Noel and Casey both start at the same place. Noel walks due south and Casey walks due east. After some time has passed, Noel is 6 miles south and Casey is 8 miles east. At this time, Noel is walking at a rate of 2 mph and Casey is walking at a rate of 1 mph. How fast is the distance between them increasing at this time

Respuesta :

Answer:

2.04 miles per hour

Step-by-step explanation:

Given

Noel

[tex]n_1 =6miles[/tex]

[tex]r_1 = 2mph[/tex]

Casey

[tex]c_1 = 8miles[/tex]

[tex]r_2 =1mph[/tex]

Required

The rate at which the distance increases

Their movement forms a right triangle and the distance between them is the hypotenuse.

At [tex]n_1 =6miles[/tex] and [tex]c_1 = 8miles[/tex]

The distance between them is:

[tex]d_1 = \sqrt{n_1^2 + c_1^2}[/tex]

[tex]d_1 = \sqrt{6^2 + 8^2}[/tex]

[tex]d_1 = \sqrt{100}[/tex]

[tex]d_1 = 10miles[/tex]

After 1 hour, their new position is:

New = Old + Rate * Time

[tex]n_2 = n_1 + r_1 * 1[/tex]

[tex]n_2 = 6 + 2 * 1 = 8[/tex]

And:

[tex]c_2 = c_1 + r_2 * 1[/tex]

[tex]c_2 = 8 + 1 * 1 = 9[/tex]

So, the distance between them is now:

[tex]d_2 = \sqrt{n_2^2 + c_2^2}[/tex]

[tex]d_2 = \sqrt{8^2 + 9^2}[/tex]

[tex]d_2 = \sqrt{145}[/tex]

[tex]d_2 = 12.04[/tex]

The rate of change is:

[tex]\triangle d = d_2 -d_1[/tex]

[tex]\triangle d = 12.04 -10[/tex]

[tex]\triangle d = 2.04[/tex]