1. 1. A police cruiser approaching a right-angled intersection from the north, is chasing a speeding car that has turned the corner and is now moving due east. When the cruiser is 0:6 miles north of the intersection and the car is 0:8 miles to the east, the police determine with radar that the distance between them and the car is increasing at a rate of 20 mph. If the cruiser is moving at 60 mph at the instant of the measurement, how fast is the car moving?
For this question, I know how to solve it because of a brief answer key, but I do not get how the value of s, or the distance between the cruiser and the car is 1. I get how x, the distance of the car, is 0.8 and how y, the distance of the cruiser is 0.6. But why is s=1?
2. A man starts walking north at a speed of 1.5 m/s and a woman starts at the same point P at the same time walking west at a speed of 2m/s. At what rate is the distance between the man and the woman increasing one minute later?
I have no idea how to do this question, and am getting confused by the concept in general. Would someone mind explaining to me how to approach this question?
I have drawn a diagram, which is a right triangle, but I have no idea how to figure out the lengths from only being given the speeds.
Thanks in advance!
I answered the first question back in 2012
http://www.jiskha.com/display.cgi?id=1326819217
See if you can follow the same way for the 2nd question.
The second question is actually easier
Be sure to use 60 seconds for 1 minute at the end
1. In order to understand why the distance between the cruiser and the car is 1, let's analyze the given information. We know that the police cruiser is moving north and the car is moving east. When we have a right-angled triangle formed by the distances traveled by each, the hypotenuse of this triangle represents the distance between them.
In this case, the police cruiser is 0.6 miles north of the intersection, and the car is 0.8 miles east of the intersection. The distance between them is increasing at a rate of 20 mph. We can define this distance as "s."
Using the Pythagorean theorem, we can see that:
s^2 = (0.6 miles)^2 + (0.8 miles)^2
Simplifying this equation gives us:
s^2 = 0.36 + 0.64 = 1
Taking the square root of both sides, we find that:
s = 1 mile
Therefore, the distance between the cruiser and the car is 1 mile.
2. To approach this question, we can consider the man and the woman as two moving points forming a right-angled triangle. The sides of the triangle represent the distances each has traveled, and the hypotenuse represents the distance between them.
Since the man is initially walking north at 1.5 m/s and the woman is walking west at 2 m/s, the sides of the triangle can be represented as follows:
The side corresponding to the man's distance = (1.5 m/s) * (1 min) = 1.5 m
The side corresponding to the woman's distance = (2 m/s) * (1 min) = 2 m
Now, we can use the Pythagorean theorem to find the distance between the man and the woman:
s^2 = (1.5 m)^2 + (2 m)^2
Simplifying this equation gives us:
s^2 = 2.25 + 4 = 6.25
Taking the square root of both sides, we find that:
s = 2.5 m
Therefore, the distance between the man and the woman is initially 2.5 meters.
To find how fast the distance is increasing, we need to differentiate the equation with respect to time. Differentiating the equation s^2 = (1.5 m)^2 + (2 m)^2 with respect to time (t) gives us:
2s * ds/dt = 2 * (1.5 m) * (0) + 2 * (2 m) * (-1.5 m/s)
Simplifying this equation gives us:
2s * ds/dt = -6 m^2/s
Now, substituting the known value of s = 2.5 m, we can solve for ds/dt:
2 * (2.5 m) * ds/dt = -6 m^2/s
5 m * ds/dt = -6 m^2/s
ds/dt = -6 m^2/s / 5 m
ds/dt = -1.2 m/s
Therefore, the distance between the man and the woman is decreasing at a rate of 1.2 m/s.