What is the average speed of the driver.

[viren_t2005]

A car travellinf at a uniform speed. The driver sees a milestone showing 2-digit number.After traveling for an hour the driver sees another milestone with same digits in reverse order. After another hour the driver sees another milestone containing the same 2 digits. What is the average speed of the driver.

 

[Ouabache]

Welcome to the Physics Forums! You are right to ask your homework questions here in the HW section. In order to receive help, (besides stating your question clearly), you have to show what kind of work you've done and where you are getting stuck. There are many here who would be pleased to steer you in a successful direction.

It would be useful to you to read this https://www.physicsforums.com/showthread.php?t=94381 which is listed in this forum.

 

[Integral]

This is more of a brain teaser then it is a physics problem. You need to start by systematicly trying different combinations of numbers. Looks like you can assume that the numbers are increasing. From there it is a matter of trial and error to come up with a pair of numbers that satisfies the conditions of the problem.

 

[Tide]

There is no trial and error involved. You know the numbers are increasing based on the third observation which must be a three digit number whose first digit is 1. Because of this, the first digit in the first observation must be less than the second digit. The spacing (miles) between observations is fixed and you can deduce the spacing directly (50 miles) giving the speed. The question does not ask what the two unknown digits are and, fortuitously, they drop out.

 

[dicerandom]

I don't think that the speed can be 50. If we take x and y to be the two digits we're concerned with the first two statements give us:

10x + y + z = 10y + x
z = 9(y-x)

Where z is the speed. So, since x,y \in \mathrm{Integers}, z must be an integer multiple of 9.

I did the rest by guessing, I couldn't get an expression for the final condition that would give me anything useful. Having gotten the answer I can see why my expressions were giving me nonsense, but I can't think of a way to come up with an algebraic condition that doesn't immediately give away the answer via my explination as to why I chose it ;)

 

[Tide]

You also know that

z = 100 + 10 x + y - (10 y + x) = 100 + 9(x-y)

Now add this to the equation you already have to find z.

 

[dicerandom]

If I do that I'm back to z=50 again:

z = 100 + 9(x-y)
z = 100 - z
z = 50

Which contradicts the first equation I substituted in. This was the same equation I was trying to use when solving the problem on my own, and I think it makes an assumption about the placement of the digits in the final number which is not necessarily true. I'll send you a PM with more details since I don't want to give away the whole solution here so that the OP will still have something to work on :)

 

[Tide]

Dicerandom,

Yes, you are correct! I made a false assumption about the placement of the digits.

 

[ehild]

I don't think that the speed can be 50. If we take x and y to be the two digits we're concerned with the first two statements give us:
10x + y + z = 10y + x
z = 9(y-x)
Where z is the speed. So, since x,y \in \mathrm{Integers}, z must be an integer multiple of 9.
I did the rest by guessing, I couldn't get an expression for the final condition that would give me anything useful. Having gotten the answer I can see why my expressions were giving me nonsense, but I can't think of a way to come up with an algebraic condition that doesn't immediately give away the answer via my explination as to why I chose it ;)

You are right up to now. The speed should be an integer multiple of 9, y>x, and the third number starts with 1. Now this 1 can be among the numbers x and y, but y cancels out as y>x. So x can be equal to 1. Try out this possibility. You'll get something reasonable I promise :)

ehild