The rate of change defines the change of a value of a variable over a unit of time
(a) The time at which the runners feet will get wet is 44 minutes
(b) The time the runner shoes get wet is after approximately 24.415 minutes
Reason:
Known parameters;
The radius of the puddle, r = 5·t
The time after which the pump broke = t
Initial distance of the runner from the intersection, d = 6 miles = 31,680 ft.
The speed of the runner = 17 ft./s
(a) Required:
The time at which the runners feet will get wet
Solution;
At the time the runner's feet get wet, we have;
5·t + 17·t = 31,680
[tex]t = \dfrac{31680}{12} = 2,640[/tex]
The time at which the runners feet will get wet is 2,640 seconds = 44 minutes
(b) Location of the runner = 6 miles East and 5,000 feet North
The direction the runner runs = Due West
Speed of the runner = 17 ft. per second
The time at which the runners feet will get wet
Solution:
At the point the runner's feet get wet, we have;
The vertical component of the radius, y = 5,000 = The direction north from the runner's path
The horizontal component of the radius, x = 31,680 - 17·t
Therefore;
The magnitude of the radius of the puddle at the point the runner's feet get wet is given as follows;
[tex]5 \cdot t = \sqrt{5000^2 + (31680 - 17 \cdot t)^2}[/tex]
Which gives;
25·t² = 5000² + (31,680 - 17·t)²
264·t² - 1077120·t + 1028622400 = 0
Solving for t
t = 1524.924 seconds = 25.415 minutes or t = 2,555.076 seconds = 42.8846 minutes
Therefore;
- The time the runner shoes get wet is after, t ≈ 24.415 minutes
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