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madam [21]
3 years ago
8

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
Mathematics
1 answer:
Dovator [93]3 years ago
5 0

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

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