Lot of students gets confused while finding out the **average speed**, when various distances are travelled with different speed. Say for example, trip to Agra from Delhi is made at an average speed of 40 km/hr and the trip back at an average speed of 60 km/hr. Find their average speed for the entire trip. (Hint: It’s Not 50 Km/hr)

Rule: If a person travels a distance at a speed of x km/hr and the same distance at a speed of y km/hr, then the average speed during the whole journey is given by 2xy/x + y km/hr.

Or, If a person travels half the distance at a speed of x km/hr and the other half at a speed of y km/hr, then the average speed during the whole journey is given by 2xy/x + y km/hr.

So answer to the above question:

(2*60*40)/(60+40) = 48 Km/hr

If a person travels three equal distances at a speed of x km/hr, y km/hr and z km/hr respectively, then the average speed during the whole journey is 3xyz/xy + yz + xz km/hr.

This problem can also be dealt with assuming some hypothetical distance.

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We can also explain this with H.P ( Harmonic progressions)

if 2 equal distance are covered with speeds a, and b

Then the avg speed is harmonic mean of a and b.

so avg speed = 2/(1/a + 1/b)

This can be extended to any number of distances

So if you have a person who is going with speeds a,b,c,d…. over equal distances, d,d,d,… upto n

then the avg speed will be :

n/ ( 1/a + 1/b + 1/c + 1/d + ……………)

and so on..

This can be used to find avg speed over 3 equal distance. if speeds are a,b,c, for 3 equal distances, then avg speed will be:

H.M. of a,b,c, ( HM means harmonic mean)

= 3/ (1/a + 1/b + 1/c )

Thanks Ravi 🙂

nice work