Velocity of a Ball Thrown into the Air The position function of an object moving along a straight line is given by \( s=f(t) \). The avera the time interval \( [a, b] \) is the average rate of change of \( f \) over \( [a, b] \); its (instantaneous) velocity at \( t=\boldsymbol{a} \) is the rate of change of \( f \) at \( a \). A ball is thrown straight up with an initial velocity of \( 128 \mathrm{ft} / \mathrm{sec} \), so that its height (in feet) after \( t \mathrm{sec} \) is given by \( s=f(t)=128 t-16 t^{2} \). (a) What is the average velocity of the ball over the following time intervals? \( [3,4] \) \( [3,3.5] \) \( [3,3.1] \) (b) What is the instantaneous velocity at time \( t=3 \) ? 32 (c) What is the instantaneous velocity at time \( t=7 \) ? (To determine this, find the AROC over smaller and smaller intervals until the ROC is cle Is the ball rising or falling at this time?

Let's analyze the motion of the ball using the given position function: \[ s(t) = 128t - 16t^2 \] This function represents the height \( s \) (in feet) of the ball at any time \( t \) (in seconds). We'll address each part of the problem step by step. --- ### **Part (a): Average Velocity Over Specified Intervals** **Average velocity** over a time interval \([a, b]\) is calculated using the formula: \[ \text{Average Velocity} = \frac{s(b) - s(a)}{b - a} \] Let's compute the average velocity for each interval: 1. **Interval \([3, 4]\):** \[ s(4) = 128 \times 4 - 16 \times 4^2 = 512 - 256 = 256 \text{ ft} \] \[ s(3) = 128 \times 3 - 16 \times 3^2 = 384 - 144 = 240 \text{ ft} \] \[ \text{Average Velocity} = \frac{256 - 240}{4 - 3} = \frac{16}{1} = 16 \text{ ft/s} \] 2. **Interval \([3, 3.5]\):** \[ s(3.5) = 128 \times 3.5 - 16 \times (3.5)^2 = 448 - 196 = 252 \text{ ft} \] \[ s(3) = 240 \text{ ft} \quad (\text{from previous calculation}) \] \[ \text{Average Velocity} = \frac{252 - 240}{3.5 - 3} = \frac{12}{0.5} = 24 \text{ ft/s} \] 3. **Interval \([3, 3.1]\):** \[ s(3.1) = 128 \times 3.1 - 16 \times (3.1)^2 = 396.8 - 153.76 = 243.04 \text{ ft} \] \[ s(3) = 240 \text{ ft} \quad (\text{from previous calculation}) \] \[ \text{Average Velocity} = \frac{243.04 - 240}{3.1 - 3} = \frac{3.04}{0.1} = 30.4 \text{ ft/s} \] **Summary of Average Velocities:** - \([3, 4]\): **16 ft/s** - \([3, 3.5]\): **24 ft/s** - \([3, 3.1]\): **30.4 ft/s** --- ### **Part (b): Instantaneous Velocity at \( t = 3 \) seconds** The **instantaneous velocity** is the derivative of the position function with respect to time \( t \): \[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(128t - 16t^2) = 128 - 32t \] Evaluating at \( t = 3 \): \[ v(3) = 128 - 32 \times 3 = 128 - 96 = 32 \text{ ft/s} \] **Instantaneous Velocity at \( t = 3 \) seconds:** **32 ft/s** --- ### **Part (c): Instantaneous Velocity at \( t = 7 \) seconds** Using the derivative of the position function: \[ v(t) = 128 - 32t \] Evaluating at \( t = 7 \): \[ v(7) = 128 - 32 \times 7 = 128 - 224 = -96 \text{ ft/s} \] **Interpretation:** - **Sign of Velocity:** The negative sign indicates the direction of the velocity is downward. - **Ball's Motion at \( t = 7 \) seconds:** Since the velocity is negative, the ball is **falling** at this time. **Instantaneous Velocity at \( t = 7 \) seconds:** **-96 ft/s** (The negative value signifies that the ball is falling.) --- ### **Conclusion** - **Average velocities** decrease as the time intervals narrow around \( t = 3 \) seconds, approaching the instantaneous velocity. - The **instantaneous velocity** provides the exact rate at which the ball's position is changing at a specific moment. - At \( t = 7 \) seconds, the ball is confirmed to be **falling** due to its negative velocity.