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Why isn't average speed equal to (speed1+speed2)/2 ?

In certain situations that equation could work, but not in general.


As I understand, we are considering a situation when something has moved with a constant (or average) speed `V_1` for some time `t_1,` and then moved with a constant (or average) speed `V_2` for a time `t_2.` The average speed `V` has to be computed.


By definition, the average speed is the total distance traveled divided by the total time taken. In this case the distance is the sum of two distances, `V_1 t_1+V_2 t_2,` and the time is the sum of two times, `t_1+t_2.` So the correct value for the total average speed is


`V=(V_1 t_1+V_2 t_2)/(t_1+t_2) = V_1 (t_1/(t_1+t_2))+V_2 (t_2/(t_1+t_2)).`



The simpler formula you suggest is `V= V_1 (1/2) + V_2 (1/2).`


If `t_1=t_2,` then `t_1/(t_1+t_2)=t_2/(t_1+t_2)=1/2` and these formulas give the same result. If no, they are generally different. We can say that the speed with greater time taken makes greater contribution to the total average speed.

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