Finding average speed (hopefully this belongs here)

A

A138

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If I run one lap around a track at a speed of 5 mph, how fast do I need to run the second lap so that my average pace over the two laps will be 10 mph?
This is a bonus (optional) question and I'd like to solve this but it does not pertain to any of the other objectives that I was taught so I'm not sure where to start.

Things I know because I've already looked it up-
  • distance=rate*time
  • average rate= total distance/total time
  • 1 lap= 400 meters
Can someone please help me? Am I solving for the time of the of the second one? Both of those formulas need the time to be able to solve it but I'm unsure as to how to write an equation to solve for the time of the second lap when I already have the time for the first.
 
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Let d = distance around the track.

r1= 5mph, your speed for the 1st lap.

r2 = q mph, your speed for the 2nd lap

average rate = (total distance)/(total time)
10 = (0.5) /(t1 + t2)

Solve for t1 + t2 ...
 
5+x
----- = 10........ this is incorrect.
2
 
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creelrj said:
5+x
----- = 10
2
Click to expand...
Hi creelrj. We can't average rates like that because the times are not equal. (Don't feel bad. Albert Einstein made a similar mistake, when he tried to answer a word problem like yours.)

Jomo gave you an equation for average rate:

average rate = (total distance) / (total time)

Jomo converted 400 meters to miles and rounded (400 meter ≈ 0.25 mile). The total distance of two laps is about 0.5 miles.

10 = (0.5) / (t1 + t2)

The total time is unknown, but we can determine the time it took to run the first lap (t1) because we know both the distance (0.25 mi) and the rate (5 mi/hr).

time = distance / rate

t1 = ?

Next, substitute your value for t1 into the red equation above, and solve for t2.

Then tell us what concerns you have about your answer.

?
 
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Okay so is this it?
 

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The average would be 0.025 which isn't correct. What am I doing wrong?
 
A138 said:
Am I correct?
Click to expand...
Please don't write that 5+15 = 20/2. You know what 5+15 equals and you know what 10/2 equal. Most importantly you know that those results are not equal. Yes, the equation you solved is x=15. Now please solve your problem.
 
Otis said:
Hi creelrj. We can't average rates like that because the times are not equal. (Don't feel bad. Albert Einstein made a similar mistake, when he tried to answer a word problem like yours.)

Jomo gave you an equation for average rate:

average rate = (total distance) / (total time)

Jomo converted 400 meters to miles and rounded (400 meter ≈ 0.25 mile). The total distance of two laps is about 0.5 miles.

10 = (0.5) / (t1 + t2)

The total time is unknown, but we can determine the time it took to run the first lap (t1) because we know both the distance (0.25 mi) and the rate (5 mi/hr).

time = distance / rate

t1 = ?

Next, substitute your value for t1 into the red equation above, and solve for t2.

Then tell us what concerns you have about your answer.

?
Click to expand...
I am embarrassed to say that I can't solve this problem. Can you please look closer at the problem. I just don't get it. Just verify if there is a problem or not. Even if there is a problem my results should show this! Thanks.
 
A138 said:
… What am I doing wrong?
Click to expand...
Hi. Your value for t1 is not correct. You wrote 5, but that's the rate, not the time.

You know the following relationship:

d = r · t

If we want to express the time, we divide each side by rate:

t = d / r

For the first lap, you know the distance (0.25) and you know the rate (5). Therefore, we have:

t1 = d / r = 0.25 / 5 = ?

Use that value, in the red equation (from post #5).

Then tell us what concerns you have about your answer.

?
 
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Otis said:
Perhaps, you're making the same error in logic as Albert Einstein. Did you solve for t2?

;)
Click to expand...
That is what I was thinking! I have t1 = (2^3-sqrt(25)) min. However I am getting that same time for 2 laps at 10mph!
 
OK, I looked at this problem from a few different angles and finally figured out what is going on.
 
I'm solving for the average of the two laps... so if the first was 5 and the second was 15, average is 5+15=20 then 20/2 because there's two numbers= 10
 
A138 said:
I'm solving for the average of the two laps... so if the first was 5 and the second was 15, average is 5+15=20 then 20/2 because there's two numbers= 10
Click to expand...

But, you're assuming here that the time spent traveling at the two speeds is the same. That's the expected pitfall.
 
Let's let \(L\) be the length of one lap, and \(v\) be the average speed for the first lap, and so the time that took is:

[MATH]t_1=\frac{L}{v}[/MATH]
Now, we want our average for 2 laps to be \(2v\):

[MATH]2L=2vt[/MATH] where \(t\) is the total time, including the first lap.

Or:

[MATH]L=vt\implies t=\frac{L}{v}=t_1[/MATH]
This means we must run the second lap instantaneously, as 0 time is left, since the total time needed is equal to the time it took to run the first lap. We must therefore conclude that it is impossible.
 
A138 said:
I'm solving for the average of the two laps …
Click to expand...
That wording is a bit sloppy, A138, but I already explained why you can't average rates as shown in your work.

Did you correct your mistake in calculating t1?

\(\;\)
 
MarkFL said:
Let's let \(L\) be the length of one lap, and \(v\) be the average speed for the first lap, and so the time that took is:

[MATH]t_1=\frac{L}{v}[/MATH]
Now, we want our average for 2 laps to be \(2v\):

[MATH]2L=2vt[/MATH] where \(t\) is the total time, including the first lap.

Or:

[MATH]L=vt\implies t=\frac{L}{v}=t_1[/MATH]
This means we must run the second lap instantaneously, as 0 time is left, since the total time needed is equal to the time it took to run the first lap. We must therefore conclude that it is impossible.
Click to expand...
I liked how you knew to see the avg velocity as twice the 1st velocity. Nicely done.
 
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