Consider a car accelerating from 0 to 100 m/s. Is there a point where his instantaneous velocity equals his average velocity?

If we assume constant acceleration, his average velocity will be 50 m/s, evidently there will come a time where his instantaneous velocity equals 50 m/s since it has to accelerate to 100 m/s and 50<100.

But what if we drop the assumption of constant acceleration? How can I show that the average velocity won't cross 100 m/s. I mean it is obvious but how can I determine that only from the definition of average velocity $$v_{avg}=\frac{\Delta x}{\Delta t}$$

How can I show that the average velocity won't cross 100 m/s?

If the 100 m/s is the end-speed, then the average can easily be larger.

• For constant acceleration, the speed grows steadily up until this end-value. And the average is smaller, yes.
• But if acceleration is unrestricted and can be anything along the way, the speed and thereby also the average speed, can as well be any value along the way - also larger than 100 m/s.

ADDITION because of @ClauseChuber's comment: If the "car [is] accelerating from 0 to 100 m/s" it never overcomes this end-speed of 100 m/s. But does the average? No, the average speed also stays below the end-value no-matter what happened up until then and the explanation is actually purely mathematical. Start with the usual average formula:

$$v_{avg}=\frac{\sum v}{n}$$

Each value of speed that the car attains while accelerating are summed together and divided with the number of values.

If we write it more expanded, it is a bit easier to grasp:

$$v_{avg}=\frac{v_1+v_2+v_3+v_4+\cdots+v_n}{n}$$

This corresponds to a computer measuring the speed many, many times and getting $n$ values during the drive. Now, ideally you will of course have infinitely many values to sum (which would then be denoted as an integral), but the following logic is quite intuitive:

• If all the values are equal (if the car reaches the max-speed right-away), then adding them all is just the same as multiplying with $n$: $$v_{avg}=\frac{v+v+v+v+\cdots+v}{n}=\frac{nv}{n}=v$$ And the average then equals this max-value.
• If just one of the values is a tiny bit less than all the others, the numerator will be a tiny bit smaller. So, the average will be a tiny bit smaller than $v$.
• If more values are a bit less, then the numerator just gets even smaller. And the average is even smaller than $v$.

In the end, this pattern illustrates the idea that the largest value that appears over the duration is the highest possible value for the average. Because, if all values were the same as this, the average would be equal to it. This is of course not the case, and the average is smaller.

• He seems to assume that the velocity remains smaller that 100m/s: "car accelerating from 0 to 100 m/s". – user130529 Oct 6 '16 at 17:21
• @claudechuber I see and I believe you are right. A mathematical explanation of this is added. Thanks – Steeven Oct 6 '16 at 18:02
• So in statistical terms average velocity would be the expectation of the velocity distribution, and since the expectation of $v=100~m/s$ is not 1, and since for all $v>100~m/s$ expectation is zero (meaning those velocities not being possible), we conclude that the average velocity can't be equal or greater than 100 m/s ? – bonehead Oct 6 '16 at 18:05
• @ahra The statistical viewpoint sounds correct, yes, but I'm not an expert there. – Steeven Oct 6 '16 at 19:01

The answer to the first question is given by the Intermediate value theorem.

Regarding the second question: the average velocity is less or equal to the maximum velocity (property of the integral), and equality occurs only if the velocity is constant except on a subset of null measure.

• That addresses the fact that the instantaneous velocity will equal the average velocity at some point. But that doesn't in any way answer the question whether the average velocity can be greater than the maximum instantaneous. – bonehead Oct 6 '16 at 16:57
• I missed your second question. The average velocity is less or equal to the maximum velocity (property of the integral), and equality occurs only if the velocity is constant except on a subset of null measure. – user130529 Oct 6 '16 at 17:05

Calculate the outputs for the beginning and end of the interval, add them and divide by the interval you will get the average slope. Take the derivative of the function, set it equal to the average slope number and solve. If you get a real number that exists within the interval then that is where the current slope of the function equals it's average slope.

• You are assuming constant acceleration whereas that is not mentioned anywhere. – bonehead Oct 6 '16 at 18:09
• Why would you say i'm assuming constant acceleration? – Yogi DMT Oct 6 '16 at 18:11
• $v_{average}=\frac{v_{final}+v_{initial}}{2}$ holds only for $a=const.$ since in that case $v(t)$ is a linear function. – bonehead Oct 6 '16 at 18:13
• Not true, it holds for when a isn't constant as well. – Yogi DMT Oct 6 '16 at 18:16
• Can you provide some source for that statement? Source for mine: pokit.org/get/?2123a0dcaaf76b619fa41a07882522ad.jpg (Serway and Jewett - Physics for Scientists and Engineers, 6th edition) – bonehead Oct 6 '16 at 18:17