## Average Speed

Consider the following situation: A train is travelling at `50 kmph` to cover a distance in between stations A and B .In its further journey, train took a speed of `70 kmph` to reach station C from station B. Find the average speed of the train in the total journey from station A to station C? You may easily reach a (**obviously wrong!!!**) conclusion; that the average speed of the entire journey `= (50+70)/2 =` ~~60 kmph~~. What is wrong with this conclusion?

In this question , the distance between stations are not specified, hence it is not possible to find a unique solution for the question. The above answer, ~~60 kmph~~, is actually **average of speeds** not **average speed**.

Suppose in the above example if distance between Station A and station B is 150 km and distance between Station B and C is 140 km. then,

Time taken to cover AB `= 150/50= 3 hrs`

Time taken to cover BC `= 140 /70= 2 hrs`

Total Distance (AB + BC) `= 150 + 140 = 290 km`

Total time taken for journey from A to C `= 3 + 2 = 5 hrs`

Average speed `= 290/5= 58 kmph`

### Average Speed of Round Trip: When distance covered in to and fro segments of journey are same.

We can consider the situation in two ways. In a to and fro (round trip) journey both the distances are equal, i.e. AB = BA. So we can consider a similar situation in the form of a series trip such as A to B then B to C. If AB = BC (distances are equal) then the situations are similar.

Let the distance between A and B = `d km`, `:.` Distance between B and C = `d km`

Time taken for trip A to B = `d/u hrs` (`u` is the speed with which A to B is covered)

Time taken for trip B to C = `d/v hrs` (`v` is the speed with which A to B is covered)

We know that, `"Average speed" = "Total distance" / "Total time"`

`-> (d + d)/ (d/u + d/v) = (2uv)/(u+v) kmph`

**Note:**Harmonic mean of any two quantities u and v `= (2uv)/(u+v)`, ie. in the above context, average speed of the entire round trip is harmonic mean of the speeds.

### Average Speed: When distance covered in two segments of the journey are different.

Let's consider following situation: Cities A, B and C situated as the three vertices of an equilateral triangle, i.e. the cities are equidistant to each other and connected by straight line roads such as the sides of an equilateral triangle. If a person starts his journey from A to B at 'x' kmph, then B to C at 'y' kmph and further C to A at 'z' kmph then stopped. What is the average speed of this entire round trip?

A car covered AB at a constant speed of 'u' kmph and BC at a constant speed of 'v' kmph. Then

`"Average speed of the entire journey from A to C via B" = ((x + y)uv)/(xv + yu) kmph`

### Average Speed and Average of Speeds

Let's look at example to understand this concept in detail.

Let the speed from B to C `= p kmph`

Average speed `= (2 xx 70 xx p)/(70 + p)`

Average of the speeds `= (70 + p)/2`

Now, requirement is `"Average speed" > "Average of the speeds"`

`-> (2 xx 70 xx p)/(70 + p) > (70 + p)/2`

`-> (2 xx 70 xx p) xx 2 > (70+p)^2`

`-> p^2 - 140p + 4900 < 0`

`-> (p - 70)^2 < 0`, which is not possible since square of a natural number is always greater than or equal to 0.

*This means, average of speeds will be always greater than or equal to average speed.*