Angela, Mary and Tiff are all standing near the intersection of University and 42nd streets. Mary and Tiff do not move, but Angela runs toward Tiff at12 ft/sec along a straight line, as pictured. Assume the roads are50 feet wide and Tiff is60 feet north of the nearest corner. Where is Angela located when she is closest to Mary and when does she reach this spot?
What picture?
18,42
Well, it seems like Angela is determined to reach Mary. To find out where Angela is located when she is closest to Mary, let's delve into some clown math.
Since Angela is running towards Tiff at 12 ft/sec and Mary and Tiff are standing still, Angela will eventually catch up to Tiff. Let's call the time it takes for Angela to reach Tiff as "t" seconds.
Within this time, Angela will have covered a distance of 12t feet. Since the roads are 50 feet wide, Angela will be closest to Mary when she is in line with the road closest to her.
Now, let's figure out where Angela will be at this point. We know Tiff is 60 feet north of the nearest corner, and since Angela is running along a straight line, she will be in line with Tiff at this point. So, Angela will also be 60 feet north of the nearest corner.
When does Angela reach this spot? Well, we can use the fact that Angela is moving at 12 ft/sec to determine this. The formula to find the time it takes for Angela to cover a distance of 60 feet is:
time = distance / speed
t = 60 / 12
t = 5 seconds
So, Angela will be closest to Mary when she is 60 feet north of the nearest corner, and she will reach this spot in 5 seconds.
Just remember to watch out for clown cars and unicycle traffic when crossing the intersection! Safety first, even in clown math.
To determine where Angela is located when she is closest to Mary, we need to analyze their positions and movements.
First, let's define the coordinate system. Let the corner of University and 42nd streets be the origin (0, 0). The positive x-axis extends along University Street, and the positive y-axis extends along 42nd Street. Therefore, Tiff's position is (0, 60) in this coordinate system.
Since Tiff is not moving, her position remains constant. Mary's position is also fixed, as mentioned in the question. We just need to determine Angela's position over time.
Let's assume t represents the time in seconds. Angela starts at position (0, 0) and moves toward Tiff at a rate of 12 ft/sec. Therefore, her position at any given time t is given by (12t, 0).
To find when Angela is closest to Mary, we need to find the point on the line connecting Mary and Tiff that is closest to Angela's position (12t, 0).
The distance between two points (x1, y1) and (x2, y2) is given by the distance formula:
d = √((x2 - x1)^2 + (y2 - y1)^2)
In our case, we want to minimize the distance between Angela's position (12t, 0) and the line connecting Mary and Tiff.
The line connecting Mary and Tiff can be represented as:
x = 0 (since Mary and Tiff share the same x-coordinate)
So our task reduces to finding the y-coordinate where Angela is closest to Mary. The distance between Angela's position (12t, 0) and the line x = 0 is simply the y-coordinate of Angela's position (12t, 0).
Therefore, we want to minimize the function:
f(t) = (12t - 0)^2 + (0 - y)^2
To minimize this function, we can take its derivative with respect to t and set it to zero:
f'(t) = 2(12t - 0)(12) + 2(0 - y)(0) = 0
24(12t) = 0
288t = 0
t = 0
The minimum point occurs at t = 0, indicating that Angela starts closest to Mary. Let's calculate the position at t = 0 to confirm.
At t = 0, Angela's position is (12 * 0, 0) = (0, 0).
Therefore, Angela is located at the same position as Mary (0, 0) when she is closest to Mary, and this occurs at t = 0 seconds.
So, Angela is closest to Mary when they are both at the intersection of University and 42nd streets, and this happens at the start of the scenario.