Question

Two children own two-way radios that have a maximum range of 3 miles. One leaves a certain point at 1:00 P.M., walking due north at a rate of 6 mi/hr. The other leaves the same point at 1:15 P.M., traveling due south at 9 mi/hr. For how many minutes after 1:00 P.M. will they be able to communicate with one another? (Round your answer to the nearest minute.)

Answer 1

Answer:

The answer is 21 minutes

Step-by-step explanation:

We use the equation Xf = Xo + vt

1) At 1:00 PM, child one leaves the starting point heading north at a constant velocity of 6 mi/hr or .1 [mi/min] (divide by 60 to convert from [mi/hr] to [mi/min])

2) He walks for 15 minutes before kid 2 starts walking. In 15 minutes he is able to cover 1.5 [mi]

• $x_{1f1} =x_{o} +v_{1} t_{1} \\x_{1f1} =0+.1(15)\\x_{1f1} =1.5 [mi]$

3) Now, child 2 starts walking and we know that when the range reaches 3 miles, they won´t be able to communicate. So the sum of the final position of child 1 and child 2 must be 3[mi]

• Child 1 final position => $x_{1f} = x_{1f1} +v_{1} t\\x_{1f} =1.5+v_{1} t$
• Child 2 final position => $x_{2f} =0+v_{2} t$

4) Sum the equations and equate to 3

• $x_{1f} +x_{2f} =3$

5) Substitute the values we already know

• $1.5+v_{1} t+v_{2}t=3\\ 1.5+.1t+.15t=3\\1.5+.25t=3\\t=\frac{3-1.5}{.25} \\t=6 [min]$

6) in 15 + 6 minutes they will be 3miles apart

7) In 21 minutes they will still be able to communicate with one another.