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I drove to the beach yesterday, at a speed of only 10 miles per hour. I drove back at a faster speed, and when I did the math, I found my average speed for the round trip was 20 miles per hour. What was my average speed home? Am I lying? Why?

2007-08-11 04:46:09 · 5 answers · asked by Gary H 6 in Science & Mathematics Mathematics

5 answers

You are lying because the average speed = total distance/total time = 2d/(t1+t2) = 2/(1/v1 + 1/v2) ≤ 2/(1/v1) = 20 miles per hour. The "=" sign in "≤" only works when your driving speed on your way back home is infinitely large.
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Here d is the one way distance, t is the time, and v is the speed.

2007-08-11 04:56:47 · answer #1 · answered by sahsjing 7 · 0 0

You're lying, because it's impossible.

The only way to double your average speed is to transport home from the beach.

Let's say the beach is ten miles away. Then the distance there and back is 20 miles. You drove at 10 mph, so your trip there took an hour. You now want your average speed to be 20 mph, which means you would drive 20 miles in an hour. But 20 miles is the total distance, and you've already driven for an hour. Unless you can get home in zero time, there is no way for this to work.

2007-08-11 05:35:23 · answer #2 · answered by TychaBrahe 7 · 0 0

this could nicely be an uncomplicated one. x km at 60 km/hr supplies y = x/60 hr (time taken) and finished distance ÷ finished time = a hundred and twenty so ( x + w) / (y +z) = a hundred and twenty and all of us understand y in terms of x so (x+w) / (x/60 + z) = a hundred and twenty or x+w = 2x + 120z or w = x + 120z and since the question is what's w/z we replace w with x + a hundred and twenty z so w/z = (x+120z)/z = x/z +a hundred and twenty occasion enable's think of x = 60 km so y = 1hr giving a hundred and twenty = (60+w)/(a million+z) enable's think of z is two hr the cost w/z is 60/2 + a hundred and twenty = a hundred and fifty examine (60 + 3 hundred)/(a million+2) = 360/3 = a hundred and twenty yup. purely ensure that on the top of two hours at a hundred and fifty kph you finally end up "back". So, i assume the respond is any arbitrary speed extra advantageous than a hundred and twenty for a time which would be extra advantageous through fact the cost methods a hundred and twenty. Gee, you weren't foolishly assuming that the gap traveled to come back could be an analogous thereby requiring countless speed? thank you for the honest warning.

2016-10-02 02:48:06 · answer #3 · answered by courcelle 4 · 0 0

You are lying because in order to double your average speed you'd have to increase your return speed exponentially, to make up for the time you've already lost on the way there. However, the amount by which you can increase your speed is limited, either by traffic laws or laws of physics (or simply by the performance of your car), so it would be impossible to double your average speed.
Very nice trap though. Could fall into thinking the answer was 30 mph very easily. Definitely worth a star in my opinion

2007-08-11 04:56:22 · answer #4 · answered by Shadow 3 · 1 1

Youre trip going back is 30mph making it less than an hour to get to my moms house....30-10=20mph lol anyways an average of 30 and 10 is 20 and if youre reading this i have no idea what im typing right now while you r playing guitar hero 2! stop messing up on guitar hero 2 and stop messing up on sweet child of mine! Use a controller! Its better!

2007-08-11 05:12:03 · answer #5 · answered by Anonymous · 0 3

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