# Could someone help us with this Alg 2 problem?

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Dd has done all of the other ones in this section with little trouble, but this problem has her stumped. In addition, this is exactly the kind of problem I had trouble with in high school (otherwise Algebra was a piece of cake, so it would be nice if I could actually remember it now! IT would have also helped if I'd a. liked the teacher I had that semester & had gone to ask her for help, which I didn't do.) We have the basic answer, which I haven't told her, but have no idea how to get there.

Two railroad workers are together in a 0.6 -mile mountain tunnel. One walks east and the other west in order to get out of the tunnel before the Bad Creek Express comes through at 30 mph. Each man reaches his respective end of the tunnel in 3 min. If he man walking east reaches the east entrance just before the train enters, and the train passes the other man .12 miles beyond the west end of the tunnel, as what rate did each man walk? :confused::001_unsure::confused1::ack2:

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Fast enough to survive?

Sorry - I'm no help. :tongue_smilie:

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Ok, it's been a really long time and I'm too lazy to get up for a pencil and paper and actually try to figure it out, but try this.

-figure out how long it takes for the train to go (at 30mph) from the tunnel entrance to the second man - so .72 miles.

-that time is how long it took the second man to go .12 miles

- now you should be able to figure out the second mans speed, using the train travel time and the distance of .12 miles

-use the three minute amount with the second mans speed to figure out how much of the tunnel he walked

-subtract that distance from the .6 total tunnel distance to get the first mans distance covered

- figure out first mans speed using the 3 mins and his distance covered.

Let me know if it works! (or if it even made any sense):001_smile:

ETA your combination of emoticons made me laugh and laugh and laugh....

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Swamped, so can't write a long reply... but here's purplemath's explanation of the d=rt problems.

Using a chart REALLY helps with these.

The other thing you need to do is be sure to use the same units, so be sure to convert the 3 min into hours (3 min = 3/60 hrs).

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Dd has done all of the other ones in this section with little trouble, but this problem has her stumped. In addition, this is exactly the kind of problem I had trouble with in high school (otherwise Algebra was a piece of cake, so it would be nice if I could actually remember it now! IT would have also helped if I'd a. liked the teacher I had that semester & had gone to ask her for help, which I didn't do.) We have the basic answer, which I haven't told her, but have no idea how to get there.

Two railroad workers are together in a 0.6 -mile mountain tunnel. One walks east and the other west in order to get out of the tunnel before the Bad Creek Express comes through at 30 mph. Each man reaches his respective end of the tunnel in 3 min. If he man walking east reaches the east entrance just before the train enters, and the train passes the other man .12 miles beyond the west end of the tunnel, as what rate did each man walk? :confused::001_unsure::confused1::ack2:

Westward speed = 5 mph ( same as 5 miles per 60 min)

Eastward speed = 7 mph (same as 7 miles per 60 min)

I will try to post detail answer later if anyone interested. I have to go.

Best regards.

mpcTutor

http://www.mpclasses.com/ContactUS.htm

US Central Time:2:28 PM 9/27/2010

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When the train enters the tunnel, both workers are at their respective ends (but since they walked at different speeds, they did not start out in the middle of the tunnel)

First we need to find out what time it takes the train to meet the second worker at the west end:

Train travels distance (0.6+0.12) miles at 30mph

time= distance/speed = 0.72 miles / (30miles/60min)= 1.44 min

During these 1.44 minutes, the west worker has walked 0.12 miles. So his speed is 0.12 miles/1.44 minutes =0.08333 miles/min = 5mph.

If the west walker was walking at 5mph inside the tunnel, he had covered during those 3 minutes a distance of 0.25 miles (d=vt).

Thus, we know the east walker started 0.35 miles from the east end of the tunnel. It took him 3 minutes to walk, so his speed was 0.35miles/3minutes=0.1166miles/min= 7mph.

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Dd has done all of the other ones in this section with little trouble, but this problem has her stumped. In addition, this is exactly the kind of problem I had trouble with in high school (otherwise Algebra was a piece of cake, so it would be nice if I could actually remember it now! IT would have also helped if I'd a. liked the teacher I had that semester & had gone to ask her for help, which I didn't do.) We have the basic answer, which I haven't told her, but have no idea how to get there.

Two railroad workers are together in a 0.6 -mile mountain tunnel. One walks east and the other west in order to get out of the tunnel before the Bad Creek Express comes through at 30 mph. Each man reaches his respective end of the tunnel in 3 min. If he man walking east reaches the east entrance just before the train enters, and the train passes the other man .12 miles beyond the west end of the tunnel, as what rate did each man walk? :confused::001_unsure::confused1::ack2:

I didn't solve this using any alg 2, so maybe I am doing something wrong (I haven't slept in months.....so that is very likely!!)

I solved it simply. If the length of the tunnel is .6 mi, then you can calculate the time it takes the train to travel the length + the extra distance by using rt=d.

First, I converted miles/hr to miles/min b/c the man is walking in terms of mins. So, 30 mi/hr=.5 mi/min

The man walks .12 more miles, so they meet after the train travels .6+.12=.72 mi

rt=d

.5 mi/min t = .72

t= .72 min/.5= 1.44 min (that is now the total time the train travels from the beginning of the tunnel to meeting the man)

I assumed that the man was at his end of the tunnel when the train enter the tunnel. So, I used 1.44 mins as the elapsed time from him exiting the tunnel to the train meeting him.

rt=d

1.44 mins r = .12 mi (the distance the man walked)

.12 mi/1.44 mins = r

.08333333 mi/min = r

multiply that by 60 to convert to hrs = 5 mi/hr

I realized my error. Out of the tunnel, if they are in the middle, the each exit at 6 miles/hr. The 2nd man reduces his speed to 5 mi/hr after leaving the tunnel. (I think....forget it.....I quit!)

Edited by 8FillTheHeart
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However, I made the assumption that they were in the middle of the tunnel and both walked out of the tunnel in 3 mins at the same rate.

That assumption is incorrect. If they were in the middle of the tunnel, the man walking to the east would cover half the tunnel, 0.3 miles, at the 5mph you calculated it would take him not 3 minutes but 3.636 minutes.

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Ok.....I see what you did.....you went back in a substituted in the 2nd guy's rate to determine the rate of the first.

I just assumed the second guy was booking it to get out of the tunnel and then reduced his speed once he knew he wasn't train bait! LOL!!

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Ok.....I see what you did.....you went back in a substituted in the 2nd guy's rate to determine the rate of the first.

I just assumed the second guy was booking it to get out of the tunnel and then reduced his speed once he knew he wasn't train bait! LOL!!

HA! If he changed speeds, wouldn't it now be a calculus problem?!

Thanks for the laugh!

Blessings,

April

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HA! If he changed speeds, wouldn't it now be a calculus problem?!

Thanks for the laugh!

Blessings,

April

Now you know why I am not teaching my high schoolers math this yr! I am blessed by having the world's most awesome math mom as my kids' teacher. Thank goodness!!

If they had come to me with my flawed logic, I would have been 1/2 asleep (like I am perpetually these days) and would have agreed that it was fine! :tongue_smilie::lol:

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Westward speed = 5 mph ( same as 5 miles per 60 min)

Eastward speed = 7 mph (same as 7 miles per 60 min)

I will try to post detail answer later if anyone interested. I have to go.

Best regards.

Mohan Pawar

http://www.mpclasses.com/ContactUS.htm

US Central Time:5:14 PM 9/27/2010

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I'm not sure enough information is given to answer this problem.

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Thanks, all for the humour & other replies. The pdf link gives us what we need. We knew some of the logic, but were not very smart today, either. Aside from this, this was one of those days where I was ready to take dd to the ps & register her today. It wasn't just that she was annoying and then cranky, but so was my other dd. I realize that this isn't unusual for 15 yos, but it was just not a good day. If I liked alcohol, I'd be having a glass of wine or something right now.

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Fast enough to survive?

Sorry - I'm no help. :tongue_smilie:

:lol::iagree:

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