Average velocity problem~From Jay Orear book.

AI Thread Summary
The discussion revolves around calculating average velocity when a car travels two distances, x1 and x2, at velocities v1 and v2, respectively. The average velocity should be weighted by the distances traveled, meaning longer distances contribute more to the average. The correct formula involves multiplying each velocity by its corresponding distance, summing these products, and dividing by the total distance. Participants clarify that "weighting factors" refer to the distances traveled at each velocity, which is crucial for this calculation. The problem is sourced from Jay Orear's book, highlighting the distinction between average velocity by time and average velocity by distance.
Biggy_G
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Homework Statement


Car passes distance x1 with velocity v1
further it passes distance x2 with velocity v2. Find average velocity (averaged by distance), at that consider x1 and x2 as weighting factors.

Homework Equations



Vaverage=(v1t1+v2t2)/(t1+t2)

The Attempt at a Solution



I used substitution t=x/v and replaced t's in the formula and tried to simplify, but couldn't get correct answer.

I will be grateful for anyone's help.
Thanks
 
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Biggy_G said:
Find average velocity (averaged by distance), at that consider x1 and x2 as weighting factors.

What do you mean by weighting factors?

Other than that averge vel=(v1*t1+v2*t2)/t1+t2 doesn't seem wrong from any angle...
 
Biggy_G said:

Homework Statement


Car passes distance x1 with velocity v1
further it passes distance x2 with velocity v2. Find average velocity (averaged by distance), at that consider x1 and x2 as weighting factors.

Homework Equations



Vaverage=(v1t1+v2t2)/(t1+t2)

The Attempt at a Solution



I used substitution t=x/v and replaced t's in the formula and tried to simplify, but couldn't get correct answer.

I will be grateful for anyone's help.
Thanks

If they want the velocity averaged by distance, then the distances become the weighting factors (so that more "weight" is given to longer distance stretches in the figuring of an average). Multiply each velocity by its "weighting factor" (which is the distance that it traveled at that speed), sum them up, and divide by the sum of the weights.
 
Frankly Gneill i did not exactly get you...
I have done some physics but have never come across this term "weighing factors" and i always thought that average velocity is always total displacement by total time...
Ok i will search this term up in google and then i will try and get back to you...
 
Thanks gentlemen for replies,
The thing is I found this problem in Jay Orear's book. Chapter 2 Excercise 5. I am reading the book in Russian and I think that I translated Weighting Factor correctly. When they say Average velocity, mostly authors mean Velocity averaged by time. However, as they say, weighted by distance average velocity is used in Hydrodynamics.
 
gneill said:
If they want the velocity averaged by distance, then the distances become the weighting factors (so that more "weight" is given to longer distance stretches in the figuring of an average). Multiply each velocity by its "weighting factor" (which is the distance that it traveled at that speed), sum them up, and divide by the sum of the weights.

Exactly Correct answer :) Thanks
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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