Problem 7 - Active Calculus

Question

A diver leaps from a 3 meter springboard. His feet leave the board at time \(t=0\), he reaches his maximum height of \(4.5 \mathrm{~m}\) at \(t=1.1\) seconds, and enters the water at \(t=2.45\). Once in the water, the diver coasts to the bottom of the pool (depth \(3.5 \mathrm{~m}\) ), touches bottom at \(t=7\), rests for one second, and then pushes off the bottom. From there he coasts to the surface, and takes his first breath at \(t=13\). a. Let \(s(t)\) denote the function that gives the height of the diver's feet (in meters) above the water at time \(t\). (Note that the "height" of the bottom of the pool is -3.5 meters.) Sketch a carefully labeled graph of \(s(t)\) on the provided axes in Figure 1.1.5. Include scale and units on the vertical axis. Be as detailed as possible. b. Based on your graph in (a), what is the average velocity of the diver between \(t=2.45\) and \(t=7 ?\) Is his average velocity the same on every time interval within [2.45,7]\(?\) c. Let the function \(v(t)\) represent the instantaneous vertical velocity of the diver at time \(t\) (i.e. the speed at which the height function \(s(t)\) is changing; note that velocity in the upward direction is positive, while the velocity of a falling object is negative). Based on your understanding of the diver's behavior, as well as your graph of the position function, sketch a carefully labeled graph of \(v(t)\) on the axes provided in Figure \(1.1 .6 .\) Include scale and units on the vertical axis. Write several sentences that explain how you constructed your graph, discussing when you expect \(v(t)\) to be zero, positive, negative, relatively large, and relatively small. d. Is there a connection between the two graphs that you can describe? What can you say about the velocity graph when the height function is increasing? decreasing? Make as many observations as you can.

Step-by-Step Solution

Answer

The average velocity between \(t=2.45\) and \(t=7\) is approximately \(-0.769 \text{m/s}\). The velocity graph reflects the rate of change of the height function. Positive \(v(t)\) implies increasing \(s(t)\) and vice versa.

Step 1 - Identify Key Time Points

List all the key moments given in the problem: - Feet leave the board at \(t=0\).- Maximum height of \(4.5 \, \text{m}\) at \(t=1.1\) seconds.- Enters water at \(t=2.45\) seconds.- Touches bottom of pool at \(t=7\) seconds.- Rests for \(1\) second.- Takes first breath at \(t=13\) seconds.

Step 2 - Sketching Height Function

Based on the key moments, plot the height function \(s(t)\). Start at \(3 \text{m}\) at \(t=0\), reach \(4.5 \text{m}\) at \(t=1.1\), then return to \(0 \text{m}\) at \(t=2.45\). At \(t=7\), reach \(-3.5 \text{m}\), remain there until \(t=8\), and then return to \(0 \text{m}\) at \(t=13\).

Step 3 - Calculate Average Velocity

Average velocity is calculated by dividing the change in height by the change in time within the interval \([2.45, 7]\). The height changes from \(0 \text{m}\) (at \(t=2.45\)) to \(-3.5 \text{m}\) (at \(t=7\)). The average velocity is \[ v_{\text{avg}} = \frac{s(7) - s(2.45)}{7 - 2.45} = \frac{-3.5 - 0}{7 - 2.45} = \frac{-3.5}{4.55} \approx -0.769 \text{m/s} \]

Step 4 - Sketching Velocity Function

We can sketch \(v(t)\) based on \(s(t)\). Remember that velocity is the derivative of position. From \(t=0\) to \(t=1.1\), \(v(t)\) is positive, peaking at \(t=0.5\). From \(t=1.1\) to \(t=2.45\), \(v(t)\) is negative, hitting a minimum as the diver enters the water. The velocity is zero while diver is at the bottom of the pool (\(t=7\) to \(t=8\)). For the ascent from \(t=8\) to \(t=13\), \(v(t)\) is positive.

Step 5 - Describe Connection Between Graphs

Observe the relationship between the graphs. When \(s(t)\) is increasing (positive slope), \(v(t)\) is positive. When \(s(t)\) is decreasing (negative slope), \(v(t)\) is negative. At the peak of \(s(t)\), \(v(t)\) is zero. Similarly, \(v(t)\) indicates the rate of change of \(s(t)\), and the sign of \(v(t)\) aligns with whether \(s(t)\) is rising or falling.

Key Concepts

average velocity

Average velocity helps us understand the speed and direction of an object over a given time interval. In our exercise, we look at the diver's motion between two key time points: from entering the water at \(t = 2.45\) seconds to touching the bottom of the pool at \(t = 7\) seconds. We find the diver's average velocity by taking the total change in height and dividing it by the total time taken. Using the given values, the average velocity is: \[ v_{\text{avg}} = \frac{s(7) - s(2.45)}{7 - 2.45} = \frac{-3.5 - 0}{7 - 2.45} = \frac{-3.5}{4.55} \approx -0.769 \text{m/s} \] This negative value tells us that the diver is moving downward during this interval. However, it's important to note that average velocity doesn't capture the variations in speed and direction at specific moments within this interval.

instantaneous velocity

Instantaneous velocity describes the rate at which an object's position changes at an exact moment in time. Unlike average velocity, it can show the diver's speed and direction changes at specific points during their dive. We use the concept of derivatives to find this value. For instance, when the diver's feet leave the board (\(t=0\)), the initial velocity is positive, indicating upward movement. As the diver descends after reaching the peak height at \(t=1.1\) seconds, the instantaneous velocity becomes negative. The function \(v(t)\) representing this velocity will be derived from the height function \(s(t)\). At any point where \(v(t) = 0\), it indicates that the diver momentarily stops before changing direction, such as at the peak height or when they touch the pool bottom.

derivative as rate of change

Derivatives play a crucial role in understanding motion, particularly when calculating velocity. The rate of change of the diver's position function \(s(t)\) is captured by its derivative, yielding the velocity function \(v(t)\). Mathematically, if \(s(t)\) represents the height of the diver at any time \(t\), then: \[ v(t) = \frac{ds(t)}{dt} \] This derivative tells us how quickly and in what direction the diver's position is changing at each moment. A positive \(v(t)\) means the diver is moving upward, while a negative \(v(t)\) indicates downward motion. Analyzing the rate of change helps us predict the diver's behavior at each point in the dive, such as when they reach the maximum height or hit the water's surface.

graphical analysis

Graphs help visualize the diver's motion along with their position and velocity. For instance, plotting \(s(t)\) on a graph allows us to see how the diver's height changes over time. The curve's slope at any point corresponds to the diver's instantaneous velocity. The steepness indicates how fast the diver is moving, and the curve's direction shows whether it's an ascent or descent. Similarly, the graph of \(v(t)\) depicts how this velocity changes. Areas where the velocity graph is above the time axis (positive) indicate upward motion, while areas below (negative) show downward motion. Zero points on this graph correspond to maxima, minima, or points where the direction changes, like when the diver reaches the peak of the jump or hits the water.

motion along a vertical line

Understanding motion along a vertical line helps us analyze the diver's behavior. The diver's movement is restricted to up and down without any horizontal displacement. As their initial motion is upward, reaching a peak, and then reversing downward to enter the water, the position function \(s(t)\) represents this vertical journey. At the peak, the velocity is zero as the diver transitions from going up to coming down. Once submerged, the diver continues downward to the pool's bottom, again hitting zero velocity as they pause. This motion is a good demonstration of classic freefall and subsequent rise after a push, with all dynamics occurring along a single vertical axis.