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When is the average speed equal to the average of the speeds?

  • Thread starter Gjmdp
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1. Homework Statement
In which conditions the average speed equals the Average of the Speeds?

2. Homework Equations
Average speed= (Xf-Xo)/(tf-to)
Average of the speeds= (vf+vo)/2
3. The Attempt at a Solution
I tried to set up a formula for vf and vo, but it would be false to determine that
vf=xf/tf, because if vf=0, xf=2 and tf=1 ->vf=2/1 which is not 0. I tried also to set
vx=a*t + vo, but I'm getting nowhere. Please help :)

I know that it is the case when there is no acceleration. But I'm trying to figure if there is other cases.
 

NascentOxygen

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Is this going to be for the case of 2 different speeds, or a more general case?

Suppose it's at 2 steady speeds, and there is no acceleration (apart from one abrupt change to a new speed partway through the journey).

So you travel at a speed of v1 for some of the trip, at at a speed of v2 for the remainder.

►► What is an expression for the trip's average speed, in terms of those speeds?
 

haruspex

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Average of the Speeds
suggests to me we are to take the average of all of the set of speeds, might be more than two. But suppose it is just two, v1 and v2. Let the times at those speeds be t1 and t2. What equations can you write relating to average speeds?
 
Average speed= (Xf-Xo)/(tf-to) is general expression which is valid for both (constant as well as non constant acceleration). Whereas Average of the speeds= (vf+vo)/2 would only be equal when the graph between velocity and time would be a straight line (or alternatively when there would be a constant acceleration). As can be seen from the equation of Average of the speeds that it is the equation of finding the mid point of line segment.

Conclusion:
The only time Average speed= (Xf-Xo)/(tf-to) and Average of the speeds= (vf+vo)/2 would be equal when there is a constant acceleration.
 

NascentOxygen

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I see you have chosen to interpret this problem differently to how I believe was likely intended. Do you understand what I wrote in post #2 in this thread?
 

kuruman

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Average speed= (Xf-Xo)/(tf-to) is general expression ...
This expression is for the average velocity, not the average speed. Average speed is equal to total distance covered (as in recorded by a car's odometer) divided by the time interval it takes to cover that distance.
 

haruspex

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This expression is for the average velocity, not the average speed. Average speed is equal to total distance covered (as in recorded by a car's odometer) divided by the time interval it takes to cover that distance.
Gjmdp did not define Xf and X0.
The only time Average speed= (Xf-Xo)/(tf-to) and Average of the speeds= (vf+vo)/2 would be equal when there is a constant acceleration
It doesn't sound from the question as though we are to consider an infinity of speeds, which rules out any smooth accelerations. As Nascent wrote, it would be a finite set of constant speeds.
Besides, there are many ways in which the two averages could happen to be the same.
 

kuruman

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Gjmdp did not define Xf and X0.
I agree, but the numerator should be the absolute value of the difference otherwise it admits the possibility of a negative ratio which cannot be a speed.
 

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