Time Question - how long did the train block the crossing?

AI Thread Summary
The discussion centers on calculating how long a commuter train blocks a crossing, given its length and varying speeds. The train is 500 meters long, initially traveling at 78.6 km/h and slowing to 15.8 km/h as it approaches the crossing. The calculations involve converting speeds to meters per second and using average velocity to determine the blockage time, resulting in approximately 38.14 seconds. Participants suggest that while the calculations appear correct, taking the average velocity before converting units could save time. Additionally, they note that the actual blockage time may be underestimated if the length of the intersection is considered.
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Time Question -- how long did the train block the crossing?

Homework Statement


A commuter train 500 m long travels on a straight track at a speed of 78.6 km/h it slows down as it approaches a crossing. The last car passes the crossing at a speed of 15.8 km/h. With a constant acceleration, how long did the train block the crossing?


Homework Equations


avg v=(v0+vf)/2
t=change x/avg v


The Attempt at a Solution


First I converted 78.6km/h*1000m/km*1h/3600 secs=78600/3600=21.83 m/s
Secondly I converted 15.8km/h*1000m/km*3600h/3600secs=15800/3600=4.39 m/s

Using avg v=21.83+4.39/2=13.11 m/s

t=500m/13.11m/s
t=38.14s

I am pretty sure I did this correctly, but it took me a long time figuring it out (like what to do first and what equations to use). Any advice?
 
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It looks like you did it correctly. When you're first learning a new subject, you've got to figure that it's going to take you longer than when you gain more experience. Nice job.

Chet
 
Looks good to me too. My only advise would be to take the average velocity BEFORE you convert it into m/s. Then you just convert the average km/h to m/s. The way you did it is just fine. This would just save a LITTLE time, since you'd only have to do that conversion once.
 
I personally shy away from 'averages' since I've been burned more than once by them. In this case it's of course OK.
But you could approach the problem more formally:
Keeping our eye on the front of the train only,

v(t) = v(0) + ∫0t a(t')dt'
= v(0) +at since a is constant.
So v(T) = v(0) + aT where T = total time of intersection blockage. ... (1)
Then, s(t) = distance traveled by head of locomotive until the intersection is cleared
= ∫0t v(t')dt' + s(0) but s(0) = 0,
= ∫0t(v(0) + at')dt'
and s(T) = v(0)T + aT2/2 = 0.5 ... (2)

Now you can solve (1) and (2) for T and a.

BTW this ignores the length of the intersection. The intersection is actually blocked from the moment the locomotive first intersects with the near end and stays blocked until the last wagon passes the far end, so our answer for T is too short for any finite length intersection. :eek:
 
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