The best part of
the Hyperloop is that it’s the coolest real-world physics problem. What’s the
Hyperloop? Who knows for sure, it is some type of transport that could get from
LA to NY in just 45 minutes. Here is my favorite infographic
on what we do and don’t know about the Hyperloop.
The Hyperloop seems to be based on some other ideas about
Evacuated Tube Transportation. Essentially, you get a big pipe and put some
people in a pod-like device that goes in the tube. Pump most or some of the air
out and then shoot the pod down the tube.
For the sake of this physics problem, let’s make some
assumptions (or guesses if you like).
- Reduced air pressure in the tube.
- Little or no friction on the rails due to magnetic levitation.
- A travel time of 45 minutes from LA to NY (a distance of 3.95 x 106 m).
- Maximum acceleration of 1 g (9.8 m/s2).
Now for the physics.
Practice With Graphs
Let me start with a graph. This shows the horizontal
acceleration of the pod as a function of time.

Here the pod accelerates at 9.8 m/s2 for 2.6
minutes and then travels at a constant speed. At the end of its trip, the pod
spends the last 2.6 minutes with an acceleration of -9.8 m/s2.
Question 1: Sketch a graph of velocity vs.
time and another graph for position vs. time for this same trip. Be very
careful. The common problem is to draw a velocity graph that LOOKS like the
acceleration graph. However, remember the definition of acceleration and average
velocity:

This says that the acceleration will be the slope of the
velocity-time graph and the velocity will be the slope of the position-time
graph. But in this case we are going backwards. It isn’t too difficult to draw
the velocity graph though. The graph should have a positive slope of 9.8
m/s2 for the first time interval, then it should have a zero slope
for the next part. Of course, the velocity graph should be continuous – that
would make the middle velocity constant (zero slope) and non-zero (so that it
matches up with the previous interval).
What about the position graph? The first part of the velocity
graph says that the slope of this position graph has to increase. This means
that it would be a parabola. Or, if you like you can use the following kinematic
equation.

Actually, that t should really be a Δt. But
let me go ahead and show the two graphs for velocity and position that go along
with that same acceleration graph above. Actually, let me change the problem. If
the acceleration part is just around 5 minutes total out of 45 minutes, the
curved part of the position graph is rather difficult to see. Instead, this pod
accelerates for 7 minutes at the beginning and 7 minutes at the end.

For the position graph, many people would want to have the
final position back at zero. Notice in this position graph, the SLOPE is zero at
the end, not the position.
By Rhett Allain
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