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?

Question
Two-way tables
asked 2021-01-15
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?

Answers (1)

2021-01-16
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.
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