Results 1 to 4 of 4

Math Help - Distance Problem

  1. #1
    Newbie
    Joined
    Aug 2007
    Posts
    12

    Distance Problem

    I'm having some trouble setting up an equation for this problem. Any suggestions?


    Two children own two-way radios that have a maximum range of 2 miles. One leaves a certain point at 1 p.m. walking due north at a rate of 4 mi/hr. the other leaves the same point at 1:15 p.m., travleing due south 6 mi/hr. When will they be unable to communicate with one another?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    After first 15 minutes, the distance between the children is 1 mile.
    After that the distance is t(4+6)=10t, where t is time in hours.
    Then the total distance is 1+10t.
    So 1+10t>2\Rightarrow t>\frac{1}{10}\Rightarrow t>6 minutes.
    Thus the children will be unable to communicate after 1:21.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2007
    Posts
    12
    Thank you red_dog. It looks like I just need to have a little more confidence in myself because this is what I originally had:

    4mph(1/4hr) + 10mph(X hrs) = 2
    4 x 1/4 = 1
    1 +10x= 2
    10x =1
    x= 1/10 hr or 6 minutes

    1:15 + 6 minutes = 1:21
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2007
    Posts
    20

    Smile Talking through the problem and solving it

    1. @ 1pm, the kid starts walking towards north at 4 miles/hr

    2. @ 1:15pm, the distance between the kids is 1 mile (4 miles/hr => 1 mile/15 minutes)

    3. @ 1:15pm, the other kid starts walking towards south @ 6 miles/hr

    4. From the above instant, since the kids are going in opposite directions, the distance between them increases at the rate of (4 + 6) miles/hr => 10 miles/hr

    5. For them to go out of range, they need to go 1 more mile apart

    6. To cover 1 more mile @ 10 miles/hr, it takes 1/10th of an hour, which is 1/10 * 60 mins = 6 mins

    7. Time when they go out of range = 1:15pm + 6 mins = 1:21pm

    Hope this helps

    Mahurshi Akilla
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. distance problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 25th 2010, 02:35 AM
  2. distance problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 26th 2010, 02:44 PM
  3. distance problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 5th 2009, 06:36 AM
  4. 3d distance problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 30th 2009, 07:52 AM
  5. Distance Problem
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 7th 2008, 05:29 PM

Search Tags


/mathhelpforum @mathhelpforum