The second video on the time and distance series covers the concepts of Average Speed and Relative Speed.
In general average speed is assumed to be the summation of two or more speeds and is calculated that way, which is incorrect.
Average speed is actually total distance travelled / the total time taken.
Let us understand this using some examples.
Say a person travels from a point to another at a speed of 30km/h and returns at a speed of 20km/h given that the distance is ‘X’ kms. We know that average speed is
Total distance travelled/total time taken
2X/(X/20) + (X/30)
Which on solving is 24. Thus the average speed is 24km/h.
An alternate way to solve this question would be to determine the distance by taking the taking the LCM of the two speeds given. The time can then be ascertained on the basis of the distance and we don’t need to use variables to solve the question.
The two speeds given are 20km/h and 30km/h. Taking LCM of 20 and 30 we get 60. Hence 60kms is the distance. Thus the time taken in going can be calculated as follows
Time = distance/ speed
=> 60/30 = 2hrs
While the time taken in returning is
=> 60/20 = 3 hrs
Average speed can thus be calculated as,
120/2+3 = 120/5
Which is 24 km/h.
In the second example we have a person who travels a triangular distance covering point A, B and C all of which are equidistant from each other. He travels from point A to B at the speed of 20 km/h from point B to C at the speed of 30 km/h and from point C to A at the speed of 40 km/h.
Let us assume the distance to be ‘X’ kms.
The average speed will be ascertained as follows:
3X/(X/20) + (X/30) + (X/40)
Which on solving results in 360/12 kms.
This can also be solved using the LCM approach.
The LCM of 20, 30 and 40 is 360.
The time taken to travel 20 kms will thus be
120/20 = 6 hrs.
Time taken to travel is 30 kms
120/30 = 4hrs
And the time taken to travel 40 kms
120/40 = 3hrs
Average speed is thus,
360/6 + 4 + 3 = 360/13 km/h.
Up till now we saw examples where the distance travelled was equal, we shall now look at problems where the distance travelled is unequal.
Say a person travels from point A to B, a stretch of 6 kms in three parts. He travels the 1st km in 30 mins, the next 2 kms in 45 mins and the last 3 kms in 45 mins.
Average speed is given by the total distance travelled / total time taken
The total time taken here is 30+45+45 = 120 mins, which on converting into hours comes to 2 hours.
=> 6/ 2 =3 km/h
Hence the average speed is 3km/h.
In this example we are given the different distances travelled at different speeds by a person. In order to ascertain the average speed we need to find the time taken in each of the distances separately. Given that the person travels a stretch of 160 kms in three parts. The first part of 60kms at a speed of 20km/hr. The second part of 80kms at the speed of 10 km/h and the last part of 20kms at 5 km/h.
The time taken in each case will be:
- 60/20 = 3hrs
- 80/10 = 8 hrs
- 20/5 = 4 hrs
Average speed can now be calculated as
160/ 3+8+4 = 160/ 15 = 32/3 km/h
Next comes the concept of relative speed. Relative speed is the speed of a body in comparison to the speed of another body.
We come across two kinds of questions in relative speed.
One is termed as an independent question where the speed of one body does not influence the speed of another body. For instance if a train A travels from a point at a speed of 30 km/h it will neither affect nor be affected by the speed of another train B travelling at a speed of 20km/h from the opposite direction. Similarly the speed of a man on the road will neither affect nor be affected by the speed of a thief running on the same road.
While in a dependent question the speed of one body affects the speed of another body. For instance a boat on a running stream is bound to be affected by the speed of the stream. Likewise a person trying to climb up an escalator will be influenced by the speed of the escalator.