SPOILER ALERT. Actually, this isn’t much of a spoiler because this scene appears in the movie trailer, and the movie is a couple years old. Here’s the deal. Quicksilver is with Magneto and they need to get past a bunch of guards in a hallway. To do this, Quicksilver runs down the hall super fast while pushing Magneto. Oh, he also holds the back of Magneto’s head so that he won’t get whiplash.
Now for the analysis. What kind of acceleration would this require? Would Magneto have to worry about whiplash?
Estimations and AssumptionsFor any analysis of a movie clip, you must start with some assumptions and estimations. Since the clip isn’t in “real time,” I really do have to make some guesses.
- The hallway is 50 feet (15.2 meters) long. You can’t see it all, so this is an estimation but I think it’s on the low end.
- The guards are thrown into the air, or perhaps lifted. I’m not sure. Either way, it’s clear that Quicksilver makes it through the hallway before they hit the ground.
- The guards reach a height of about 1.5 meters. I will use this to get an estimate of the time for Quicksilver.
Finding the TimeSuppose these guards were launched into the air to a height of 1.5 meters. How long would this take? Assuming there is only the gravitational force acting on the guards, it is a simple projectile motion problem (actually, it’s just like hang time in basketball). I could of course just look up the “hang time formula,” but then I’d have to change the name of this blog from Dot Physics to Dot Just-Look-It-Up.
Let’s start with half of the motion—the part where the guard moves up (assuming that’s what he does) to his highest point. The velocity at the highest point is zero and the time this motion takes is the same time it takes him to fall back down (so finding the total time would just be twice this value). I know the acceleration is –g (-9.8 m/s2) so that I can use the definition of acceleration (in just one dimension):
Now that I have an expression for the “launch” velocity, I can use the two definitions of average velocity.
With two expressions for the initial velocity, I can set them equal to each other and eliminate v1.
Remember, this is the time for half of the “jump”. The total time the guard is in the air would be twice this value. This value is important because during this time Quicksilver has to start from rest, run down the hall, then stop. Actually, it would probably be less than this time since the guards probably didn’t actually jump and Quicksilver probably got to the end of the hallway before they fell.
Using a height of 1.5 meters means that the maximum run time would be 1.1 seconds (MAX).
Accelerating Down the HallwayQuicksilver has to start from rest, run and increase in speed and then slow down and stop. There are many ways he could do this, but I am going to assume he increases speed with a constant acceleration and then slow down with the same acceleration (except negative). In this case, he would increase speed half the distance and then decrease speed the other half. The motion can be broken into two equal times.
Now instead of a problem with two different accelerations, I have a simpler problem with just constant acceleration. In this problem, Quicksilver starts from rest and runs half the length of the hall in half the time. I will again start with the definition of acceleration (in one dimension).
I am still using Δt from above. Remember that in both cases this is half the total time, so it’s OK. Let me also call the total length of the hallway as s so that half the hallway will be s/2. As before, I can now use the definition for average velocity (this only works if the acceleration is constant).
Now with two expressions for the final velocity, I can set them equal to each other and solve for the acceleration.
OK, now for a comment. You are probably thinking, “Wouldn’t this be easier to just plug values into that one kinematic equation?” Well, that might take less time but it skips all the fun steps. The thing I like to point out is that you can do a bunch of cool stuff just using a few fundamental definitions for acceleration and average velocity.
If I use my values for s and Δt, I get an acceleration of 12.56 m/s2 (just 1.28 G’s). That’s not so bad, but that uses the maximum estimated time. What if Quicksilver wants to do it in half that time (which is more likely since the clip shows all the guards still in the air). With a time of 0.55 seconds, the acceleration is 50.2 m/s2 (5.1 G’s). OK, one more time. If he does it in just a fourth of the total time, the acceleration jumps up to 201 m/s2 (20.5 G’s). That’s still not too bad (it’s just a little bit bad).
But I really think the time is much shorter than that. You actually get a few frames in which you can see the blur of Quicksilver (with Magneto). It’s only 3 frames, but it’s difficult to determine how long of a time interval this corresponds to since it’s clearly in “slow motion mode.” If it wasn’t in slow motion, these three frames would be just 0.066 seconds for an acceleration of 3489 m/s2 (356 G’s). Now that’s a serious acceleration. Magneto wouldn’t get whiplash, he would be dead (assuming that beyond his magnetic super powers he’s mostly human).
Yes, I know there are still many problems with my estimations, in particular the length of the hallway and the time of run. But even in my “best case scenario” I think Magneto would die from the acceleration.
Modeling the Two Acceleration ProblemI said that we could break this running problem into two parts—a part where Quicksilver increases speed and a part where he slows down. I also said that the time for these two parts would be the same. Let’s make sure that’s true.
I can easily model the motion of an accelerating Quicksilver (both positive and negative acceleration) with a numerical calculation. Breaking the motion into small time intervals, I can calculate the position and velocity changes for each step. Putting all the steps together I will get a graph of position vs. time.
I’m not going into all the details, but you can see something very similar in this numerical solution to the xkcd velociraptor problem.
Now for the Quicksilver run—feel free to look at the code by clicking the “pencil” to switch to edit mode.
Notice that I cheated just a little bit. I ran the simulation until the position is 0.98 times the length of the hallway. If you use the entire length, Quicksilver stops before the end of the hallway and then the program runs for ever. You could fix this in a number of ways, but I wanted to do something simple.
The cool thing about the position plot is that it shows two parabolas. The first parabola is for constant and positive acceleration and the second is for constant negative acceleration. Here are some things you can try.
- What happens if you increase the value of the acceleration (increase the magnitude).
- Sketch a graph of velocity vs. time. Now check your answer with a plot of velocity vs. time.
- Come up with a different motion in which Quicksilver accelerates, moves at a constant speed and then slows down and stops. Plot both position vs. time and velocity vs. time.