Optimum Speed

irish life

I got the following question:

• Calculate the optimum speed for cars in the tunnel.

Suggested approach: Try to calculate the number of cars per minute leaving the tunnel at speeds between 5 to 70 mph.

Basically there is a tunnel being built (one lane in each direction) and there has to be a safe stopping distance between cars in the tunnel. I won't waste my time explaining everything as I've worked out most.

I've worked out a formula to calculate the number of cars that will pass through the tunnel (a fixed point) at a certain speed. It is:

____1760v____
v^2 + 20v + 260

So, at 5mph there will be 22.86 cars passing through per minute. At 30mph there will be 30 cars passing through per minute. At 50mph there will be 23.4 cars per minute.

How do I calculate the optimum speed for cars in the tunnel? I can't go through every speed from 5-70 using the formula as there in an infinite amount of speeds.

Thanks.

irish life

I set up the formula and ran it in Hugs and used the hot or cold approach. The answer is 16.125mph (a total of of 33.685 cars pass through per minute). I still have to work out how I would get that 16.125mph though.

Sev

irish life wrote:

How do I calculate the optimum speed for cars in the tunnel

You mean find the speed that maximizes the number of cars that go through the tunnel per minute?

Well.. say y = 1760v/(v^2 + 20v + 260)


Well, you just differentiate y with respect to v and equate it with 0. You should get a quadratic in v, then solve for v. The answer you get represents the value for v at which the slope of the plot of y against v is at 0, ie a maximum.

irish life

Sev wrote:

Well, you just differentiate y with respect to v and equate it with 0. You should get a quadratic in v, then solve for v. The answer you get represents the value for v at which the slope of the plot of y against v is at 0, ie a maximum.

Thanks for the help Sev but I really don't know what any of that means. It does bring back memories of the Leaving Cert which I did 18 months ago but I forget it all. This question is for a Math's Modelling class in Computer Science btw.

What exactly does this mean: Well, you just differentiate y with respect to v and equate it with 0?

Thanks again.

Sev

irish life wrote:

Thanks for the help Sev but I really don't know what any of that means. It does bring back memories of the Leaving Cert which I did 18 months ago but I forget it all. This question is for a Math's Modelling class in Computer Science btw.

What exactly does this mean: Well, you just differentiate y with respect to v and equate it with 0?

Thanks again.

Oh right ok.. well

y = 1760v/(v^2 + 20v + 260)

If you drew a graph of y against x from like 0 to 70, youd get this curve, something like the one ive drawn here. So given any given velocity v, the height of the curve gives the number of cars that will come out of the tunnel per minute.



You want to find dy/dv, that quantity represents the slope of the graph at any given v. You know what I mean by slope right, its the tan of the angle that the line makes with the x axis at the point, the steeper the slope, the greater the magnitude of dy/dx.

So we need to differentiate y with respect to v. To do that im gonna use the quotient rule

dy/dv = d(a/b)/dv= [ b(da/dv) - a(db/dv) ]/b^2

Where a = 1760v and b = (v^2 + 20v + 260)

da/dv = 1760
db/dv = 2v + 20

so dy/dv = [ (v^2 + 20v +260)(1760) - (1760v)(2v + 20) ] / [v^2 + 20v + 260]^2

And we get...

dy/dv = -1760v^2 + 457600 / [v^2 + 20v + 260]^2

With this value.. dy/dv, you can plug any value of v into this equation and get the slope on that graph at that point.

As the graph suggets, we want to find the value of v for which dy/dv = 0

so set

dy/dv = 457600 -1760v^2 / [v^2 + 20v + 260]^2 = 0

and solve..

we can multiply both sides by [v^2 + 20v + 260]^2

which makes things a lot easier,
then we get

457600 -1760v^2 = 0

so v^2 = 457600/1760 = 260

so v = sqrt(260) = 16.1245

which is our answer.

irish life

Thanks a million Sev, thats a great help.