Calculating Train Speeds and Time with MST3K: A Fun Experiment

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Homework Help Overview

The discussion revolves around calculating the time it takes for a train, specifically one mile long, to pass a point, considering both its constant speed and the time required for acceleration from a stop. The subject area includes concepts from kinematics and basic physics related to motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the time taken for a train to pass based on its length and speed, while also considering acceleration. Some participants question the validity of the equations used and the assumptions made regarding acceleration and speed.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the calculations. There is recognition of potential errors in the original approach, and some guidance has been offered regarding unit checks and the formulation of equations.

Contextual Notes

The original poster expresses uncertainty about their understanding of the concepts involved, indicating they are not formally studying the subject. This context may influence the depth of discussion and the nature of the questions raised.

TomFoolery
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I was watching Mystery Science Theater 3k and they said something about a train a mile long. I decided to make up a problem to see if this was a rational length for a train by determining how much time it might take a train like that to pass.

Train Length (L) = 1 mile (5280 feet)
Velocity (V) = 50 mph (264,000 feet per hour)
Time (T) = Time in Seconds

I've found, through various experimenting, that:

T = V/L

As in Time = 50mph/1 mile = 50 trains per hour or 1/50*60 = 1.2 minutes per train.

So, then I figured, what if it were stopped and required an acceleration (A) period? I know that I must then account for a) the extra time to accelerate and b) the length that passes during acceleration.

I think that: A = the average acceleration velocity (v/2) * acceleration time.

If it took 5 seconds to accelerate to full speed, then A = 5 sec * v/2 right?
or A = 5 sec * (73.3333 feet per second / 2) or 183.3333 feet

Anyhow, assuming that I've been correct up until now. Then the final calculation should be:

T = (73.333333 feet per sec / 5096.6666 feet) + 5 seconds

I got the 5096.6666... by Total Length - Length passed during acceleration period. and the other numbers were achieved using regular conversions of hours to minutes, miles to feet, etc.

Finally, T = 5.014388489 seconds (I hope)



** I am new to this but enjoy doing stuff like this. I am not studying it in school so I'm not sure if it's calculus, or trigonometry, or what. Until now (the discovery of this forum), I've been unable to find out if the answers that I get are correct or not.
 
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Actually, I just realized that this answer is completely messed up. Where did I go wrong? I'm I way off the equation that I'm using in the first place? There is no way that it can go faster from zero than from a continuous full speed. Plus, it's way too fast. Now I'm really lost...
 
TomFoolery said:
Anyhow, assuming that I've been correct up until now. Then the final calculation should be:

T = (73.333333 feet per sec / 5096.6666 feet) + 5 seconds

Check your units. feet per sec/feet = 1/sec.

I think you want 5096.666 feet/73.33333 feet per sec instead.
 
Yeah, you're right, thanks. So, it turns out that from a dead stop, it takes only about 2.5 seconds longer than from at full speed (given my rate of acceleration). Which makes sense since it averages half the velocity for 5 seconds.

So, just let me know that this is correct.

time = [ velocity / length (less length passed during acceleration period) ] + acceleration period time

or

T = V/(L-La) + 5 sec

and La = V/2 * 5 seconds
 

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