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
\(Since Distance=Speed \cdot time\)
Distance travelled by first children \((d_1)\) at any time t after 1:00 P.M will be 4t and distance travelled by second children \((d_2)\) at any time t after 1:00 P.M will be
\(d_2=\begin{cases} 0& 0\leq t\leq \frac{1}{4}\\ 6(t-\frac{1}{4}) & t\geq\frac{1}{4} \end{cases}\)
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 \(0\leq t\leq \frac{1}{4}\) 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 \(t \geq\frac{1}{4}\) they will be unable to communicate when distance between them will be greater than 2 that is,
\(4t+6\bigg(t-\frac{1}{4}\bigg)>2\)

\(10t-\frac{3}{2}>2\)

\(10t>\frac{7}{2}\)

\(t>\frac{7}{20} \text{hour}\) and \(\frac{7}{20} \text{hr}=\frac{7}{20} \cdot 60\text{ minutes}\)

\(=21\text{ minutes}\) Hence 21 minutes after 1:00 P.M that is after 1:21 P.M they will be unable to communicate.

\(10t-\frac{3}{2}>2\)

\(10t>\frac{7}{2}\)

\(t>\frac{7}{20} \text{hour}\) and \(\frac{7}{20} \text{hr}=\frac{7}{20} \cdot 60\text{ minutes}\)

\(=21\text{ minutes}\) Hence 21 minutes after 1:00 P.M that is after 1:21 P.M they will be unable to communicate.