formula for average of two velocities on same distance


by pc2-brazil
Tags: average, distance, formula, velocities
pc2-brazil
pc2-brazil is offline
#1
Apr22-08, 01:56 PM
P: 205
1. The problem statement, all variables and given/known data
(translated from Portuguese) during the performance test of a new automobile model, the pilot goes through the first half of a track with an average velocity of 60 km/h and the second half with 90 km/h. what is the average velocity during the complete test, in km/h?
2. Relevant equations
[tex]v = \frac{2v_1v_2}{v_1 + v_2}[/tex]
3. Attempt at solution
initially, we weren't sure about how we would solve it; we thought it was through arithmetic average (average v between v1 and v2 = v = (v1+v2)/2).
but it was solved in class, and the teacher said we were supposed to use that special formula, 2*v1*v2/(v1 + v2) in all situations similar to this, that is, when the problem asks us to find the average of velocities when the distance traveled is the same, but with different velocities. look:
since the pilot goes through the same distance (half of the path) with different velocities, v1 = 60 km/h and v2 = 90 km/h, the general formula for this situation is:
[tex]v = \frac{2v_1v_2}{v_1 + v_2} \Rightarrow v = \frac{2(60)(90)}{60 + 90} = \frac{10800}{150} = 72 km/h.[/tex]
our question is: how do we obtain this formula, 2*v1*v2/(v1+v2)? why do we need to use this formula specifically, that is, why can't we just calculate the arithmetic average between 60 and 90, which would be (60 + 90) / 2 = 150 / 2 = 75?
thank you in advance.
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
lzkelley
lzkelley is offline
#2
Apr22-08, 02:02 PM
P: 277
anytime you see something of the form a*b / a+b -> its from adding things inversely (in my experience).
lets think of it in terms of time. total time t = t1 + t2; by v=d/t we can transform this to d/v = d1/v1 + d2/v2 (where v is average, and v1 and v2 are the 2 different velocities)
you also know that d1 = d/2 = d2
so you can solve for v.
pc2-brazil
pc2-brazil is offline
#3
Apr22-08, 02:22 PM
P: 205
so, from your reply, we can think of 2v1v2/(v1+v2) as:
2 * inverse of [tex]\frac{1}{v_1} + \frac{1}{v_2} = 2\frac{1}{\frac{1}{v_1} + \frac{1}{v_2}}=[/tex]
[tex]= \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}}[/tex]
but wouldn't this be the harmonic mean?

SheldonG
SheldonG is offline
#4
Apr22-08, 02:58 PM
P: 50

formula for average of two velocities on same distance


Quote Quote by pc2-brazil View Post
our question is: how do we obtain this formula, 2*v1*v2/(v1+v2)? why do we need to use this formula specifically, that is, why can't we just calculate the arithmetic average between 60 and 90, which would be (60 + 90) / 2 = 150 / 2 = 75?
thank you in advance.
Average velocity is the displacement divided by the time it takes to travel that displacement. Your problem doesn't give the distance or the time, but it does say that 1/2 of the displacement is traveled at each velocity. Since velocity is constant, D/2 = v1*t1 and D/2 = v2*t2. You want the average velocity, that is v_avg = D/total_time. So v_avg = D/(t1 + t2). If you work out the algebra from here, you will see where your teacher got the formula.
lzkelley
lzkelley is offline
#5
Apr22-08, 03:04 PM
P: 277
I'm not super familiar with harmonic means, but i think that in this case -> the harmonic mean of the distances will be the mean of the velocities? ... something to that extent.
1/2v1 + 1/2v2 = 1/v = (v2/2*v1*v2)+(v1/2*v1*v2) = ... combine fractions and invert both sides.
SheldonG
SheldonG is offline
#6
Apr22-08, 03:06 PM
P: 50
Quote Quote by pc2-brazil View Post
our question is: how do we obtain this formula, 2*v1*v2/(v1+v2)? why do we need to use this formula specifically, that is, why can't we just calculate the arithmetic average between 60 and 90, which would be (60 + 90) / 2 = 150 / 2 = 75?
thank you in advance.
Sorry, forgot to answer this bit. There is a period where the car is traveling at 60 and a period where it is traveling at 90. But since the change is not instantaneous, there will be a 3rd period where the car is traveling between 60 and 90. So the average of the end points will be too high in this case, since it ignores the middle part.
lzkelley
lzkelley is offline
#7
Apr22-08, 03:26 PM
P: 277
Quote Quote by SheldonG View Post
Sorry, forgot to answer this bit. There is a period where the car is traveling at 60 and a period where it is traveling at 90. But since the change is not instantaneous, there will be a 3rd period where the car is traveling between 60 and 90. So the average of the end points will be too high in this case, since it ignores the middle part.
Thats incorrect. The problem specifies that half the distance is average 60, and half is average 90. We can treat this as no acceleration; and velocities are constant in each domain and then change instantaneously inbetween.
The reason you can't take a direct average is because we want to know the average velocity over all time, while 60+90/2 would be the average over distance.
I.e. if the car spent half of the TIME going 60 and half of the TIME going 90, then the average would be 75.
Because the car spends more time going 60 (to cover the same distance as going 90), the true average will be lower than 75.
Does that make sense?
SheldonG
SheldonG is offline
#8
Apr22-08, 03:29 PM
P: 50
Yes, you're right. Thanks.
pc2-brazil
pc2-brazil is offline
#9
Apr22-08, 03:50 PM
P: 205
salutations,

lzkelley: working on your suggestion, we found:
[tex]\frac{d}{v} = \frac{d_1}{v_1} + \frac{d_2}{v_2}[/tex]
then, as [tex]d_1 = d_2 = \frac{d}{2}[/tex],
[tex]\frac{d}{v} = \frac{d}{2v_1} + \frac{d}{2v_2}[/tex]
then,
[tex]\frac{1}{v} = \frac{1}{2v_1} + \frac{1}{2v_2}[/tex]
summing the fractions:
[tex]\frac{1}{v} = \frac{v_2 + v_1}{2v_1v_2}[/tex]
and inverting:
[tex]v = \frac{2v_1v_2}{v_1 + v_2}[/tex] <-- solution.
a question: does it make sense to say this:
[tex]\frac{d}{v} = \frac{d}{2v_1} + \frac{d}{2v_2} = \frac{v_2d_1}{v_1v_2} = \frac{d_2v_1}{v_1v_2} = \frac{v_2d}{2v_1v_2} = \frac{v_1d}{2v_1v_2}[/tex]?
SheldonG: we tried to work on your suggestion:
[tex]\frac{d}{2} = \frac{v_1t_1}{1} = \frac{v_2t_2}{1}[/tex]
then:
[tex]d = 2v_1t_1 = 2v_2t_2; v_1t_1 = v_2t_2[/tex]
so:
[tex]v_1t_1 + v_2t_2 = \frac{d}{t_1 + t_2}[/tex]
is it right? we got stuck here.
thank you in advance.
SheldonG
SheldonG is offline
#10
Apr22-08, 03:57 PM
P: 50
Quote Quote by pc2-brazil View Post
salutations,

SheldonG: we tried to work on your suggestion:
[tex]\frac{d}{2} = \frac{v_1t_1}{1} = \frac{v_2t_2}{1}[/tex]
then:
[tex]d = 2v_1t_1 = 2v_2t_2; v_1t_1 = v_2t_2[/tex]
so:
[tex]v_1t_1 + v_2t_2 = \frac{d}{t_1 + t_2}[/tex]
is it right? we got stuck here.
thank you in advance.
You want to eliminate t. [tex] v_{avg} = D/(t_1 + t_2) [/tex]

And for each half: [tex] D/2 = v_1t_1\quad D/2 = v_2t_2 [/tex]

Solve these last equations for time, and then substitute them in the first equation.
pc2-brazil
pc2-brazil is offline
#11
Apr22-08, 04:14 PM
P: 205
SheldonG: from your reply,
[tex]t_1 = \frac{d}{2v_1}[/tex] and [tex]t_2 = \frac{d}{2v_2}[/tex]
then:
[tex]v = \frac{d}{\frac{d}{2v_1} + \frac{d}{2v_2}} = \frac{d}{\frac{v_2d + v_1d}{2v_1v_2}} = \frac{d \cdot 2v_1v_2}{v_2d + v_1d} = \frac{d \cdot 2v_1v_2}{d(v_2 + v_1)} = [/tex]
[tex] = \frac{2v_1v_2}{v_1 + v_2}[/tex] <-- solution.
thank you, all answers were very helpful.


Register to reply

Related Discussions
[SOLVED] Average speed with an unknown distance Introductory Physics Homework 2
average velocity with respect to distance Introductory Physics Homework 2
average distance at STP Introductory Physics Homework 5
Average power formula General Math 2
average distance between points on a circle General Math 2