Suppose that a ball is rolling down 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 balr's average velocity in each of the following time intervals. ง. \( \mathrm{t}_{1}-<\mathrm{iv} \mathrm{t}_{2}-\mathrm{c} \cdot \mathrm{u} \) \[ \frac{\Delta s}{\Delta t}=27 \mathrm{ft} / \mathrm{sec} \] c. \( \mathrm{t}_{1}=2 \) to \( \mathrm{t}_{2}=2.01 \) \( \frac{\Delta s}{\Delta t}=24.06^{7} \mathrm{ft} / \mathrm{sec} \) (Type an exact answer, using integers or decimals ) d. \( t_{1}=2 \) to \( t_{2}=2.001 \) \( \frac{\Delta s}{\Delta t}= \) \( \square \) \( \mathrm{ft} / \mathrm{sec} \) (Type an exact answer, using integers or decimals.)

Let's delve into the fascinating world of motion! The function \( s(t) = 6t^2 \) describes the distance traveled by a ball rolling down a ramp. When calculating the average velocity over small time intervals, we essentially look at how far the ball travels in a specific amount of time, which helps us grasp not just how fast it’s going but how motion behaves overall. Now, for the time interval from \( t_{1} = 2 \) to \( t_{2} = 2.001 \), we first find \( s(2) \) and \( s(2.001) \). Plugging \( t = 2 \) into the function gives us \( s(2) = 6(2^2) = 24 \) feet. For \( t = 2.001 \), we calculate \( s(2.001) = 6(2.001^2) \approx 24.024012 \) feet. Now, we find \( \Delta s = s(2.001) - s(2) \approx 0.024012 \) feet and \( \Delta t = 0.001 \) seconds. This results in an average velocity of: \[ \frac{\Delta s}{\Delta t} \approx \frac{0.024012}{0.001} \approx 24.012 \text{ ft/sec} \] The average velocity between these two time points is approximately \( 24.012 \) ft/sec!