# Deriving the Formula for Average Speed (Same distance).

Let me start of by specifying the question:

A and B are two towns. Kim covers the distance from A to B on a scooter at 17Km/hr and returns to A on a bicycle at 8km/hr.What is his average speed during the whole journey.

I solved this problem by using the formula (since the distances are same):

$$\text{Average Speed (Same distance)} = \frac{2xy}{x+y} = \frac{2\times17\times8}{17+8} =10.88 \text{Km/hr}$$

Now I actually have two questions:

Q1- I know that $$Velocity_{Average}= \frac{\Delta S }{\Delta T}$$ Now here does $$\Delta S$$ represent $$\frac{S_2+S_1 }{2} \,\text{or}\, S_2-S_1 ?$$

Where S2 is the distance covered from point A to point B and S1 is the distance covered from point B to point A

Q2. How did they derive the equation: $$Velocity_{Average(SameDistance)} = \frac{2xy}{x+y}$$

Could anyone derive it by using $$Velocity_{Average}= \frac{\Delta S }{\Delta T}$$

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If one traveled distance $d_k$ at speed $v_k$, this took time $t_k=\dfrac{d_k}{v_k}$. It took time $T=\sum\limits_kt_k$ to travel distance $D=\sum\limits_kd_k$ and the average speed $V$ solves $D=VT$, hence $$V=\frac{\sum\limits_kd_k}{\sum\limits_k\frac{d_k}{v_k}}.$$ In the particular case when there are $n$ distances which are all equal, one gets $V=\dfrac{n}{\sum\limits_{k=1}^n\frac1{v_k}}$, or $$\frac1V=\frac1n\sum\limits_{k=1}^n\frac1{v_k}.$$

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average speed= total distance/total time taken lat take 1 km up and down journey. spped while going x kmper hour, while coming y km per hour: then tot dist= up+down=1+1=2 km total time taken= (1/x) + (1/y) so avg speed=total distance/total time taken=2/((1/x) + (1/y) )=2xy/(x+y)

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Question 1 is impossible to answer since you haven't told us what you mean by the symbols $S_1$ and $S_2$.

For question 2, make believe the distance from $A$ to $B$ is $xy$ kilometers (where I take it $x$ and $y$ are the two velocities, although you didn't tell us that, either). Then the total distance travelled is $2xy$. The journey at speed $x$ takes $y$ hours, and the journey at speed $y$ takes $x$ hours, so the total time taken is $x+y$. Thus, the average speed is distance/time which is $${2xy\over x+y}$$

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Sorry for the lack of Info. I just fixed it.S2 is the distance covered from point A to point B and S1 is the distance covered from point B to point A – Rajeshwar Jul 1 '12 at 11:02
Good. Aren't $S_1$ and $S_2$ the same? Anything to say about my answer to Question 2? – Gerry Myerson Jul 1 '12 at 11:20
Yes they are at least in this scenario. I wanted to know what would delta applies to in general. I know delta is the difference. – Rajeshwar Jul 1 '12 at 11:28

Let the distance from the point A to point B be 100 miles. And if the car traveled at a speed of 40 M/hr. and reached back to the point A traveled from point B at speed of 60 M/hr. the average speed is 48 M/hr. 2ab/a+b. Here a and b are the speed. Therefore, 2x40x60/ 40+60.

I would like to explain it some other way. In the above example I told you let the distance be 100 Mile, from A to B. So,if 40 M/hr, it'll take 150 minutes to cover 100 Miles. Like, the same distance, from B to A it'll take 100 minutes if the seed is 60 M/hr. It 'll take 250 minutes to travel a total distance of 200 Miles. So, we can figure it out by an equation to get the average like this. 200x60/250 = 48.

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What's the simple derivation of average speed traveled in equal distance ? – Rajendran Karunagath Raman Apr 10 '13 at 18:31
1. Here $\Delta s$ means the total distance travelled and it is $ab+ba=2ab$. Since the distance between the town will remain same. If distance between A and B is $w$ the $\Delta s=2w$.
2. Yes, this equation can be derived by, $v_{avg}=\dfrac{\Delta s}{\Delta t}$. Let speeds are $x$ while going and y while returning. Time taken while going is $\dfrac{d}{s}=\dfrac{w}{x}$, and time taken while returning is $\dfrac{w}{y}$.

So, using $v_{avg}=\dfrac{\Delta s}{\Delta t}$,

$v_{avg}=\dfrac{2w}{\dfrac{w}{x}+\dfrac{w}{y}}=\dfrac{2xy}{x+y}$

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