a little confused INeedSomeHelp

INeedSomeHelp said:
View attachment 31241
i am an older student trying to understand my college algbra class. any insights
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Let [imath]D_1[/imath] be the distance train 1 traveled, [imath]D_2[/imath] be the distance train 2 traveled, and [imath]t[/imath] be the time elapsed (in hours) elapsed since train 2 traveled. Since train 1 had a 45 minutes head start, we have
[math]D_1=45+60t\\ D_2=75t[/math]Solve for [imath]t[/imath] when the distance is the same. Once you have the time when the trains meet, the distance will follow.
 
INeedSomeHelp said:
View attachment 31241
i am an older student trying to understand my college algbra class. any insights
Click to expand...
There are several rather different ways you could solve this, using algebra or not. I'd like to see any attempt you have made, so we can use your own ideas to proceed. Also, what topic are you currently covering, which could give a hint as to how you are expected to solve it.

I would tend to make time the variable. But be careful: Use either hours or minutes, but not a mix of both!

(It is also possible to express both times in terms of a variable representing the distance traveled. That will lead to a rather different equation.)
 
INeedSomeHelp said:
any insights
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Hi INSH. Draw a picture, to help organize the given info. Let time begin when the second, faster train (B) departs. Let time equal t hr, when the trains meet.

You could use a dotted line to represent the first, slower train's head-start distance. Then use the given rates to label the three distances. (Convert 45min to hour, before expressing the head-start distance because the rates are both given in terms of hours.)

Do you understand how BigBeachBanana determined that train B's distance is 75t (miles)?

3BAC1422-F231-4C77-8367-92DB31F023F2.jpeg

:)

[imath]\;[/imath]
 
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Organize your data first. cross out superfluous stuff that doesn't bring you anything towards solution.
Important data
Train travels at 60 km/h
45 minutes later another train leaves the station.
it is traveling in the same direction
it is traveling at 75 km/h
your FIND: how far from the station the faster train will catch the slower one. So you are looking for distance
but it will do you good to find in how much time the faster train will catch up with the slower train, and for this I would do it like this
As I see it, the way I have been trained here by the tutors is to tag things, give names to these facts given.
Example:
my velocity 1 (because there are two in the problem)
V1 (velocity)=60km/h
V2=75 km/h
there is the time involved in the problem so t will stand for time
so t= time
remember that you were given this fact: 45 minutes later another train leaves the station.
what are you gonna do with that?
there's is a time difference of 45 minutes between the two trains.
how much of an hour is that?
45/60 hours =0.75 h ( that is how much it is)
now, with all these organized facts, you can set up an equation

you have one velocity 1.
you are looking for the time in which the last train to leave will catch up with the first one, plus you have a time difference of 0.75 hours between the two of them, and all of this can be equated to the velocity the second train is traveling at times the time(which is your unknown here)
so you can set it up like this
[math]60 * (t + 0.75)=75t[/math]
you have to solve for t now to find the time needed for one to catch up with the other and having the time, you can find the distance.[/I]
 
eddy2017 said:
Organize your data first. cross out superfluous stuff that doesn't bring you anything towards solution.
Important data
Train travels at 60 km/h
45 minutes later another train leaves the station.
it is traveling in the same direction
it is traveling at 75 km/h
your FIND: how far from the station the faster train will catch the slower one. So you are looking for distance
but it will do you good to find in how much time the faster train will catch up with the slower train, and for this I would do it like this
As I see it, the way I have been trained here by the tutors is to tag things, give names to these facts given.
Example:
my velocity 1 (because there are two in the problem)
V1 (velocity)=60km/h
V2=75 km/h
there is the time involved in the problem so t will stand for time
so t= time
remember that you were given this fact: 45 minutes later another train leaves the station.
what are you gonna do with that?
there's is a time difference of 45 minutes between the two trains.
how much of an hour is that?
45/60 hours =0.75 h ( that is how much it is)
now, with all these organized facts, you can set up an equation

you have one velocity 1.
you are looking for the time in which the last train to leave will catch up with the first one, plus you have a time difference of 0.75 hours between the two of them, and all of this can be equated to the velocity the second train is traveling at times the time(which is your unknown here)
so you can set it up like this
[math]60 * (t + 0.75)=75t[/math]
you have to solve for t now to find the time needed for one to catch up with the other and having the time, you can find the distance.[/I]
Click to expand...
If you expand your brackets you should get the expression that BBB was pointing you towards in post #3 (q.v.).
So what answer do you now get for how far from the station does the faster train catch up with the slower one?
 
The Highlander said:
If you expand your brackets you should get the expression that BBB was pointing you towards in post #3 (q.v.).
So what answer do you now get for how far from the station does the faster train catch up with the slower one?
Click to expand...
The thread is not mine Highlander. I just was trying to give the poster a lead. Can't give the full answer. Forum rules.
 
My apologies. I've done this before, Doh! :oops::rolleyes:
I thought you were the OP, showing all your working (just not following it through to a final answer, lol).
Well, s/he has everything they need now, maybe we'll hear from them again to see if all our efforts have been worthwhile. :eek::D
 
Subhotosh Khan said:
Assume that those meet 'd ' km away. How much time did the slow train take to cover the distance ? Continue .......
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The slow train will be at the meeting point after travelling for (d/60) hours

The slow train will be at the meeting point after travelling for (d/75) hours

d/75 + 3/4 = d/60 ...............................[edited]

Calculate 'd'
 
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