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We have a car which has cruise control which can be set to a whole number of miles per hour. It can also inform you what your average speed was over a given journey.
My wife drove to her mother's house about 2 weeks ago, she texted me when she got there ands said: "I got here by driving at just 2 different speeds on the cruise control, and the average speed the car displayed was a whole number of miles per hour equal to the sum of the two speeds divided by two."
After her return along the same route she said: "for my return journey I used the same two speeds as before but I had to add a third cruise speed to go through some roadworks, which was lower than the other two. My average speed was a whole number of miles per hour again, equal to the sum of the three speeds, divided by three.".
What are the minimum, and the maximum time (in whole minutes) that her return journey might have taken? (ignore acceleration and deceleration)
Originally posted by @iamatigerThere is a common trap here.Mph is a ratio in itself so for example 1 hr at 20mph and 1hr at 10mph is not an average speed of 15mph.
We have a car which has cruise control which can be set to a whole number of miles per hour. It can also inform you what your average speed was over a given journey.
My wife drove to her mother's house about 2 weeks ago, she texted me when she got there ands said: "I got here by driving at just 2 different speeds on the cruise control, and the average ...[text shortened]... whole minutes) that her return journey might have taken? (ignore acceleration and deceleration)
The formula is 2ab/(a + b) which is 13.3mph.
For 3 speeds the formula is 3abc/(ab + bc + ca).
Not sure how to work out the answer to the poser but I expect you have to assume a certain distance travelled and time taken
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The necessary info is there. You do not have to guess anything. The equation that I use for average speed S is S = (t1.s1 + t2.s2)/(t1 + t2)
where t1 is the time for which the car was travelling at s1 etc.
The wording of the question is intended to mean that the car calculates the average speed as above and by coincidence, in the first case, it was equal to the sum of the two speeds divided by 2. similarly by coincidence in the second case it was equal to the sum of the three speeds divided by 3
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Originally posted by @vendaHmm, if you go at 20 mph for 1 hour, and 10mph for 1 hour, you have travelled 30 miles in two hours, so your average speed IS 15mph in that case, because the times are equal.
There is a common trap here.Mph is a ratio in itself so for example 1 hr at 20mph and 1hr at 10mph is not an average speed of 15mph.
The formula is 2ab/(a + b) which is 13.3mph.
For 3 speeds the formula is 3abc/(ab + bc + ca).
Not sure how to work out the answer to the poser but I expect you have to assume a certain distance travelled and time taken
Originally posted by @iamatigerYes that sounds right.
Hmm, if you go at 20 mph for 1 hour, and 10mph for 1 hour, you have travelled 30 miles in two hours, so your average speed IS 15mph in that case, because the times are equal.
I must be forgetting something.
I'll have to dig out my old books and see what I thought remembered refers to.
Probably something sililar
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There is a bit of an old teaching trick puzzle that says..
A race track is one mile around. If you do the first lap at 30 mph, what speed must you do the 2nd lap to have a 60 mph average?
The 'obvious' answer is 90 mph but, if course, it is infinity, as it took 2 mins to do lap 1, and that is all the time you have to do 2 miles at 60 mph
Interesting puzzle. Will try to find time to look into it
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Originally posted by @vendaHi Venda, thanks to blood on the track’s post I have realised your formula gives the average speed when the same distance is driven at each speed. The formula I gave is for the more general case when different distances may be driven at each speed: Average_speed = total_distance/total_time
There is a common trap here.Mph is a ratio in itself so for example 1 hr at 20mph and 1hr at 10mph is not an average speed of 15mph.
The formula is 2ab/(a + b) which is 13.3mph.
For 3 speeds the formula is 3abc/(ab + bc + ca).
Not sure how to work out the answer to the poser but I expect you have to assume a certain distance travelled and time taken
Originally posted by @iamatigerThanks.I'd not got round to digging the books out so you've saved me a job.
Hi Venda, thanks to blood on the track’s post I have realised your formula gives the average speed when the same distance is driven at each speed. The formula I gave is for the more general case when different distances may be driven at each speed: Average_speed = total_distance/total_time
I suppose i could have looked on the net as that seems to be the main source of info.these days
Originally posted by @iamatigerAs I see things:
We have a car which has cruise control which can be set to a whole number of miles per hour. It can also inform you what your average speed was over a given journey.
My wife drove to her mother's house about 2 weeks ago, she texted me when she got there ands said: "I got here by driving at just 2 different speeds on the cruise control, and the average ...[text shortened]... whole minutes) that her return journey might have taken? (ignore acceleration and deceleration)
D=s1*t1+s2*t2 (D distance, s Speed, t time, 1 and 2 respective Speeds)
with sa=(s1+s2)/2
yielding sa*(t1+t2)=s1*t1+s2*t2 and thus t1=t2
This time we will call ta
Then we have
D=s1*t3+s2*t4+s3*t5
with sb=s1+s2+s3, which can only be true with t3=t4=t5
This is tb
So we have at constant D:
sa*ta = sb*tb or one equation with four unknowns...
applying the whole number thingie we obtain Minimum Speeds: s1=6 mph, s2= 2 mph, s3= 1 mph (since s1+s2+s2 should be disible by 3 and s1+s2 by 2 and s1>s2>s3).
This would yield: 4ta=3tb.
Thiking of whole numbers again (and minutes being asked) we can say that the minimim numbers are ta=3 and tb=4 minutes. Maximum is of course not defined (I can always calculate a bigger ta to tb, regardless of starting conditions)
The question is why woud any bofy drive 3 minutes at 4 Miles per hour instead of Walking faster 😉
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Originally posted by @iamatigerYou are right with the first equation.
Nice work
However there is a problem in the quoted bit.
t3=t4=t5 does not follow
One way of calculating distance equals the other way of calculating distance
So
(s1+s2)(t1+t2)/2 = s1t1 +s2t2
Which rearranges and factorises to.
(s1-s2)(t1-t2) = 0
So either s1 = s2 (which the question excludes) or t1 = t2 (which is what you got)
However for the three speed case it is not so simple, we get
(s1+s2+s3)(t1 + t2 +t3)/3 = s1t1 + s2t2 + s3t3
This rearranges to
s1t2 +s1t3 + s2t1 + s2t3 + s3t1 + s3t2 = 2(s1t1 + s2t2 + s3t3)
But this doesn’t factorise like the two speed case did.
Originally posted by @iamatigerThat is exactly the same for calculating parallel resistance in electronics or very close. That is R1 * R2/ R1+R2 (for two resistors) for more it would be R total = 1/Rtotal= 1/r1+1/r2+1/r3 and so forth so 3 ohms + 2 ohms+1 ohm = 0.923 ohms ( you invert the answer of 1/r1,etc.) which in this case is about 1.083 inverted to 0.923.
The necessary info is there. You do not have to guess anything. The equation that I use for average speed S is S = (t1.s1 + t2.s2)/(t1 + t2)
where t1 is the time for which the car was travelling at s1 etc.
The wording of the question is intended to mean that the car calculates the average speed as above and by coincidence, in the first case, it was eq ...[text shortened]... larly by coincidence in the second case it was equal to the sum of the three speeds divided by 3
Originally posted by @vendaThe harmonic mean?
Yes that sounds right.
I must be forgetting something.
I'll have to dig out my old books and see what I thought remembered refers to.
Probably something sililar
http://www.statisticshowto.com/harmonic-mean/
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Should I give some hints? Only normal algebra is required for this. Equations and inequalities. Is anyone still trying it? A reminder:
My wife’s first journey took 1 hour. She drove at two different speeds. Her average speed was the sum of her two speeds divided by two.
My wifes second journey (over the same distance) was at 3 different speeds, two of them were the same as last time, and the new speed was the lowest. Her average speed was the sum of her three speeds divided by 3.
What are the minimum and maximum times my wife’s second journey could have taken?