# Two children own two-way radios that have a maximum range of 2 miles. One leaves a certain point at 1:00 P.M., walking due north at a rate of 4 . The other leaves the same point at 1:15 P.M., traveling due south at 6 . When will they be unable to communicate with one another?

Two children own two-way radios that have a maximum range of 2 miles. One leaves a certain point at 1:00 P.M., walking due north at a rate of 4 . The other leaves the same point at 1:15 P.M., traveling due south at 6 . When will they be unable to communicate with one another?
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Step 1 Consider that the maximum range of the radio is 2 miles. One leaves at 1:00 P.M with rate 4 miles/hr due north and other leaves at 1:15 P.M with rate eaves at 1:00 P.M with rate 4 miles/hr due south. Find at what time they will be unable to communicate. Let t denote the time in hour. after 1:00 P.M. Since 60 minutes =1 hour therefore, $15\text{minutes}=\frac{60}{4}\text{minutes}=\frac{1}{4}\text{hour}$ Step 2 $SinceDistance=Speed\cdot time$ Distance travelled by first children $\left({d}_{1}\right)$ at any time t after 1:00 P.M will be 4t and distance travelled by second children $\left({d}_{2}\right)$ at any time t after 1:00 P.M will be ${d}_{2}=\left\{\begin{array}{ll}0& 0\le t\le \frac{1}{4}\\ 6\left(t-\frac{1}{4}\right)& t\ge \frac{1}{4}\end{array}$ Children will unable to communicate when the distance between them will be greater than 2 miles and since they are travelling in opposite direction therefore the distance between them will be the sum of the distance they have travelled. For $0\le t\le \frac{1}{4}$ the distance between the is maximum $4\cdot \frac{1}{4}=1$ mile which is less than 2 mile hence they will be able to communicate. After $t\ge \frac{1}{4}$ they will be unable to communicate when distance between them will be greater than 2 that is, $4t+6\left(t-\frac{1}{4}\right)>2$
$10t-\frac{3}{2}>2$
$10t>\frac{7}{2}$
$t>\frac{7}{20}\text{hour}$ and
Hence 21 minutes after 1:00 P.M that is after 1:21 P.M they will be unable to communicate.