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Train Speeds and MST3K

  1. Dec 29, 2009 #1
    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.
  2. jcsd
  3. Dec 29, 2009 #2
    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....
  4. Dec 29, 2009 #3
    Check your units. feet per sec/feet = 1/sec.

    I think you want 5096.666 feet/73.33333 feet per sec instead.
  5. Dec 29, 2009 #4
    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


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

    and La = V/2 * 5 seconds