Suppose a police car drives towards an intersection at a speed of 60 km/h. A red Ferrari drives along a perpendicular road towards the same intersection at a speed of 80 km/h. (a) At 1:00pm the police car is 0.4 km away from the intersection, and the red Ferrari is 0.65 km away from the intersection. At that moment, the police officer in the police car uses a radar to measure how fast the distance between the two cars is changing. What number does the radar show? What are the units? Explain your reasoning and round your final answer to 2 decimal points. (b) Is the radar measurement in the previous question an accurate estimate of the speed of the red Ferrari? If instead the police car was 0.1 km away from the intersection (but all other quantities remained the same), then would the radar measurement provide a more accurate or less accurate estimate of the Ferrari's speed? How would the police officer know which estimate is more accurate? You may assume that the radar works perfectly to measure how fast the distance between the cars is changing, regardless of the magnitude of that distance, and that the police officer knows that the radar works perfectly.
(a) To find the rate at which the distance between the two cars is changing, we can take the derivative of the distance function with respect to time. Let's set up a coordinate system where the intersection is the origin, and the police car is initially at (-0.4, 0) and the red Ferrari is initially at (0, -0.65).
Let the position of the police car be given by the function P(t) = (-0.4, 0) + (60t, 0) = (60t - 0.4, 0).
Let the position of the red Ferrari be given by the function F(t) = (0, -0.65) + (0, 80t) = (0, 80t - 0.65).
The distance between the two cars can be found using the distance formula:
D(t) = sqrt((x2 - x1)^2 + (y2 - y1)^2)
= sqrt((0 - (60t - 0.4))^2 + ((80t - 0.65) - 0)^2)
= sqrt((0 - (60t - 0.4))^2 + (80t - 0.65)^2).
Note that we squared all the terms inside the square root to eliminate the square root.
Now, we can find the derivative of D(t) with respect to t using the chain rule:
D'(t) = (1/2)*(2*(60t - 0.4)*60 + 2*(80t - 0.65)*80)
= 120*(60t - 0.4) + 160*(80t - 0.65).
At t = 1:00pm, we have t = 0, so:
D'(0) = 120*(60(0) - 0.4) + 160*(80(0) - 0.65)
= 120*(-0.4) + 160*(-0.65)
= -48 + (-104)
= -152.
Therefore, the radar shows that the distance between the two cars is changing at a rate of -152 km/h. The negative sign indicates that the distance between the two cars is decreasing.
(b) The radar measurement is an accurate estimate of the speed of the red Ferrari. Since the radar measures how fast the distance between the two cars is changing, and the distance between the two cars is always changing due to the motion of both cars, the radar measurement will be accurate regardless of the specific values of the distance, time, or speeds involved.
If the police car was 0.1 km away from the intersection, all other quantities being the same, the radar measurement would provide a more accurate estimate of the Ferrari's speed. This is because the closer the police car is to the intersection, the smaller the distance between the two cars will be, and therefore the radar measurement will be more sensitive to changes in the distance. As a result, the radar measurement will provide a more precise estimate of the speed of the red Ferrari when the police car is closer to the intersection.
The police officer would know which estimate is more accurate by comparing the magnitudes of the radar measurements. A larger magnitude of the radar measurement would indicate a faster change in the distance between the two cars and therefore a higher speed for the red Ferrari. Conversely, a smaller magnitude of the radar measurement would indicate a slower change in the distance and a lower speed for the red Ferrari.
(a) To find the radar measurement of how fast the distance between the two cars is changing, we can calculate the relative velocity of the red Ferrari with respect to the police car.
The relative velocity is the sum of their individual velocities, considering they are moving towards each other.
Relative velocity = velocity of the red Ferrari - velocity of the police car
Relative velocity = 80 km/h - 60 km/h
Relative velocity = 20 km/h
The radar measurement will show the relative velocity between the two cars, which is 20 km/h. The units for the radar measurement will also be in kilometers per hour (km/h).
(b) The radar measurement in the previous question is NOT an accurate estimate of the speed of the red Ferrari. The radar measurement only provides the relative velocity between the two cars, not the individual speeds of each car.
If the police car was 0.1 km away from the intersection, the radar measurement would provide a more accurate estimate of the Ferrari's speed. This is because the radar can accurately measure the relative velocity between the cars, and a smaller distance between the cars would reduce the impact of any potential error.
The police officer would know which estimate is more accurate by comparing the radar measurement with the actual measured speed of the red Ferrari using another method (e.g., a speed gun). By comparing these two measurements, the police officer can determine if the radar measurement is consistent and reliable.