Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Too Easy?

  1. Nov 13, 2006 #1
    My teacher gave me problems to do related to Optimization. Well, I did one of the questions, using basic geometry, and didn't involve calc at all. I am 100% I am wrong. Can you please tell me how to do this problem.

    John wants to get to the bus stop as quickly as possible. The bus stop is across a grassy park, 2000 feet west and 600 feet north of her starting position. Jane can walk west along the edge of the park on the sidewalk at a speed of 6 ft/sec. She can also travel though the grass in the park, but only at the rate of 4 ft/sec. What path will get her to the bus stop the fastest?

    The method i used was below to solve this problem

    x = vt

    2000 + 600 = 6t
    t = 433.3333

    I got 2088 below by using the phythoregeon theorm.

    2088 = 4t
    t = 522
  2. jcsd
  3. Nov 13, 2006 #2
    Create an equation for the total time taken including both of the ways. Also, make an equation relating one of the ways to the other...
  4. Nov 13, 2006 #3
    I am really confused. What equation is there to create when you just plug in the numbers like I did below?
  5. Nov 13, 2006 #4
    Well, you don't know how long the person will walk along the sidewalk, and how long they will walk along the grass.

    Here are a few possibilities on how the person could get across.

    Point is, you don't know how the person will distribute the amount walked on grass and the amount walked on sidewalk. So with two variables, say S and G (representing sidewalk distance and grass distance), can you make an equation which will symbolically tell you how long it takes to move along those paths?

    EDIT: Notice how going straight across the grass is the shortest way there. However, the person walks slower across the grass than across the sidewalk. So, it will be to the person's advantage to walk part of the way along the sidewalk, and then the rest diagonally through the grass.
  6. Nov 13, 2006 #5
    Wow, thanks, a freind of mines gave me a similar idea.

    So lets say the person walks x amount of feet west and then walks though the grass, so I am guessing this is the equation

    t = (x/6) + (600^2+(2000-x)^2)^(1/2)

    So then I take the derivitive of that equation and set it equal to 0. I only get 1 value of x when t' = 0. What do I do with that x value?
  7. Nov 13, 2006 #6
    Ok, what does that x value mean? What did you say it stood for? Could you possibly use it to calculate the total time? Where would you plug that in?

    EDIT: BTW, you may want to take a second look at your equation. The second part is the distance through the grass right? So make sure to take into account the 4m/s thing. hint hint
  8. Nov 13, 2006 #7
    Oh yea, lol the 4 I forgot to divide.

    Since, I only get 1 x value for t'=0, then I am guessing it has to be minimum. So, I plug the x into the original equation and I get 455 for time. But how does this answer help me choose which path the boy take? Walking on the sidewalk or across the grass?
  9. Nov 13, 2006 #8
    Oh I'm sorry. I didn't mean to write the thing about getting the total time. For some reason I thought that was what the question was asking for.

    So the question asks
    So what does x stand for? You could say how many feet west, and then say that you go straight from there to the bus stop, through the grass ___ feet.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook