# Homework Help: Physics Finding Total Distance?

1. Mar 26, 2012

### RichardT

1. The problem statement, all variables and given/known data
John had 5 dogs pulling his sled at full speed for 24 hours. 2 dogs than escaped. however, the journey continued at 3/5 the original velocity and reached his destination 48 hours later. if the two dogs did not escape for an additional 50km he would have reached 24 hours late instead. What is the distance he covered??

3. The attempt at a solution
Well so far heres what i got:
I think i can assume(not sure) that with 5 dogs he traveled 50km in 24 hours according to the last bit of information.

So than that means in the previous 24 hours he would have went 50km with 3 dogs - this is what i dont understand, how can 5 dogs travel 50km in 24 hours while 3 dogs travel 50km in 24 hours as well??

And therefore in the first 24 hours the dogs traveled (50/3) x 5 = 83.333km

and for my total I get 50 + 83.333 = 133.33

Please explain to me the part where 3 dogs travel the same distance in the same time as 5 dogs...
Thanks!

The possible answers are 100, 133, 167, 200, or 267 (all kilometers)

Last edited: Mar 26, 2012
2. Mar 26, 2012

### Staff: Mentor

No, you don't know how long it took the five dogs to cover the 50km. All you know is that it takes a total of 24 hours to cover the "second stretch" if the first 50km are done with a full team of 5 dogs, and 48 hours if done with 3 dogs.

So if v is the "top speed" with 5 dogs, then to cover the second stretch:
$$\frac{50 km}{v} + \frac{x}{\frac{3}{5}v} = 24 hours$$
where x is the currently unknown portion of the second stretch: d2 = 50 km + x

You should be able to write another equation for d2 for the case where only three dogs are employed for the whole second stretch. Substitute for x and d2 accordingly, solve for v.

3. Mar 26, 2012

### RichardT

Ok, so that means i would have 3 equations:

Equation 1. 50km/v + x/3/5v = 24hours
Equation 2. d2 = 50km + x
Equation 3. d2(with 3 dogs) = 3/5v x 48hours

Correct?
If correct, than:
Equation 3: d2 = (144/5)v
And plug that into Equation 2 which is d2=50km + x
So sub in (144/5)v for d2 and solve for x which means: x = (144/5)v - 50km
And than sub the new equation 2 into equation 1 and solve for v?

So.. 24= ( 144/5v - 50km)/(3/5)v + 50/v

Which turned out to be 1.3888km/h

So than in the first 24 hours he traveled 1.3888 x 24 = 33.333km?
and than in the 48 hours with 3/5 dogs he traveled 1.38888x48 = 39.999
so in total it equals 73.333km??

I know im wrong but i dont know where? Please reply back..

4. Mar 26, 2012

### Staff: Mentor

Well, the results look reasonable to me. Why do you think you're wrong? can you find a contradiction between the results and the given information?

5. Mar 26, 2012

### RichardT

Its just that i have the possible answers and 73.333km isnt one of them, is there anywhere i could have gone wrong??

6. Mar 26, 2012

### Staff: Mentor

I don't think so, your calculations looked okay to me (at first glance at least!). I'll give it some thought...

7. Mar 26, 2012

### Staff: Mentor

Perhaps I've made a misinterpretation of the problem text? When you wrote:
Did you mean "late" or "later"? I had assumed a typo, given the previous sentence In other words, would he have arrived in the next 24 hours, or would he arrive 24 hours AFTER he would have arrived with a full team (24 hours "late")?

8. Mar 26, 2012

### RichardT

Ok thanks!, but can you tell me why i cannot assume the last part as the dogs travelling 50km in 24 hours?

9. Mar 26, 2012

### RichardT

No its meant to be he arrived only 24 hours late rather than the 48 hours he would have been late

So if he had 5 dogs for an additional 50km he would have been ONLY 50km late

Hope that clears up some misinterpretations

10. Mar 26, 2012

### Staff: Mentor

Because it wasn't stated as such All you know is that at one speed it takes a given time, and doing a portion at another speed it takes a different time. You can't assume that the distance was half just because the time was halved.

11. Mar 26, 2012

### Staff: Mentor

Really, that only helps a bit "50km late"?

12. Mar 26, 2012

### RichardT

Oops sorry i meant ONLY 24 hours late... Sorry...

13. Mar 26, 2012

### Staff: Mentor

So, just to be clear, in the first case he is 48 hours late, and in the second case he would be only 24 hours late. This would be based on the expected arrival time of T = D/v where D is the total distance and v the velocity with a full team for the entire trip. Is that the scenario?

14. Mar 26, 2012

### RichardT

Ya thats what i believe the scenario is.

15. Mar 26, 2012

### Staff: Mentor

Okay, then write equations for elapsed times. Let T be the expected time, v be the normal velocity, d1 the first distance using all the dogs, d2 the remaining distance. D is the total distance.

Normal travel would yield an "expected" elapsed time of T = D/v. How would you write equations for the described cases? One elapsed time is T + 48hr, and the other is T + 24hr. What distance and velocity expressions would correspond to them? Hint: dealing with the 50km is the tricky bit

16. Mar 26, 2012

### RichardT

and honestly i have no idea for the 50km but is it d1 + 50km =T??

17. Mar 26, 2012

### PeterO

A picture/graph might help.

With all 5 dogs, you would have a flat v-t graph for the entire trip - ending at some time t.

For the trip with the 2 dogs running away, it would be flat at the original speed for 24 hours, then drop to 3/5 ths of that speed for extra time - finishing at t + 48

Draw those two graphs on the same axes, and you see given a common trip, the area under each graph has to be the same.
The extra 48 hours at 60% of original speed, makes up for the missing section from 24 hours to t - which amounts to 40% [the top bit] of the distance covered if all dogs had stayed.

Now draw in the third possibilty - with the 5 dogs staying for an extra 50km. But that extra 50 k is done at full speed.

The distance covered in 24 hrs = only 40% of 50km.

The

Last edited: Mar 26, 2012
18. Mar 26, 2012

### Staff: Mentor

No, d1 is the distance covered in the first 24 hours. So d1 = v*24hr. You could also say that the time to cover distance d1 is d1/v = 24hr.
You can't equate distance with time...the units don't match.

If the two sections of the trip described in the first scenario are d1 and d2, the first traveled at speed v and the second at (3/5)v, what are the elapsed times for each (symbolically)?

If d1 is made 50km longer, then d2 must be made 50km shorter. What are the new elapsed times for each (symbolically)?

19. Mar 26, 2012

### RichardT

So, would the graph look like the picture i included? and Are my equ ations correct?

20. Mar 26, 2012

### PeterO

In case you are not familiar with solving this type of problem with v-t graphs, the attached file shows you them.
I find them particularly useful.

File size:
78 KB
Views:
95