Activity 2.
Each of the following questions concern \(s(t) = 64 - 16(t-1)^2\text{,}\) the position function from the Preview Activity.
(a)
Compute the average velocity of the ball on the time interval \([1.5,2]\text{.}\) What is different between this value and the average velocity on the interval \([0,0.5]\text{?}\)
Hint.
Remember to use the formula for average velocity from above: \(AV_{[a,b]} = \frac{s(b)-s(a)}{b-a}\text{.}\) Think carefully about whether certain quantities are positive or negative.
Answer.
\(AV_{[1.5,2]} = -24\) ft/sec, which is negative.
Solution.
\(AV_{[1.5,2]} = \frac{s(2)-s(1.5)}{2-1.5} = -24\) ft/sec. We note that this average velocity is negative, and in fact is the opposite of the average velocity of 24 ft/sec on the interval \([0,0.5]\text{.}\)
(b)
Use appropriate computing technology to estimate the instantaneous velocity of the ball at \(t = 1.5\text{.}\) Likewise, estimate the instantaneous velocity of the ball at \(t = 2\text{.}\) Which value is greater?
Hint.
To estimate the instantaneous velocity at \(t = 1.5\text{,}\) consider average velocities on the intervals \([1.499,1.5]\) and \([1.5,1.501]\text{.}\)
Answer.
The instantaneous velocity at \(t = 1.5\) is approximately \(-16\) ft/sec; at \(t = 2\text{,}\) the instantaneous velocity is about \(-32\) ft/sec, and \(-16>-32\text{.}\)
Solution.
Since \(AV_{[1.499,1.5]} = -15.984\) and \(AV_{[1.5, 1.501]} = -16.016\text{,}\) it appears that the instantaneous velocity of the ball at \(t = 1.5\) is approximately \(-16\) ft/sec. Similar computations show that at \(t = 2\text{,}\) it appears the instantaneous velocity is about \(-32\) ft/sec. Note that \(-16>-32\text{,}\) so the instantaneous velocity at \(t = 1.5\) is greater because it is “less negative.” Asking which number is “greater” is different from asking which number is “more negative.”
(c)
How is the sign of the instantaneous velocity of the ball related to its behavior at a given point in time? That is, what does positive instantaneous velocity tell you the ball is doing? Negative instantaneous velocity?
Hint.
Think about whether the ball is rising or falling.
Answer.
When the ball is rising, its instantaneous velocity is positive, while when the ball is falling, its instantaneous velocity is negative.
Solution.
When the ball is rising, its instantaneous velocity is positive, while when the ball is falling, its instantaneous velocity is negative.
(d)
Without doing any computations, what do you expect to be the instantaneous velocity of the ball at \(t = 1\text{?}\) Why?
Hint.
What is the average velocity of the ball on small intervals that contain \(t = 0\text{?}\)
Answer.
Zero.
Solution.
Note that \((1,s(1))\) is the vertex of the parabola given by \(s(t)\text{.}\) At this point, the ball is neither rising nor falling. On intervals of the form \([a,1]\text{,}\) where \(a \lt 1\text{,}\) the average velocity of the ball is positive; on intervals of form \([1,b]\text{,}\) where \(b > 1\text{,}\) the average velocity is positive. Hence we expect the instantaneous velocity of the ball at the moment \(t = 1\) to be zero.
