Average speed with an unknown distance

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The discussion centers on calculating average speeds for a trip from Place A to Place B and back, with varying speeds during each leg. The average speed from A to B is determined to be 80 km/h, calculated by averaging the speeds for equal time segments. For the return journey from B to A, the average speed is found to be 75 km/h, based on substituting a distance value and calculating time taken at each speed. The overall average speed for the entire trip can also be derived using the average speeds from both legs of the journey. The calculations demonstrate the importance of understanding how distance and time interact to determine average speed.
yorick
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[SOLVED] Average speed with an unknown distance

You travel from Place A to Place B, half the time at a speed of 60km/h and the other half at 100km/h. On the return journey, you travel half the distance at 60km/h, and the other half at 100km/h.

a) What is your average speed from A to B?

b) What is your average speed from B to A (return journey)?

c) What is your average speed for the entire trip?



a) \overline{Sp} = (60 + 100)/2 = 80km/h... Easy enough.


b&c)
I just don't know where to start with this.
I tried a displacement vs. time graph but without any actual values for the displacement I didn't know what to do.

I'm not necessarily after an answer for this, but a nudge in the right direction would be greatly appreciated.

Thanks,
Yorick.
 
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Okay so I've been playing around a bit more, substituting values for displacement.

This is for b)

Letting x=100km,
I travel 50km @ 60km/h = 50min = 5/6 h
and 50km @ 100kmh/h = 30min = 1/2 h

Then \bar{Sp} = Distance traveled / (t_{2} - t_{1}) right?

So t2 would be 5/6 + 1/2, t1=0

then \bar{Sp} = 100 / (4/3)
=75 km/h




Let x=200km
100km @ 60km/h = 100min = 10/6 h
100 km @ 100km/h = 1 h

then \bar{Sp} = 200 / (16/6)
=75km /h

Valid reasoning?
PS sorry for bad Latexing, my subscripts came out as superscript so I gave up.
 
Indeed, it is possible to calculate out .
t= 1/2s/60 +1/2s/100= s/120+s/200
v=s/t = s/(s/120+s/200) = 75km h^-1

c) just easy , A->B = the same distance.
you have the average speeds for both sides...
 
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