[ s(2) = 2(2^2) = 2(4) = 8 \text{ ft} \] 2. Calculate \(s(201)\): \[ s(201) = 2(201^2) = 2(40401) = 80802 \text{ ft} \] 3. Now find \(\Delta s\) and \(\Delta t\): \[ \Delta s = s(201) - s(2) = 80802 - 8 = 80894 \text{ ft} \] \[ \Delta t = 201 - 2 = 199 \text{ sec} \] 4. Now calculate the average velocity: \[ \frac{\Delta s}{\Delta t} = \frac{80894}{199} \approx 406.56 \text{ ft/sec} \] So, the average velocity from \(t_1=2\) to \(t_2=201\) is approximately \(406.56 \text{ ft/sec}\).