# Thread: motion word problem: At 9:45 a bus left Point Reyes, CA, going north at 40 mph....

1. ## motion word problem: At 9:45 a bus left Point Reyes, CA, going north at 40 mph....

At 9:45 a bus left Point Reyes, CA, going north at 40 mph. 2 hours later, a second bus left Point Reyes going south at 35 mph. At what time will the buses be 230 miles apart?
I have set this up thus: first I divided 35 by 40 (.875), and devised this equation: 35t + 40(t-.875)=230 with the solution being:3.8 hours. So, at 9:45 plus 3 hours and 48 minutes or at 1:33 the buses will be 230 miles apart. Does this look right? Is this how to solve this problem? If so, why do we need to know that one bus left 2 hours before the other?

Something tells me I am overlooking something.

2. Originally Posted by allegansveritatem
… first I divided 35 by 40 (.875) … Does this look right? Is this how to solve this problem? If so, why do we need to know that one bus left 2 hours before the other?
The ratio of speeds (35/40) is not going to help.

When the northbound bus travels while the southbound bus does not, the distance between the buses is increasing. Therefore, we need to know how long the northbound bus traveled (while the southbound bus sat at the depot), so that we can determine the distance between the buses at 11:45.

What was the distance between the buses at 11:45?

How much more distance is then required, to reach a total of 230 miles?

That additional distance comes from 40t + 35t.

Can you try again, or do you still have questions?

PS: Thanks for showing your work.

3. Originally Posted by allegansveritatem
At 9:45 a bus left Point Reyes, CA, going north at 40 mph. 2 hours later, a second bus left Point Reyes going south at 35 mph. At what time will the buses be 230 miles apart?
I have set this up thus: first I divided 35 by 40 (.875), and devised this equation: 35t + 40(t-.875)=230 with the solution being:3.8 hours. So, at 9:45 plus 3 hours and 48 minutes or at 1:33 the buses will be 230 miles apart. Does this look right? Is this how to solve this problem? If so, why do we need to know that one bus left 2 hours before the other?

Something tells me I am overlooking something.
To correct your method, all you need to do is to replace -.875 with +2. This is correct because the time the first bus is traveling is 2 hours more than the time (t) that the second is traveling. The ratio of speeds, as you've been told, is irrelevant -- in fact, it isn't a time, so it makes no sense to subtract it from t!

But you also have to take note of the meaning of t: you used it in the equation as the time the second bus travels, but then added it to the time the first bus started!

4. Originally Posted by mmm4444bot
The ratio of speeds (35/40) is not going to help.

When the northbound bus travels while the southbound bus does not, the distance between the buses is increasing. Therefore, we need to know how long the northbound bus traveled (while the southbound bus sat at the depot), so that we can determine the distance between the buses at 11:45.

What was the distance between the buses at 11:45?

How much more distance is then required, to reach a total of 230 miles?

That additional distance comes from 40t + 35t.

Can you try again, or do you still have questions?

PS: Thanks for showing your work.
I like this way. I did actually figure the answer out by the diagram method but did not want to do it that way and so I discarded my results ...this supposed to be an algebra problem. It did not occur to me to do the thing in two stages. So thanks for this method.

5. Originally Posted by Dr.Peterson
To correct your method, all you need to do is to replace -.875 with +2. This is correct because the time the first bus is traveling is 2 hours more than the time (t) that the second is traveling. The ratio of speeds, as you've been told, is irrelevant -- in fact, it isn't a time, so it makes no sense to subtract it from t!

But you also have to take note of the meaning of t: you used it in the equation as the time the second bus travels, but then added it to the time the first bus started!
Well, I don't know why I didn't see that I was mixing apples and oranges with the term with the ratio....I can give some account of my reasoning however: I had just dealt with another problem similar to this The other poblem stated:a bus and a car travel the same route both leaving from the same point. The bus goes 72kmh, the car goes 80kmh. If the bus leaves one hour before the car, how long will it take the car to catch up to the bus. In solving this problem I used the ratio method and it worked! Here is the equation I devised: 72t = 80(t-.9). The solution is : 9. The back of the book confirmed my result. I don't know how the author solved this problem but at least in this instance the ratio method hit the target. Maybe this was just coincidence. I haven't had time to test it. Obviously it doesn't work in the present porblem.

I think my reasoning in this case was something like: the ratio of speeds indicates the lag between the speeds and this indication can be used somehow to sub for a time value. I don't know. I guess what killed me was the memory of how well this method worked at least once before in a similar problem.

I like ratios. I like to find ways to use them. And obviously, to abuse them as well.

6. Originally Posted by allegansveritatem
… I did actually figure the answer out by the diagram method but did not want to do it that way and so I discarded my results ...this supposed to be an algebra problem …
I think it's a good idea to do both. Start with a labeled diagram, to organize the given information. Then use the diagram to come up with relationships (i.e., equations) or algebraic expressions for stuff. Starting with a good diagram often helps us "see" an algebraic approach.

7. Originally Posted by Denis
Same problem:

h-2 hours<----------(@35mph)A(@40mph)-------------------->h hours

Bus leaves point A, travelling East at 40 mph.
2 hours later, another bus leaves point A, travelling West at 35 mph.
When the buses are 230 miles apart,
how many hours has the 1st bus travelled?

Let h = hours travelled by 1st bus (then 2nd bus travelled h-2 hours.
Using distance = speed * time:
40h + 35(h - 2) = 230

Take over...
Thank you for this reply. This is the third method I have been presented with. That is why I love mathematics....there are so many ways to skin a cat (to use a rather ugly analogy). I bet if we tried we could come up with a couple more.

8. Originally Posted by mmm4444bot
I think it's a good idea to do both. Start with a labeled diagram, to organize the given information. Then use the diagram to come up with relationships (i.e., equations) or algebraic expressions for stuff. Starting with a good diagram often helps us "see" an algebraic approach.
I agree. Pictures help keep us in the bounds of reality.

9. Originally Posted by allegansveritatem
Well, I don't know why I didn't see that I was mixing apples and oranges with the term with the ratio....I can give some account of my reasoning however: I had just dealt with another problem similar to this The other poblem stated:a bus and a car travel the same route both leaving from the same point. The bus goes 72kmh, the car goes 80kmh. If the bus leaves one hour before the car, how long will it take the car to catch up to the bus. In solving this problem I used the ratio method and it worked! Here is the equation I devised: 72t = 80(t-.9). The solution is : 9. The back of the book confirmed my result. I don't know how the author solved this problem but at least in this instance the ratio method hit the target. Maybe this was just coincidence. I haven't had time to test it. Obviously it doesn't work in the present porblem.

I think my reasoning in this case was something like: the ratio of speeds indicates the lag between the speeds and this indication can be used somehow to sub for a time value. I don't know. I guess what killed me was the memory of how well this method worked at least once before in a similar problem.

I like ratios. I like to find ways to use them. And obviously, to abuse them as well.
The traditional algebraic way to solve this one would be to define t as the time after the car leaves, so that the bus takes t+1 hours. Then 72(t + 1) = 80t. This is equivalent to your equation. I think it was just luck (and the fact that the difference is 1 hour) that your work gave the right answer.

For non-algebraic methods, what Denis did is what I would do. But you are probably in a class where you are expected to learn algebra, not general problem-solving skills, as valuable as the latter are.

10. Originally Posted by Denis
Quickest way for this problem is pretend the 2 buses are leaving at
same time, distance 80 miles apart (the 1st 2 hours bus#1 travelled).

Left to travel is 230-80 = 150 miles.
At combined speed of 75mph, so another 2 hours.
Game over!
Yes...short and sweet.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•