A man starts walking north at 4ft/s from a point P. Five minutes later a woman starts walking south at 5ft/s from a point 500 ft due east of P. At what rate are the people moving apart 15 min after the woman starts walking?
I'm confused about the diagram and what to do with time
if the man starts at t=0, his position at time t is (0,4t)
The woman is at (500+5(t-5),0)
the distance d is
d^2 = (500+5(t-5))^2 + (4t)^2 = 41t^2 + 4750t + 225625
at t=15,
d^2 = 306100, so d=553.26
2d dd/dt = 82t + 4750
1106.52 dd/dt = 5980
dd/dt = 5.4 ft/s
makes sense, since the 500' lead makes most of the distance due to the woman's motion, at 5 ft/s
Rats. I thought she was walking east.
she is at (500,-5(t-5))
d^2 = 500^2 + (4t+5(t-5))^2 = 81t^2 - 450t + 250625
at t=15, d = 512
2d dd/dt = 162t - 450
1024 dd/dt = 1980
dd/dt = 1.93 ft/s
To solve this problem, let's break it down step-by-step:
Step 1: Create a diagram:
Start by drawing a diagram to represent the situation. Label the point where the man starts walking north as P, and the point where the woman starts walking south as Q. Also, mark the distance between P and Q as 500 ft.
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| Q
500 |
| .
| .
| .
P | .
| .
| .
|
```
Step 2: Find the distance each person has traveled after 15 minutes:
The man starts walking 5 minutes earlier than the woman. Therefore, after 15 minutes, the man has traveled for a total of 20 minutes (15 minutes after the woman starts walking plus the original 5 minutes). Since the man walks at a speed of 4 ft/s, his distance can be calculated as: distance = speed × time = 4 ft/s × (20 minutes × 60 seconds/minute) = 4 ft/s × 1200 s = 4800 ft.
The woman walks for 15 minutes after she starts. So, her distance can be calculated as: distance = speed × time = 5 ft/s × (15 minutes × 60 seconds/minute) = 5 ft/s × 900 s = 4500 ft.
Step 3: Calculate the distance between them after 15 minutes:
To find out the distance between them after 15 minutes, we need to use the Pythagorean theorem because they are moving in perpendicular directions. The distance between P and Q after 15 minutes is the hypotenuse of a right triangle with sides of 4800 ft (the distance traveled by the man) and 500 ft (the initial distance between them).
Applying the Pythagorean theorem:
Distance^2 = (side1)^2 + (side2)^2
Distance^2 = 4800^2 + 500^2
Distance = √(4800^2 + 500^2)
Distance ≈ 4815.51 ft
Step 4: Calculate the rate at which they are moving apart:
To find the rate at which they are moving apart, we need to consider their relative velocity. The man is moving north at 4 ft/s, while the woman is moving south at 5 ft/s. Thus, their relative velocity can be calculated as: relative velocity = 4 ft/s + 5 ft/s = 9 ft/s.
Therefore, the rate at which they are moving apart is 9 ft/s.
So, the rate at which the people are moving apart 15 minutes after the woman starts walking is approximately 9 ft/s.
To solve this problem, let's break it down step by step.
1. Draw a diagram: Start by drawing a coordinate plane, with point P as the origin (0,0). Mark the north direction as the positive y-axis and the east direction as the positive x-axis.
2. Determine the initial positions of the man and woman: The man starts at point P, while the woman starts 500 ft due east of P. So, the man's initial position is (0,0), and the woman's initial position is (500,0).
3. Determine the velocities of the man and woman: The man walks north at a rate of 4 ft/s, which means his velocity vector is (0, 4). The woman walks south at a rate of 5 ft/s, which means her velocity vector is (0, -5).
4. Determine the positions of the man and woman after 15 minutes: Since the man starts walking 5 minutes earlier, at the 15-minute mark, he would have been walking for 20 minutes, and the woman would have been walking for 15 minutes. Convert these time values into seconds to match the velocity units.
The man's position after 20 minutes is given by:
(0, 0) + (0, 4 ft/s) * (20 min * 60 s/min) = (0, 0) + (0, 4 ft/s) * (1200 s) = (0, 4800 ft).
The woman's position after 15 minutes is given by:
(500, 0) + (0, -5 ft/s) * (15 min * 60 s/min) = (500, 0) + (0, -5 ft/s) * (900 s) = (500, -4500 ft).
5. Determine the distance between the man and woman after 15 minutes: This can be calculated using the distance formula d = sqrt((x2 - x1)^2 + (y2 - y1)^2), where (x1, y1) and (x2, y2) are the positions of the man and woman, respectively.
Therefore, the distance between them is:
d = sqrt((0 - 500)^2 + (4800 - (-4500))^2) = sqrt(500^2 + 9300^2) = sqrt(250000 + 86490000) = sqrt(86740000) ≈ 9316.8 ft.
6. Determine the rate at which the man and woman are moving apart: Since we are interested in finding their rate of separation after 15 minutes, we need to consider their velocities. The rate at which they are moving apart is the magnitude of the difference between their velocities.
Thus, the rate at which they are moving apart is |4 ft/s - (-5 ft/s)| = |4 ft/s + 5 ft/s| = 9 ft/s.
So, 15 minutes after the woman starts walking, the man and woman are moving apart at a rate of 9 ft/s.