A turtle and a rabbit engage in a footrace over a distance of 4.00 km. The rabbit runs 0.500 km and then stops for a 90.0-min nap. Upon awakening, he remembers the race and runs twice as fast. Finishing the course in a total time of 1.75 h, the rabbit wins the race. (a) Calculate the average speed of the rabbit. (b) What was his average speed before he stopped for a nap? Assume no detours or doubling back.

Question

nincompoop · Answer

Speed_ave = L/T 
Speed_ave = 4.00km/1.75hr 
Speed_ave = 2.29km/hr

nincompoop · Answer

can you help me out with b?

e.mccormick · Answer

If you think about it, he travelded .5 at normal speed.  And 2x speed he would have completed an additional .5, right?

nincompoop · Answer

I have T_1 + T_2 = 0.250 hr

e.mccormick · Answer

The .25 hours is important, but the $T_1$ and $T_2$ don't really matter.

e.mccormick · Answer

There is a way of getting $V_1$ in terms of $V_2$ which lets you solve for $V_2$, which is then easily converted to $V_1$.

That is why I asked: If you think about it, he travelded .5 at normal speed. And 2x speed he would have completed an additional .5, right?

nincompoop · Answer

how'd you get that?

e.mccormick · Answer

$V=\dfrac{D}{T}$  or $S=\dfrac{L}{T}$ using your formula... that is why I used V and D.

In the problem is says he travled .5km at $V_1$.  Then he travled the rest at $V_2$.  It adds, $2V_1=V_2$ right?

nincompoop · Answer

yes

e.mccormick · Answer

$D_1=.5\text{ and } D_T=4.0 \; \therefore \; D_2=3.5$

nincompoop · Answer

right

e.mccormick · Answer

But, in $T_1$ if the rabet was going $V_2$ it would have travled twice the distance because it is going twice as fast.  Correct?

nincompoop · Answer

yes

e.mccormick · Answer

OK, so I use that fact.  It means that in $T_T$ the rabbit would have travled $2D_1+D_2=4.5$.

nincompoop · Answer

right

e.mccormick · Answer

Well, now you have $V_2=\dfrac{\text{Simuated Total}}{T_T}$ when you solve that with numbers, you have $V_2$ and since $V_1$ is half of that...

e.mccormick · Answer

Basically you put $D_1$ in terms of $D_2$ so you can use the totals to find $V_2$.

e.mccormick · Answer

Because $D_1$ in terms of $D_2$ makes the simulated total distance.  And I hope that makes sense... I am sleeeepy.

nincompoop · Answer

Same here, I am sleepy, but I am bugging that I can't solve this atm. I am trying to catch up with what you're saying. I am using my poor iphone to read all what you typed. 
there's something that is not connecting in my brain

e.mccormick · Answer

If I take the $T_T=2.5$ you found (and I agree with) and the $D_S=4.5$ we agreed on, I get:
$V_2=\dfrac{4.5}{.25}$

nincompoop · Answer

and since V_2 = 2V_1 
therefore: V_1 = 9

e.mccormick · Answer

Yes.

nincompoop · Answer

you mean T_t = .25 correct?

e.mccormick · Answer

$T_T=.25$, $V_2=18$, and $V_1=9$

e.mccormick · Answer

I like all of those and they all work mathematically.

nincompoop · Answer

thank you.
I tried doing algebraic manipulation first but I end up with 
9V_1 = 14 
don't ask me why 
I had it as 
0.5/V_1 + 3.5/V_2 = 0.25, where V_2 = 2V_1

e.mccormick · Answer

Hmmm... you would need a good second equation for doing it that way.  Hmm... wich you do have since V_1+V_2=V_A

e.mccormick · Answer

0.5/V_1 + 3.5/V_2 = 0.25
V_1+V_2= 2.29

Hmmm...but would all the units work out....

e.mccormick · Answer

We did this unitless....  well... I made sure I was working in proper unit comparisons so that I could drop them.  But that...

e.mccormick · Answer

I need sleep...  so I'll let you think about it.  Hehe.

nincompoop · Answer

your way was neat to look at. 
I'll revisit this again tomorrow,  my brain hasn't been making any sense

nincompoop · Answer

thank you again

e.mccormick · Answer

Yah, I made some logical jumps in there that... I'm just too sleepy to explain.  It all depends on the fact that double speed would make double distance and therefore you have a simulated total distance if the trip was made entirely at double speed.  Because you know total time, this simulated total distance and total time can give the average speed at the double speed.  Once the double speed is known, the regular speed is found with /2.  Wheee... that was the concept in a nutshell.

e.mccormick · Answer

Oh, and I think it is neat comparing our smart scores.  Since we are both 85, it shows a bit of what causes what.  You did more fans and have teammate.  I did more answers and got Mentor.