Physics Finding Total Distance?

[RichardT]

Homework Statement

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??
Please explain...

The Attempt at a Solution

Well so far here's 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 don't 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)

 

[gneill]

Homework Statement

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??
Please explain...

The Attempt at a Solution

Well so far here's 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.

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.

 

[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 I am wrong but i don't know where? Please reply back..

 

[gneill]

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?

 

[RichardT]

Its just that i have the possible answers and 73.333km isn't one of them, is there anywhere i could have gone wrong??

 

[gneill]

Its just that i have the possible answers and 73.333km isn't one of them, is there anywhere i could have gone wrong??

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

 

[gneill]

Perhaps I've made a misinterpretation of the problem text? When you wrote:

if the two dogs did not escape for an additional 50km he would have reached 24 hours late instead.

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")?

 

[RichardT]

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

 

[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

 

[gneill]

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

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.

 

[gneill]

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

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

 

[RichardT]

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

 

[gneill]

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?

 

[RichardT]

Ya that's what i believe the scenario is.

 

[gneill]

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

 

[RichardT]

sooo.. not so sure about this but: d1 = v(T+24)? and d2 = 3/5v(T+48)?
and honestly i have no idea for the 50km but is it d1 + 50km =T??

 

[PeterO]

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

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 possibility - 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

 

[gneill]

sooo.. not so sure about this but: d1 = v(T+24)? and d2 = 3/5v(T+48)?

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.

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

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)?

 

[RichardT]

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

ations correct?

 

[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 possibility - 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.

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.

 

[RichardT]

Ok, i can understand all the graph. However, how will that give me the total distance when i do not have all the times??

 

[PeterO]

Ok, i can understand all the graph. However, how will that give me the total distance when i do not have all the times??

From that one area value of 40% of 50 (ie 20) you can work out all the times and distances - just evaluate carefully as you go.

 

[RichardT]

How did you find out that it was 40% of 50? I am not sure but basically the graph is saying that the man traveled more than 50km in the first 24 hours and than he traveled 20km?? How would i find out the area under that graph with no proper velocity?? And i have the velocity for area G and J i have the velocity but no time? Please help me read the graph

 

[PeterO]

How did you find out that it was 40% of 50? I am not sure but basically the graph is saying that the man traveled more than 50km in the first 24 hours and than he traveled 20km?? How would i find out the area under that graph with no proper velocity?? And i have the velocity for area G and J i have the velocity but no time? Please help me read the graph

We were told the speed after the 2 dogs left was 3/5 the initial velocity, so the second, lower velocity is 60% of the first.

In graph 2, where area D = Area C, we know from that 3/5 velocity, that area C is 40% of the total (B+C)

In Graph 3, we similarly know F is 40% of the total (E+F) ; and we are told that that extra bit (E+F) is the extra 50km the full team stayed on - or could have stayed on
"if the two dogs did not escape for an additional 50km he would have reached 24 hours late instead"

That gives us the way in.

F = 20 so E = 30 also K = 20, but J = K so J = 20 and we are off. You next get H then G. You can then establish Area A [which is not a whole number!]

EDIT: by comparing areas J and K to areas E and G you can establish the time interval to t

 

[RichardT]

Would G be 30 as well and H 20? I am sorry that I am not really understanding this...

 

[PeterO]

Would G be 30 as well and H 20? I am sorry that I am not really understanding this...

Yes it is:

Here is the logic.

In Diagram 2, area D = Area C, and area D is "48hrs wide"

In diagram 3, area J is "24hrs wide", so Area K is also "24hrs wide" ; they are equal in size.

As shown in Graph 2, area B is 0.6V high, so Area C must be 0.4V high

Same height relation for E & F and G & H.

Area J and Area H have the same sized area. Area H has height 2/3 of Area J, so area H has width 1.5 times Area J. But Area J is "24 hrs wide" so Area H is "36 Hours wide". Area E is the same width.

Areas E&F together total 50 km, and are "36hrs wide". Area A is the same height, but only "24hrs wide" and thus must have area 33.333 km

 

[RichardT]

Ok so the total distance is 50 + 33.333+ 20 +20 +20 = 143.333km?
I can understand that but I am given the possible answers and they are: 100, 133, 167, 200, and 267.
Is there anything I am missing?

 

[PeterO]

Ok so the total distance is 50 + 33.333+ 20 +20 +20 = 143.333km?
I can understand that but I am given the possible answers and they are: 100, 133, 167, 200, and 267.
Is there anything I am missing?

You missed a bit there:
Area is [one interpretation] A + 50 + G + J = 33.33 + 50 + 30 + 20 = 133.333

 

[RichardT]

ok, one more thing, how do you assume that the area of H and G equal the area of F and E if only you know that they have the same height. You don't know the widths right? Because since i know that the width of H and G = 36 and the width of F and E equal 24, wouldn't the distances be different??
EDIT: wait that's only assuming he travels 50 km in 24 hours which you can't so how did you get H and G equal to E and F?

 

[PeterO]

ok, one more thing, how do you assume that the area of H and G equal the area of F and E if only you know that they have the same height. You don't know the widths right? Because since i know that the width of H and G = 36 and the width of F and E equal 24, wouldn't the distances be different??
EDIT: wait that's only assuming he travels 50 km in 24 hours which you can't so how did you get H and G equal to E and F?

Compare second graph to 3rd graph: Area D = J + K and Area C = F + H

Also Area F = Area K

Since D is "48hrs" wide, and J is "24hrs" wide J & K are the same size.
That means F & H are the same size, and thus E & G are the same size.

Indeed width of F is "36hrs" it is area A that is "24hrs" wide

so working along the bottom.

A is 24hrs wide
E is 36hrs wide
G is 36hrs wide
J is 24hrs wide
K is 24hrs wide

up to graph 2
B is 72 hours wide,
D is 48hrs wide

up to graph 1

t = 96hrs.

 

[PeterO]

ok, one more thing, how do you assume that the area of H and G equal the area of F and E if only you know that they have the same height. You don't know the widths right? Because since i know that the width of H and G = 36 and the width of F and E equal 24, wouldn't the distances be different??
EDIT: wait that's only assuming he travels 50 km in 24 hours which you can't so how did you get H and G equal to E and F?

Note that using the full team for an extra 50km, made section K 24hrs wide, not the time taken to cover the 50km.(that was 36hrs it turns out).

 

[RichardT]

Ok so area G and H are equal to area of E and F because area of J = the area of K?

 

[PeterO]

Ok so area G and H are equal to area of E and F because area of J = the area of K?

Correct! and I suspect you find it surprising - but it was all in the original information.

when the dogs left - 48 hours late
If the dogs stayed on for an extra 50km, only 24 hours late so the area J is one half of D, and thus J = K