It would be a shame for a Calculus student to get behind the wheel of a car and not even realize that their favorite subject is on full display right there on the dashboard. The speedometer gives velocity as a function of time while the odometer gives displacement (i.e. change in position) as a function of time. As long as velocity[math]\ge[/math]0 over the entire trip, "displacement" may also referred to as "distance traveled."[br][br]Okay, technically a car's speedometer gives a reading of speed [math]\left|v\left(t\right)\right|[/math], not velocity [math]v\left(t\right)[/math]. Check "allow neg v(t)" to let this virtual speedometer lower into negative values. This activity is not particularly attentive to the distinction between speed and velocity at times, but the difference is worth a deeper conversation in a Calculus class.[br][br]In the interactive dashboard below, press "start," drag the [color=#ff0000]red point[/color] on the speedometer needle throughout the car trip, and press "stop" when finished with your trip. Examine how displacement may be viewed as the "area" bounded by the velocity curve and the t-axis. If you allow the velocity to become negative, then equal quantities of "negative area" and "positive area" will offset each other. As class activities, consider:[br][list][*]Have one person narrate a story while another person drags the speedometer needle accordingly (e.g. "You drive through a residential neighborhood, you see a stop sign, you enter the highway,...").[/*][*]See who can come up with the best estimate of the distance traveled by approximating the area between the velocity curve and the t-axis. Check the "odometer" box and compare estimates.[/*][/list][br]Use "guess average" to predict what might be considered the [i]average velocity[/i] over the time interval. Drag the [color=#ff00ff]pink point[/color] up/down to make your guess. You may alternatively select the [color=#ff00ff]pink point[/color] and press up/down arrows on a keyboard. Press Ctrl at the same time as the arrow key to increase the increment.[br][br]Check the "actual average" box to show what the true average velocity is over the course of the trip. Consider leaving the "actual average" displayed during a car trip. Discuss how the instantaneous velocity (current speedometer reading) influences the average velocity at any given time.[br][br]In older devices, the construction may run poorly inside a web browser. In such cases, consider downloading the GeoGebra file and opening in the standalone GeoGebra app.
The first paragraph mentions that "displacement" is the same as "distance traveled" when velocity [math]\ge[/math] 0 over an entire trip. How does "displacement" differ from "distance traveled" when velocity [math]\le[/math] 0 for at least part of the trip? If you don't know for sure, please make your best guess.
"Displacement" allows motion in the positive direction to be offset my motion in the negative direction. "Distance traveled" increases with all motion regardless of direction. [br][br]On a velocity vs. time graph, "displacement" allows "negative area" beneath the t-axis (and above a negative velocity curve) to cancel out "positive area" above the t-axis (and beneath a positive velocity curve). On that same graph, all "area" bounded by the velocity curve and the t-axis contributes positively to "distance traveled" regardless of whether that area is above or below the t-axis.
The text above mentions that displacement may be viewed as the "area" bounded by the velocity curve and the t-axis. Restate this point using the vocabulary of Calculus. Make a similar statement about distance traveled using the vocabulary of Calculus. On paper you should be able to write corresponding equations using Calculus symbols. (Unfortunately you are not allowed to enter all the necessary symbols here.)
Displacement is equal to the definite integral of [u]velocity[/u] over the interval of time in question.[br]Distance traveled is equal to the definite integral of the [u]absolute value of velocity[/u] (i.e. speed) over the interval of time in question.[br][math]Displacement=\int_a^bv\left(t\right)dt[/math] and [math]Distance\ Traveled=\int_a^b|v\left(t\right)|dt[/math] for the interval [math]a\le t\le b[/math].
When using the [color=#ff00ff]pink point[/color] to approximate average velocity, explain in your own words how you estimated where the horizontal average velocity curve should be positioned.
As hinted by the "actual average" slider labels, we may think of the area under the (instantaneous) velocity curve as a block of ice, and the average velocity would be where the water line would settle if all that ice melted.[br]Yes chemistry nerds, we're ignoring the fact that the density of water changes as it freezes, but let's not ruin a worthy metaphor with pedantry.[br][br]Of course, if the the odometer box is checked, one may simply calculate average velocity by dividing the odometer reading (in km) by time elapsed (in hours). But don't let that keep you from developing an intuitive understanding from the graph alone.
At the point in a Calculus course where you might encounter this activity, you are probably already familiar with the Mean Value Theorem (MVT) for Derivatives. Something about the slope of a tangent line equaling the slope of a secant line if certain conditions are met, right? It yields this equation:[br][math]F'\left(c\right)=\frac{F\left(b\right)-F\left(a\right)}{b-a}[/math][br]Have you been introduced yet to the Mean Value Theorem (MVT) for Integrals? If so, what does it say? How is it related to the MVT for Derivatives?
The MVT for Integrals states that if [math]f\left(x\right)[/math] is continuous on the closed interval [math]\left[a,b\right][/math], then there exists at least one number [math]c[/math] in the closed interval [math]\left[a,b\right][/math] such that [math]f\left(c\right)=\frac{\int_a^bf\left(x\right)dx}{b-a}=average\ value[/math]. Or using the ice/water metaphor, the contour of the ice [math]f\left(x\right)[/math] intersects the water line (average value) in at least one location [math]x=c[/math] on the interval. Accordingly, notice in the interactive dashboard that the velocity curve always intersects the "average value" line at least one time.[br][br]To connect the two versions of MVT, we recognize that by convention, [math]F'\left(c\right)[/math] is often written as [math]f\left(c\right)[/math]. Accordingly, the Fundamental Theorem of Calculus (FTC) states that if [math]F[/math] is an antiderivative of a function [math]f[/math] that is continuous on the closed interval [math]\left[a,b\right][/math], then [math]F\left(b\right)-F\left(a\right)=\int_a^bf\left(x\right)dx[/math]. Starting with the MVT for derivatives, substituting in [math]f\left(c\right)[/math] for [math]F'\left(c\right)[/math] and [math]\int_a^bf\left(x\right)dx[/math] for [math]F\left(b\right)-F\left(a\right)[/math] yields the formula given by the MVT for integrals.
As technical point: It's worth noting that in this model, cusps ("sharp points") occur on the velocity curve every time velocity changes. Why would it be conceptually problematic in the "real world" for a velocity curve to have cusps?
In the real world, velocity is continuous and differentiable for all values of time t. Curves are not differentiable at cusps. Since acceleration is the derivative of velocity ([math]\left(a\left(t\right)=v'\left(t\right)\right)[/math], the acceleration curve would become undefined and experience a jump discontinuity for any t value where the velocity curve has a cusp. Acceleration is never undefined for objects in motion in "the real world."