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#1 Aug 1 2013 12:13pm
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The subway train offers access to all twelve intersections downtown. If every intersection is .75 miles apart and the train travels at 20 mph and spends 45 seconds at every stop to unload and onboard new customers, how long will it take to get from the first intersection to the last intersection?

5 minutes
22 minutes
33 minutes
47 minutes

how lnog does it take to travel 3/4 mil in 20 seconds all ineed to know lol im dumb

This post was edited by jppark21 on Aug 1 2013 12:17pm
#2 Aug 7 2013 02:24am
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I'm going to go into more detail than you probably need, to try to answer any questions you have in one go. I don't frequent this site anymore so I probably won't see this thread again.


The time will be composed of two parts - the travel time and the wait time.

Travel time is solved with the following commonplace equation: Distance = (Time)(Velocity)

The total distance is equal to the Distance from station 1 to station 2 + Distance from station 2 to station 3 + ... + Distance from station 11 to station 12, that is
.75+.75+...+.75 = 11(.75) = 8.25 miles of travel

Then we use the equation above, 8.25miles = (Time)(20miles per hour), thus Time = 8.25miles/20 miles per hour [*note: this will give us an answer in terms of hours, not minutes, so multiply by 60!]
8.25/20 = .4125 hours = 24.75min which is the total time spent traveling from station to station.

Now we need to add in the wait time. Which stations will it wait at? We don't need to count the first station, because that's where it's starting (we start the watch when the train starts moving). We also don't need to count the last station because we're only interested in how long it takes to get there (we stop the clock as soon as the train pulls into station 12). So we are going to count the wait time at the 10 stations in the middle.

10(45 sec) = 450 sec = 7.5min which is the total time spent waiting at stations before getting to station 12.

Add these together, 24.75min + 7.5min = 32.25min

I assume whoever wrote your test mistakenly added in the waiting time for the first or last station to yield an even 33 minutes, or perhaps they rounded up. But according to the question you typed, in the way it is worded, 32 minutes 15 seconds is the correct answer.
#3 Aug 11 2013 03:16pm
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Paying fg for this?
#4 Aug 11 2013 03:31pm
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Quote (flussp @ Aug 11 2013 04:16pm)
Paying fg for this?

the fuck?
#5 Aug 12 2013 10:25am
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33 mins
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