Question

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

Answer 1

Answer:

2.04 miles per hour

Step-by-step explanation:

Given

Noel

$n_1 =6miles$

$r_1 = 2mph$

Casey

$c_1 = 8miles$

$r_2 =1mph$

Required

The rate at which the distance increases

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

At $n_1 =6miles$ and $c_1 = 8miles$

The distance between them is:

$d_1 = \sqrt{n_1^2 + c_1^2}$

$d_1 = \sqrt{6^2 + 8^2}$

$d_1 = \sqrt{100}$

$d_1 = 10miles$

After 1 hour, their new position is:

New = Old + Rate * Time

$n_2 = n_1 + r_1 * 1$

$n_2 = 6 + 2 * 1 = 8$

And:

$c_2 = c_1 + r_2 * 1$

$c_2 = 8 + 1 * 1 = 9$

So, the distance between them is now:

$d_2 = \sqrt{n_2^2 + c_2^2}$

$d_2 = \sqrt{8^2 + 9^2}$

$d_2 = \sqrt{145}$

$d_2 = 12.04$

The rate of change is:

$\triangle d = d_2 -d_1$

$\triangle d = 12.04 -10$

$\triangle d = 2.04$