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)

We begin with the distance function   s(t) = 2t² To find the average velocity over an interval [t₁, t₂], we use   Average Velocity = (s(t₂) – s(t₁)) / (t₂ – t₁) Notice that because s(t) = 2t², we have   s(t₂) – s(t₁) = 2t₂² – 2t₁² = 2(t₂² – t₁²) and t₂² – t₁² factors as (t₂ – t₁)(t₂ + t₁). Thus,   Average Velocity = [2(t₂ – t₁)(t₂ + t₁)] / (t₂ – t₁) = 2(t₁ + t₂). Now let’s apply this formula to each interval. a. For the interval from t = 1.0 to t = 4.0:   Average Velocity = 2(1 + 4) = 2(5) = 10 ft/sec. b. For the interval from t = 2 to t = 2.5:   Average Velocity = 2(2 + 2.5) = 2(4.5) = 9 ft/sec. c. For the interval from t = 2 to t = 20:   Average Velocity = 2(2 + 20) = 2(22) = 44 ft/sec. Thus, the average velocities are:  a. 10 ft/sec  b. 9 ft/sec  c. 44 ft/sec.