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Problem involving rates of change of distance.

  1. Jul 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Two cars start moving form the same place. One goes north at a rate of 50 mi/hr, while the other heads east at a rate of 30 mi/hr. At what rate is the distance between the two cars changing exactly two hours later?



    2. Relevant equations



    3. The attempt at a solution

    I thought that the equation for the velocity vectors is:

    r(x) = 30xi + 50yj

    and thus distance d two hours later is given by:

    d = √(30x² + 50x² )

    and

    dd/dt = ∂x/dt + ∂y/dt

    and after solving the equation I just plug in 2 for x and y to get what time the distance changes at the two hour mark.

    Is this right?
     
  2. jcsd
  3. Jul 11, 2011 #2
    Were you asked to do this using vectors ? You don't really need to use vectors to solve this.
     
  4. Jul 11, 2011 #3
    No, I wasn't. How would you solve it without vectors?
     
  5. Jul 12, 2011 #4
    You can do it with and without vectors. To do it without vectors you would set it up as a standard related rate problem. Lets say you let x = the distance traveled by the car heading east and let y = the distance traveled by the other car heading north. Let s = the distance between the two cars. Since the cars are traveling at right angles to each other we have the relationship

    (*) [tex]
    s^{2} \, = \, x^{2} \, + y^{2}
    [/tex]

    Since x, s, and y change with respect to time t, can't we differentiate the above expression with respect to t ?

    Furthermore we are given that

    [tex]
    \frac{dx}{dt} \, = \, 30 mi/hr \mbox{ and } \frac{dy}{dt} \, = \, 50 mi/hr
    [/tex]

    We want to find

    [tex]
    \frac{ds}{dt}
    [/tex]

    So now all you need to do is implicitly differentiate (*) , determine the appropriate values for x, y, and s when t = 2 and you should be able to solve and get the desired result.

    You could also use vectors. Using vectors is in this case extremely easy since both cars travel at uniform speeds and depart from the same point and travel in directions perpendicular.
     
  6. Jul 12, 2011 #5
    Thank you.
     
  7. Jul 12, 2011 #6
    Merci.. Glad to be of help.
     
  8. Jul 12, 2011 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Now that you already have the solution, let's look at an easier way. Measuring time in hours, the positions t hours after the start are x = 30*t and y = 50*t, so the distance between the cars is s = c*t, where c = sqrt(30^2 + 50^2). So, for *any* t the rate is ds/dt = c.

    RGV
     
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