Average of 2 Averages

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Randyyy

Junior Member
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May 14, 2020
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113
Hello Mathhelp,
This question has tripped me up a bit:
Uncle bob drives a car from point A to point B with an average velocity of 39km/h. At location of point B he makes some adjustments to his car that will allow it to now go at a faster pace. His trip back from point B to point A he now manages to keep an average velocity of 59km/h. What's his average velocity for the entire trip?
My thinking is that the displacement for both will be the same regardless of velocity because he will still be driving the length AB. I know the average velocity for both trips but I have no given time nor do I know anything about the displacement except that it is the same. How do I go about solving this issue?
 
Randyyy said:
Hello Mathhelp,
This question has tripped me up a bit:
Uncle bob drives a car from point A to point B with an average velocity of 39km/h. At location of point B he makes some adjustments to his car that will allow it to now go at a faster pace. His trip back from point B to point A he now manages to keep an average velocity of 59km/h. What's his average velocity for the entire trip?
My thinking is that the displacement for both will be the same regardless of velocity because he will still be driving the length AB. I know the average velocity for both trips but I have no given time nor do I know anything about the displacement except that it is the same. How do I go about solving this issue?
Click to expand...
Average velocity = (total distance)/(total time).
You need to assign variables to various items involved. Then express total distance and total time using those variables and data given in the problem. Some things will cancel and you should get a number as the answer.
 
lev888 said:
Average velocity = (total distance)/(total time).
You need to assign variables to various items involved. Then express total distance and total time using those variables and data given in the problem. Some things will cancel and you should get a number as the answer.
Click to expand...

I get the solution to be V=2301/49 which is about ~~ 47km/h which seems reasonable. Thanks!
 
S=v1t1,S=v2t2,V=s/t so V=2s/(t1+t2)
If we want we can now solve for t2 which will simply become 39t1/59 and S=39t1
We can continue to substitute to solve for V. t1 and t2 can both be rewritten to s/v1 and s/v2 and so then we have V=t1v1+t2v2/((s/v1)+(s/v2)) which can be simplified to 2301(39t1+59t2/98s and if we continue to substitute variables we end up at 2301(39s/v1)+(59s/v2)/98s which then gives us V= 2301/49 ~~ 47 km/h
 
Randyyy said:
S=v1t1,S=v2t2,V=s/t so V=2s/(t1+t2)
If we want we can now solve for t2 which will simply become 39t1/59 and S=39t1
We can continue to substitute to solve for V. t1 and t2 can both be rewritten to s/v1 and s/v2 and so then we have V=t1v1+t2v2/((s/v1)+(s/v2)) which can be simplified to 2301(39t1+59t2/98s and if we continue to substitute variables we end up at 2301(39s/v1)+(59s/v2)/98s which then gives us V= 2301/49 ~~ 47 km/h
Click to expand...
Seems too complicated. How I would do it:
V = 2S/((S/39)+(S/59)) = 2*2301/98
 
lev888 said:
Seems too complicated. How I would do it:
V = 2S/((S/39)+(S/59)) = 2*2301/98
Click to expand...
You´re right, why I did not jump to that is beyond me, I actually missed it. Your solution saves so much time and is way quicker. Nice catch indeed and thanks again for the guidance :)
 
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