Average speed with an unknown distance

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SUMMARY

The discussion focuses on calculating average speeds for a journey from Place A to Place B and back, with specified speeds of 60 km/h and 100 km/h. The average speed from A to B is determined to be 80 km/h, calculated using the formula for average speed based on time spent at each speed. For the return journey from B to A, the average speed is calculated to be 75 km/h using a displacement of 100 km and the respective travel times. The overall average speed for the entire trip is confirmed to be 75 km/h, as both segments of the journey cover the same distance.

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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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