# Six Minute Solutions #35 Transpo



## Blu1913 (Oct 2, 2006)

If so, when they find the time for the running speed, why arent they multiplying the 2498 X 2, as the equations gives: t = 2s/(v0-v)???

WTH?!!?

For those of you who dont have the problem here you go:

Train

Stations are 1 mi apart.

Top speed is 80 mph

Acceleration is 5.5 ft/sec^2

Decel is 4.5 ft/sec^2

What is the average running speed of the train?

Show you work plz. Let see if the problem is wrond and Im getting POed over nothing...


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## Road Guy (Oct 2, 2006)

(This is really prob #34?)

I did it a little differently from the solution

1. Determine time to go from 0 to 80 mph

Vf=Vi+a*t = 21.33 sec

2. Determine the time to stop

(Same Formula)

t= 26.07 sec

3. Determine distance traveled during acceleration &amp; deceleration

3a. d= (Vf^2 - Vi^2) / (2*a) = (117.336^2 - 0 ) / 2(5.5) = 1252 ft

* I converted 80 mph into 117.336 ft/s

3b. d= 0-117.336^2 ) / 2(4.5) = 1529.75 ft

4. Determine how much distance is traveled at full speed 80 mph

5280 - 1259+1529.75 = 2498.63 ft

5. Determine the time it takes to go 2498.63 ft

2498.63/117.336 = 26.07 seconds

Add the times together:

21.29+21.33+26.07 = 68.69

1mi/(68.69sec)/(60*60) = 52.41 mph

the first time I did it I included the 20 secon "dwell time" which incorrectly gives you 20 seconds, to me that is bullshit, but I guess that what we deal with on the exam.

Somewhere I thought I posted a good velocity/acceleration/time formula crib sheet, if I havent yet I will do it Tuesday


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