Which way is correct and why?

  • #1
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Homework Statement:

This week's question

A race car driver travels x km at 40km/h and y km at 50 km/h. The total time taken is 2.5 hrs. If the time to travel 6x km at 30km/h and 4y km at 50km/h is 14 hrs. Find x and y

Relevant Equations:

Speed = distance /time
Ave speed = total distance/ total time
Ave speed= (speed 1 + speed 2 ) /2
IMG_20200229_085438.jpg

I get two different answers even the area of the graph.
I think there is something wrong with the equation I constructed for average speed part.
I know the method 1 is correct. But I'd like to know why the average speed cannot be used. And why I get two different areas for the graph
Thank you
 

Answers and Replies

  • #3
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False.
Why? An average can be the total / the number of things added to make the total
Could you explain a bit more
 
  • #4
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You need to take into account the times travelled at each speed. Useful sites for help in caculating average speeds are inly a simple search away.
 
  • #5
haruspex
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Why? An average can be the total / the number of things added to make the total
Could you explain a bit more
It is a matter of definition.
Average speed is defined to be (total distance travelled )/( total time taken). This does not give the same number as (v1+v2)/2 unless the two times happen to be the same.
 
  • #6
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It is a matter of definition.
Average speed is defined to be (total distance travelled )/( total time taken). This does not give the same number as (v1+v2)/2 unless the two times happen to be the same.
I understand very well

Could you give me an example to two situations where it works and it doesn't. I just want to wrap my head around it
 
  • #7
haruspex
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I understand very well

Could you give me an example to two situations where it works and it doesn't. I just want to wrap my head around it
As an analogy, consider a group of four people, one of height x, one of height y and two of height z. There are three heights, so in one sense the average height is (x+y+z)/3, but we find it more useful to define the average height of the four people as (x+y+2z)/4.

The number of people of a given height in the analogy corresponds to the time spent at a given speed.
 
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  • #8
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As an analogy, consider a group of four people, one of height x, one of height y and two of height z. There are three heights, so in one sense the average height is (x+y+z)/3, but we find it more useful to define the average height of the four people as (x+y+2z)/4.
I actually meant an analogy where average speed = v1+v2/2 and average speed not equal to v1+v2/2
 
  • #9
haruspex
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I actually meant an analogy where average speed = v1+v2/2 and average speed not equal to v1+v2/2
Oh, you mean an example of where the answers are different?
If I walk at a steady 5kph for 59 minutes then stand still for one minute, is it sensible to say that my average speed is 2.5kph?
 
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  • #10
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Oh, you mean an example of where the answers are different?
If I walk at a steady 5kph for 59 minutes then stand still for one minute, is it sensible to say that my average speed is 2.5kph?
Hmmm that is true. Then an a case where someone walks for 5kph for 59mins and 6kph for 15 mins, can we say the average speed is 5.5kph?
 
  • #11
PeroK
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Hmmm that is true. Then an a case where someone walks for 5kph for 59mins and 6kph for 15 mins, can we say the average speed is 5.5kph?
The average of the numbers ##5## and ##6## is ##5.5##. But that's not how average speed is defined.
 
  • #12
symbolipoint
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Ioric (lioric?)
The system of equations shown in the left-side rectangle-contained spot of the image of your paper are the correct ones for the system. You have two linear equation in two variables. The rest of YOUR work is difficult to follow. I am guessing by this time you have your solution straightened-out. Otherwise, you have the correct system. Just simplify and use Elimination Method to find x and y.

The simplified system I find is this:
5x+4y=500
AND
5x+2y=350
Simple use of Elimination Method will quickly give you value for y.
 
  • #13
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The average of the numbers ##5## and ##6## is ##5.5##. But that's not how average speed is defined.
This what I'm trying to clarify. In which situations does the summing and dividing of speeds not fit the average speed
 
  • #14
PeroK
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Hmmm that is true. Then an a case where someone walks for 5kph for 59mins and 6kph for 15 mins, can we say the average speed is 5.5kph?
To see this, suppose you have the following data:

5km/h for 15 mins
5km/h for 15 mins
6 km/h for 15 mins
5 km/h for 15 mins

Would the average speed be 5.5km/h? Why not?
 
  • #15
PeroK
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This what I'm trying to clarify. In which situations does the summing and dividing of speeds not fit the average speed
The average speed is defined to be the total distance divided by the total time.
 
  • #16
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Ioric (lioric?)
The system of equations shown in the left-side rectangle-contained spot of the image of your paper are the correct ones for the system. You have two linear equation in two variables. The rest of YOUR work is difficult to follow. I am guessing by this time you have your solution straightened-out. Otherwise, you have the correct system. Just simplify and use Elimination Method to find x and y.

The simplified system I find is this:


Simple use of Elimination Method will quickly give you value for y.
Yes it's Lioric.
I know the solution to this problem. There is no doubt about that. X is 40 y is 75
But I'm just trying to make sense why the average speed version doesn't work. In my gut I sort of know why it cannot be that average speed is not V1+v2 /2. But I'd like to be able to explain it too. I just cannot find the words
 
  • #17
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To see this, suppose you have the following data:

5km/h for 15 mins
5km/h for 15 mins
6 km/h for 15 mins
5 km/h for 15 mins

Would the average speed be 5.5km/h? Why not?
For one the unit is per hour.
 
  • #18
PeroK
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For one the unit is per hour.
What is the average speed in the case I mentioned?
 
  • #19
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What is the average speed in the case I mentioned?
5+5+5+6/1
Since 4x15 min is 1 hour
So 21
 
  • #20
PeroK
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5+5+5+6/1
Since 4x15 min is 1 hour
So 21
Well that's just absurd. I can't help you if you are going to give daft answers.
 
  • #21
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Well that's just absurd. I can't help you if you are going to give daft answers.
Sorry
Well that was total distance over total time. I miss took the speed for distance. My bad. Very sorry

5+5+5+6 /4
Which is 5.25

Then the other is
1.25+1.25+.1.25+1.5/1

=5.25

1.25 is the distance walked at 5kph for 15min
And 1.5 is 6kpm for 15min
 
  • #22
PeroK
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Sorry
Well that was total distance over total time. I miss took the speed for distance. My bad. Very sorry

5+5+5+6 /4
Which is 5.25
It has to be 5.25km/h. The key is that the durations must be the same. You can't treat 5km/h for 1 minute with the same weight as 6 km/h for 59 mins. Because the second speed is sustained for 59 times longer.

The answer to your question is that the average of 5km/h and 6km/h is only 5.5 km/h when the two speed are sustained for the same duration.

You can see that from the definition of total distance/total time. Which you need to understand.
 
  • #23
robphy
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"Average velocity" and "Average speed" are really time-weighted averages (not straight averages)

Here's average velocity... I just pasted the equations here. For details, read my posts.

https://www.physicsforums.com/threads/a-question-about-average-velocity.41732/post-303642 (calculus) from what should be the definition... then special case of constant acceleration
\begin{align*}
v_{avg}
&\equiv\frac{\int_{t_1}^{t_2} v\ dt}{\int_{t_1}^{t_2}\ dt}=\frac{\Delta x}{\Delta t}\\
&=\frac{\int_{t_1}^{t_2} (at)\ dt}{\int_{t_1}^{t_2}\ dt}\\
&=\frac{a\frac{1}{2}(t_2^2 - t_1^2)}{t_2-t_1}\\
&=\frac{1}{2}a(t_2+t_1)\\
&=\frac{1}{2}(v_2+v_1)
\end{align*}



https://www.physicsforums.com/threa...anics-kinematics-comments.813456/post-5106894 (algebra) from what should be the definition... then special case of constant acceleration

\begin{align*}
\vec v_{avg}
&\equiv \frac{\vec v_1\Delta t_1+\vec v_2\Delta t_2+\vec v_3\Delta t_3}{\Delta t_1+\Delta t_2+\Delta t_3} \\
&=
\frac{\Delta \vec x_1+\Delta\vec x_2+\Delta \vec x_3}{\Delta t_1+\Delta t_2+\Delta t_3}\\
&=
\frac{\Delta \vec x_{total}}{\Delta t_{total}}\\
\end{align*}

\begin{align*}
\vec v_{avg}
&\equiv
\frac{\Delta \vec x_{total}}{\Delta t_{total}}\\
&\stackrel{\scriptsize\rm const \ a }{=}\frac{\frac{1}{2}\vec a(\Delta t)^2 +\vec v_i\Delta t}{\Delta t}\\
&\stackrel{\scriptsize\rm const \ a }{=}\frac{1}{2}\vec a(\Delta t) +\vec v_i\\
&\stackrel{\scriptsize\rm const \ a }{=}\frac{1}{2}(\vec v_f -\vec v_i) +\vec v_i \\
&\stackrel{\scriptsize\rm const \ a }{=} \frac{1}{2}(\vec v_f+ \vec v_i)
\end{align*}
 
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  • #24
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It has to be 5.25km/h. The key is that the durations must be the same. You can't treat 5km/h for 1 minute with the same weight as 6 km/h for 59 mins. Because the second speed is sustained for 59 times longer.

The answer to your question is that the average of 5km/h and 6km/h is only 5.5 km/h when the two speed are sustained for the same duration.

You can see that from the definition of total distance/total time. Which you need to understand.
Yep totally
It was what was going in my head the but couldn't explain it. Thank you.

So to sum it up
If the speeds are of the same duration, then we can get a mean value out of it.

But if it's from different durations then the mean value cannot be equal to Ave speed
 
  • #25
symbolipoint
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Yes it's Lioric.
I know the solution to this problem. There is no doubt about that. X is 40 y is 75
But I'm just trying to make sense why the average speed version doesn't work. In my gut I sort of know why it cannot be that average speed is not V1+v2 /2. But I'd like to be able to explain it too. I just cannot find the words
lioric
I could not find any reason to look specifically for any average speed in analyzing and solving the problem. I did not actually finish trying to solve it; maybe in a further step, an expression for average speed may appear, or maybe not. What do you find when you take from the simplified system and solve for x and for y? Does an expression for average speed show?
 

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