A body travels from A to B at 40 m/s and from B to A at 60 m/s. Calculate the average speed.

Can't we just add the given speeds and divide by 2 to get the average speed? This way answer would be 50 m/s In my book, however, they calculated total time taken and distance covered separately and then determined the average speed. The solution they reached is 48 m/s. Can someone please explain the difference?

  • $\begingroup$ Look up the definition of average speed in kinematics. $\endgroup$
    – nasu
    Jul 24 at 15:29
  • 3
    $\begingroup$ Note to homework close voters: this question is technically asking the conceptual question of why average speed is not the average of the given speeds. $\endgroup$ Jul 24 at 15:33

You cannot just take the average of $40$ and $60$ because the body spends longer going from A to B than it does returning from B to A. You can only use the average of the speeds if the time spent at each speed is the same.

Suppose A is $240$ metres from B. Then the time taken to go from A to B is $6$ seconds and the time taken to return from B to A is $4$ seconds. So the body has travelled a total distance of $2 \times 240 = 480$ metres in a total time of $10$ seconds. So its average speed is ...

If the body travelled at $40$ m/s for 6 seconds and then at $60$ m/s for 6 seconds then it would have travelled $600$ metres in $12$ seconds and its average speed would be $50$ m/s. But now it has not covered equal distances in those two time periods.


You have think about what you are adding/averaging. You will generally find that the process can be linear, geometric, or harmonic. Rates are typically linear, so if a nucleus can decay in 2 ways, say 40Bq to "A" and 60 Bq to "B", the total decay rate is just the linear sum:

$$ \omega_{tot} = \omega_A + \omega_B = 40{\,\rm Bq} + 60{\,\rm Bq}= 100{\,\rm Bq}$$

Now I can rewrite the rate, $\omega$, as an inverse of the decay time ($\tau$):

$$ \frac 1 {\tau_{tot}} = \frac 1 {\tau_A} + \frac 1 {\tau_B} = 40{\,\rm Bq} + 60{\,\rm Bq}= 100{\,\rm Bq}$$

So, had I phrased the question in terms of time: "The state decays to A every 25 ms and to state B every $16\frac 2 3$ ms, what's the average time for any decay?", you would add the times harmonically:

$$ \tau_{tot} = \frac{1}{{\frac 1 {\tau_A} + \frac 1 {\tau_B}}}=\frac{\tau_A\tau_B}{\tau_A+\tau_B}= \frac{0.000416\ldots\,{\rm s^2}}{0.0416\ldots\,{\rm s}}=0.01\,{\rm s}$$

This comes up in elementary circuits, where series resistors and parallel capacitors add linearly (which is clear from their physical construction), while parallel resistors and series capacitors add harmonically.

So what's speed? Well, it's usually called a rate of change of position, and sometimes redundantly used in "...traveling at a high rate of speed" in accident reports, so one may think it's a rate and should add linearly. The problem is the units we use, and the fact that it is a rate of distance, not time.

Had the problem been stated as a time required to cover a fixed distance:

"An object travels from A to B, covering a meter in $t_{AB}=\frac 1 {40}^{\rm th}$ of second, and B to A in $t_{BA}=\frac 1 {60}^{\rm th}$ of a second, what's the average time required to cover 1 meter?"

we could average linearly:

$$ t_{av} = \frac 1 2 (t_{AB}+t_{BA}) = \frac 1 2 \left(\frac 1{40}{\,\rm s/m}+\frac 1{60}{\,\rm s/m}\right) =\frac 1 {48}{\,\rm s/m}$$

but that is not how speed is commonly discussed. Rather, we talk about the distance covered in a fixed time, and the velocities must be averaged harmonically:

$$ \frac 1 {v_{av}} = \frac 1 2\left(\frac 1{v_{AB}} + \frac 1{v_{BA}}\right)=\frac 1 2 \left(\frac 1{40{\,\rm m/s}}+\frac 1{60{\,\rm m/s}}\right)=\frac 1 {48{\,\rm m/s}}$$


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