Two people start walking from the same point. Person A is walking south at a rate of 3 feet per second and Person B is walking east at a rate of 4 feet per second. At what rate is the distance between them increasing when they have walked for 5 seconds

5 ft / s.

Step-by-step explanation:

We have that person A has a speed of 3 ft / s, that is dA / dt is 3.

Person B has a speed of 4 ft / s, that is, dB / dt is 4.

Now, in 5 seconds, the distance of each one would be:

A = 3 * 5 = 15 ft

B = 4 * 5 = 20 ft

We know that the distance between A and B, finishes completing a right triangle. So this distance "C" would be equal to:

 C ^ 2 = A ^ 2 + B ^ 2

Replacing:

C ^ 2 = 15 ^ 2 + 20 ^ 2

C ^ 2 = 625

C = 25

25 ft away is A and B, to calculate the speed it would be, dC / dt

If we derive the previous equation with respect to time, we are left with:

2 * C * (dC / dt) = 2 * A * (dA / dt) + 2 * B * (dB / dt)

We know all the values, replacing:

2 * 25 * (dC / dt) = 2 * 15 * 3 + 2 * 20 * 4

(dC / dt) = 250/50 = 5

Therefore the speed at which the distance between them increases when they have walked for 5 seconds is 5 ft / s.

answer: 7270

step-by-step explanation:

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