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An object is traveling from a point $M$ to a point $N$ with a velocity $v_1 = 600 \,\rm km/hr$ and comes back with a velocity $v_2 = 400 \,\rm km/hr$. What is the average speed of the object?

Can we find the average speed, $v$, by using $$v= \frac12(v_1+v_2) $$ If not, why can't I?

The answer is different if I use the $v=s/t$ formula.

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    $\begingroup$ Hint: does the object spend the same amount of time travelling at those two speeds? or does one part of the journey take longer? $\endgroup$
    – innisfree
    Feb 11, 2015 at 16:48
  • $\begingroup$ The average speed is the total distance divided by the total time. If you write down the equation for the total time then do the division you'll find that in general the average velocity is not just the average of $v_1$ and $v_2$. $\endgroup$
    – John Rennie
    Feb 11, 2015 at 16:55
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    $\begingroup$ If $v_2$ were 0, would the average speed for the round trip be $300$ km/hr? $\endgroup$
    – DJohnM
    Feb 11, 2015 at 17:06
  • $\begingroup$ @innisfree of course it takes different amount of time. $\endgroup$
    – Atiqur Rahman
    Feb 12, 2015 at 4:09

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When you take the average of two quantities, you need to consider the "weighting". In this case, the time spent at each of the velocities matters, and becomes this "weight". In general, when you have a weighted average you multiply each value by its weight, and divide by the sum of the weights. When all the weights are $1$, that reduces to the familiar equation $\bar{x} = \sum{x}/n$

For this case, you could do

$$V_{av} = \frac{v_1t_1 + v_2t_2}{t_1+t_2}$$

This simplifies to $\frac{s}{t}$.

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  • $\begingroup$ i think u mean the frequency of the values. in this case its time, right?? $\endgroup$
    – Atiqur Rahman
    Feb 12, 2015 at 4:16
  • $\begingroup$ "Weight" as I used it above is sometimes "frequency". If you like here it is "the number of seconds that you travel at a velocity". If you observed the motion many times at random intervals you would end up with a similar equation (the object spends more time going slowly so that is the speed you will observe more frequently). Make sense? $\endgroup$
    – Floris
    Feb 12, 2015 at 6:03

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