# Distance Problem

• Sep 3rd 2007, 11:44 AM
uglygreencouch
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?
• Sep 3rd 2007, 12:36 PM
red_dog
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.
• Sep 3rd 2007, 12:40 PM
uglygreencouch
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
• Sep 4th 2007, 11:51 AM
mahurshi
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