The AM-HM Inequality in Average Speed Calculations

Problem: The AM-HM Inequality in Average Speed Calculations

In kinematics, average speed is often misunderstood as the simple arithmetic mean of two speeds. However, the correct average depends on the condition of the journey. If a vehicle covers equal distances at two different speeds, the average speed is not the arithmetic mean. It is the harmonic mean.

Consider a vehicle traveling from point A to point B. It covers the first half of the total distance with speed ##v_1## and the second half with speed ##v_2##.

(a) Derive the expression for the average speed ##v_{\text{avg}}## over the entire journey.

(b) Using the Arithmetic Mean-Harmonic Mean inequality, prove that ##v_{\text{avg}}## is strictly less than the arithmetic mean

###v_{\text{mean}}=\frac{v_1+v_2}{2}###

provided ##v_1 \neq v_2##.

Worked Solution and Step-by-Step Explanation

Part (a): Deriving the Average Speed

The average speed is defined as:

###v_{\text{avg}}=\frac{\text{Total Distance}}{\text{Total Time}}###

Let the total distance be ##2d##. Then each half of the journey has distance ##d##.

The time taken for the first half is:

###t_1=\frac{d}{v_1}###

The time taken for the second half is:

###t_2=\frac{d}{v_2}###

Therefore, the total time is:

###t_{\text{total}}=t_1+t_2###

Substituting the two time expressions:

###t_{\text{total}}=\frac{d}{v_1}+\frac{d}{v_2}###

Taking ##d## common:

###t_{\text{total}}=d\left(\frac{1}{v_1}+\frac{1}{v_2}\right)###

Combining the fractions:

###t_{\text{total}}=d\left(\frac{v_1+v_2}{v_1v_2}\right)###

Now substitute this into the definition of average speed:

###v_{\text{avg}}=\frac{2d}{d\left(\frac{v_1+v_2}{v_1v_2}\right)}###

Canceling ##d## and simplifying:

###v_{\text{avg}}=\frac{2v_1v_2}{v_1+v_2}###

Part (b): Applying the AM-HM Inequality

The arithmetic mean of the two speeds is:

###\text{AM}=\frac{v_1+v_2}{2}###

The harmonic mean of the two speeds is:

###\text{HM}=\frac{2}{\frac{1}{v_1}+\frac{1}{v_2}}###

Simplifying the harmonic mean:

###\text{HM}=\frac{2v_1v_2}{v_1+v_2}###

But from Part (a),

###v_{\text{avg}}=\frac{2v_1v_2}{v_1+v_2}###

Therefore, the average speed for equal distances is the harmonic mean:

###v_{\text{avg}}=\text{HM}###

The AM-HM inequality states that for positive real numbers ##v_1## and ##v_2##,

###\text{AM}\geq \text{HM}###

Therefore,

###\frac{v_1+v_2}{2}\geq \frac{2v_1v_2}{v_1+v_2}###

Hence,

###v_{\text{mean}}\geq v_{\text{avg}}###

Proving the Strict Inequality When ##v_1 \neq v_2##

To prove that the inequality is strict when ##v_1\neq v_2##, consider the difference between the arithmetic mean and the harmonic mean:

###\text{AM}-\text{HM}=\frac{v_1+v_2}{2}-\frac{2v_1v_2}{v_1+v_2}###

Taking the common denominator ##2(v_1+v_2)##, we get:

###\text{AM}-\text{HM}=\frac{(v_1+v_2)^2-4v_1v_2}{2(v_1+v_2)}###

Expanding the numerator:

###\text{AM}-\text{HM}=\frac{v_1^2+2v_1v_2+v_2^2-4v_1v_2}{2(v_1+v_2)}###

Simplifying:

###\text{AM}-\text{HM}=\frac{v_1^2-2v_1v_2+v_2^2}{2(v_1+v_2)}###

Therefore,

###\text{AM}-\text{HM}=\frac{(v_1-v_2)^2}{2(v_1+v_2)}###

Since speeds are positive, ##v_1+v_2>0##. Also, if ##v_1\neq v_2##, then ##(v_1-v_2)^2>0##. Hence,

###\text{AM}-\text{HM}>0###

Thus,

###\text{AM}>\text{HM}###

Since ##v_{\text{avg}}=\text{HM}##, we finally get:

###v_{\text{mean}}>v_{\text{avg}}, \qquad \text{when } v_1\neq v_2###

Conceptual Significance for JEE and NEET

Students often calculate average speed incorrectly as:

###\frac{v_1+v_2}{2}###

This is only correct when the object travels at ##v_1## and ##v_2## for equal time intervals. It is not correct when the distances covered at the two speeds are equal.

For equal distances, the slower speed has a greater effect on the average because the object spends more time traveling at the lower speed. This is why the harmonic mean appears naturally.

Key Takeaways

1. Average speed must always be calculated from total distance and total time.

###v_{\text{avg}}=\frac{\text{Total Distance}}{\text{Total Time}}###

2. For equal distances, the average speed is the harmonic mean.

###v_{\text{avg}}=\frac{2v_1v_2}{v_1+v_2}###

3. For unequal speeds over equal distances, the harmonic mean is strictly less than the arithmetic mean.

###v_{\text{avg}}<\frac{v_1+v_2}{2}, \qquad v_1\neq v_2###

4. The equality case occurs only when both speeds are equal.

###v_{\text{avg}}=\frac{v_1+v_2}{2} \quad \text{only when} \quad v_1=v_2###

Final Conclusion

For a vehicle covering the first half of a journey at speed ##v_1## and the second half at speed ##v_2##, the average speed is:

###\boxed{v_{\text{avg}}=\frac{2v_1v_2}{v_1+v_2}}###

This is the harmonic mean of the two speeds. By the AM-HM inequality,

###\frac{v_1+v_2}{2}\geq \frac{2v_1v_2}{v_1+v_2}###

and the inequality is strict whenever ##v_1\neq v_2##. Therefore,

###\boxed{v_{\text{avg}}<v_{\text{mean}} \quad \text{for } v_1\neq v_2}###

Thus, when a body travels equal distances at different speeds, the correct average speed is always less than the arithmetic mean of those speeds.