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 \( _{2}=25 \) \( \frac{\Delta s}{\Delta t}=\square \mathrm{tt} / \mathrm{sec} \)

We are given that the distance traveled by the ball is   s(t) = G · t² where t is the time (in seconds) and s(t) is in feet. The average velocity over an interval [t₁, t₂] is given by   v_avg = (s(t₂) – s(t₁)) / (t₂ – t₁) Step 1. Write down s(t₂) and s(t₁):   s(t₂) = G · (t₂)²   s(t₁) = G · (t₁)² Step 2. Find the difference:   s(t₂) – s(t₁) = G(t₂² – t₁²) Notice that t₂² – t₁² factors as (t₂ + t₁)(t₂ – t₁) so that   s(t₂) – s(t₁) = G(t₂ + t₁)(t₂ – t₁) Step 3. Then, the average velocity becomes:   v_avg = [G(t₂ + t₁)(t₂ – t₁)] / (t₂ – t₁) = G(t₂ + t₁) Now, we can calculate the answers for the given intervals. ───────────────────────────── (a) For t₁ = 2 sec and t₂ = 3 sec:   v_avg = G(3 + 2) = 5G The answer provided is 30 ft/sec, so we have:   5G = 30 ⇒ G = 30/5 = 6 ft/sec² ───────────────────────────── (b) For t₁ = 2 sec and t₂ = 25 sec:   v_avg = G(25 + 2) = G(27) Since we already found G = 6:   v_avg = 6 × 27 = 162 ft/sec ───────────────────────────── Thus, the average velocity for the time interval from 2 to 25 seconds is 162 ft/sec.