a bridge is 10 m. above a canal. A motorboat going 3 m/sec passes under the center of the bridge at the same instant that a woman walking 2 m/sec reaches that point. How rapidly are they separating 3 sec. later?
At a time of t seconds
let the distance covered by the boat be 3t m
let the distance covered by the woman be 2t m
If D is the distance between them,
D^2 = (3t)^2 + 10^2 + (2t)^2
D^2 = 13t^2 + 100
2D dD/dt = 26t
dD/dt = 13t/D
at 3 seconds D^2 = 81 + 100 + 16 = 197
D = √197
so at 3 seconds , dD/t = 13(3)/√197 = 2.7786
= appr 2.8 m/s
Well, it seems like we have a classic case of boats and pedestrians here. Let's see if we can solve this with some comedic math!
So, the motorboat is cruising along at 3 m/sec, and the woman is strolling at 2 m/sec. If they both reach the center of the bridge at the same time, that's quite a coincidence! Maybe they were planning a surprise party for each other? Who knows!
Now, let's fast forward 3 seconds. The motorboat will have traveled 3 seconds times 3 m/sec, which is 9 meters. Meanwhile, the woman will have walked 3 seconds times 2 m/sec, which is 6 meters.
To find out how rapidly they are separating, we just need to subtract the distances they've traveled. So, 9 meters minus 6 meters equals... you guessed it, 3 meters! So, they are separating at a rate of 3 meters per second.
Now, if only they could stop and exchange pleasantries, they might realize what a fantastic story this would make. Truly a tale of unexpected encounters and high-speed separations!
To find out how rapidly the motorboat and the woman are separating 3 seconds later, we can use the concept of relative motion.
Let's assume that the point where the motorboat passes under the center of the bridge is the origin (0,0) on a coordinate system. The height of the bridge above the canal would be 10 meters.
The position of the motorboat after t seconds can be represented as (0, -3t), where -3t represents that the motorboat is moving downward at a rate of 3 meters per second. The position of the woman after t seconds can be represented as (0, 2t), where 2t represents that the woman is moving upward at a rate of 2 meters per second.
To find the separation between the motorboat and the woman after 3 seconds, we need to calculate the distance between their respective positions.
The distance formula between two points (x₁, y₁) and (x₂, y₂) is given by:
d = √((x₂ - x₁)² + (y₂ - y₁)²)
For the motorboat and the woman, the distance formula becomes:
d = √((0 - 0)² + (2(3) - (-3(3)))²)
= √(0² + 6²)
= √(0 + 36)
= √36
= 6 meters
Therefore, 3 seconds later, the motorboat and the woman are separating at a rate of 6 meters per second.
To find how rapidly the motorboat and woman are separating 3 seconds later, we can first calculate their respective positions after 3 seconds.
Let's assume the bridge is located at the origin (0,0) on a Cartesian coordinate system, with the canal positioned horizontally below the bridge. Hence, the motorboat starts at (-3t, -10) and the woman starts at (2t, -10), where t represents time in seconds.
After 3 seconds, the motorboat's position will be (-3(3), -10) = (-9, -10) and the woman's position will be (2(3), -10) = (6, -10).
The distance between two points can be calculated using the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Therefore, the distance between the motorboat and the woman after 3 seconds is:
d = sqrt((6 - (-9))^2 + (-10 - (-10))^2)
= sqrt((15)^2 + (0)^2)
= sqrt(225)
= 15 meters
Hence, after 3 seconds, the motorboat and the woman are separating at a rate of 15 meters per second.