Time & Distance Tutorial I: Basic Concepts, Average Speed & Variation of Parameters

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.

Average Speed
"Average speed of the entire journey" = "Total distance covered"/"Total time taken"

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

Average speed when distance covered is same for the round trip
If a car travel from A to B at a constant speed of 'u' kmph and the return journey from B to a at a constant speed of 'v' kmph. Then, "Average speed of this roundtrip" = (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?

Average speed when distance covered in two segments of the journey are different
If the ratio of the distances between A and B, B and C is x : y respectively.
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
Sourav ride his bike at a constant speed of 40 kmph for a trip from Mumbai to Thane and the return journey from Thane to Mumbai he keep a constant speed of 60 kmph. Find his average speed for the entire round trip?
Here it's a round trip journey and average speed = (2uv)/(u+v) = (2 xx 40 xx 60)/(40 + 60) = 48kmph

Average Speed and Average of Speeds

Relationship between average speed and average of speeds
When the segment wise distances are equal, "Average of the speeds " >= " Average speed"

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

Joy is always driving his car as per speed calculation. One day he selected three neighboring cities A,B and C for his trip Distance between A and B is same as that between B and C. He has driven from A to B at a speed of 70 kmph and he is planning to cover the distance between B and C at a speed such that the 'average speed' of the entire trip should be greater than the 'average of the speeds'. What should be his least required speed for B to C for satisfying the condition?
Speed from A to B = 70 kmph
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.
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