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Using the data from the first 20 seconds, we compute the velocity and acceleration of both the runaway ($X,V,A$) and rescue ($x,v,a$) hovercrafts:
| Time | Runaway Pos. ($X$) | Rescue Pos. ($x$) | Runaway Vel. ($V$) | Rescue Vel. ($v$) | Runaway Accel. ($A$) | Rescue Accel. ($a$) |
|---|---|---|---|---|---|---|
| 0 s | 2536.40 m | 10.47 m | ||||
| 53.64 m/s | 3.141 m/s | |||||
| 10 s | 3072.80 m | 41.88 m | 0 m/s/s | 0.2093 m/s/s | ||
| 53.64 m/s | 5.243 m/s | |||||
| 20 s | 3609.20 m | 94.22 m |
Tutor Questions
- Question: At what time(s) are the average velocities for the Rescue hovercraft equal to the observed velocity of the hovercraft? How do you know?
- Expected Answer: These velocities occurs at 5s and 15s. Because this craft moves under constant force motion, we know that the average velocity and the arithmetic average velocity are the same, and the instantaneous velocity occurs halfway between the two time points because it changes linearly.
Now, to make things as simple as possible, we assume that the rescue cart not only starts from rest, but also accelerates uniformly (constant force from the hovercraft fan) throughout the duration of its rescue mission. Accordingly, the positions in time for both crafts are given by $$X(t)=X_{0{\rm s}}+V t\qquad\mbox{and}\qquad x(t)=x_{0{\rm s}}+\frac{1}{2}at^{2}.$$ Since we want to find the time at which the carts are at the same location, we set the above equations equal and solve for the time: $$t_{\rm jump}=\frac{V\pm\sqrt{V^{2}+2a(X_{0{\rm s}}-x_{0{\rm s}})}}{a}\approx 555.979\,{\rm s}.$$
It then follows, plugging this time back into either position equation, that the location of interest is $$x_{\rm jump}\approx 32359\,{\rm m,}$$ quite a ways away.
Tutor Questions
- Question: What assumptions did you make about the motion of the hovercrafts?
- Expected Answer: That the runaway craft has a constant velocity, and the rescue craft starts from rest with a constant acceleration.
- Question: Are these velocities and accelerations calculated from the numbers given exact?
- Expected Answer: No, we would need to take measurements a lot closer together.
- Question: Is the predicted position of the rescue craft a good one?
- Expected Answer: Not really, basing the trajectory off the first 20 seconds of data is probably not the best – but it is all we have to work with.
- Questions: Can you draw a plot of position vs. time for both crafts? What are the important features of this graph?
- Expected Answer: The point where the two curves cross is when we should jump. One should be linear, the other quadratic.
- Questions: Can you draw a plot of velocity vs. time for both crafts? What are the important features of this graph?
- Expected Answer: The acceleration is the slope of each curve (constant in both cases).
Main Points
- Students should complete this entire problem on Tuesday as they need to start working Part B. Historically, all groups have completed this problem on Tuesday and been able to move onto Part B.
- Students should be able to recognize and justify the types of motion that the two hovercrafts are executing (constant velocity vs. constant force).
- Students should be able to set up that relate the position and time of each hovercraft and solve them as simultaneous equations for when they are in the same location.
- Students should be able to draw the motion of both hovercrafts (position/velocity vs time) and explain how these graphs represent the motion given their assumptions. It is very important that students discuss and sketch the motion of the two hovercrafts, so please ask these tutor questions even if you are pressed for time.
Common Difficulties
- Some groups struggle to extract information from the table to determine what kind of motion is being executed by each hovercraft.
- Solving these simultaneous equations can provide a bit of a challenge for some groups. You may remind them that WolframAlpha is useful tool to check (or do) that work.
- Students should be pushed to graph a fairly quantitatively correct graph of the motion of each hovercraft. Extract meaning from their equations to do this can prove challenging.