Answer:
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.
