Consider two ships which are joined by a cable attached to each ship at the water line. Suppose the two ships are 200 metres apart, with the cable stretched tight and attached to a pulley which is anchored halfway between the ships at a depth of 45 metres. If one ship moves away from the other at 3 km/h,how quickly is the other ship moving after one minute?
I'm totally stuck...
Draw a diagram. Label as P the point on the ocean directly above the pulley. If the ship being pulled is x meters from P, and the pulling ship is at distance y,
√(x^2+45^2) + √(y^2+45^2) = 200
3 km/hr = 50m/min, so we want
dx/dt when t=1 and dy/dt=50.
at t=0, x=y=89.30
after 1 minute, y=139.30, so x=29.14
x/√(x^2+45^2) dx/dt + y/√(y^2+45^2) dy/dt = 0
29.14/53.61 dx/dt + 139.30/146.39 * 50 = 0
dx/dt = -87.53 m/s = 5.25 km/hr
Note that this cannot continue for long, as the ship will be pulled under by the cable.
To solve this problem, we can use the concept of related rates. Let's break down the given information step by step.
1. The two ships are initially 200 meters apart.
2. The cable is stretched tight between the ships and attached to a pulley anchored halfway between the ships at a depth of 45 meters.
Now, let's introduce some variables:
Let:
- x be the horizontal distance between the moving ship and the pulley.
- y be the vertical distance between the pulley and the fixed ship (i.e., the depth of the pulley).
- z be the total length of the cable (measured along the slanted path).
Based on this, we can write the following equation using the Pythagorean theorem:
x^2 + y^2 = z^2
We are interested in finding how quickly the other ship (the fixed one) is moving. Let's call the distance it moves in time t as y1.
Now, let's differentiate the equation with respect to time:
2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)
Since the pulley is anchored, its vertical position does not change, so dy/dt = 0. Also, dx/dt = 3 km/h since the other ship moves away at a rate of 3 km/h.
Plugging in these values, we get:
2x(3 km/h) = 2z(dz/dt)
Now, we need to find x, y, and z in terms of each other. From the given information, we know:
x = 200 meters - y
z = sqrt(x^2 + y^2)
Substituting these values into the equation:
2(200 - y)(3 km/h) = 2 sqrt((200 - y)^2 + y^2) (dz/dt)
Simplifying further:
600 - 2y = 2 sqrt((200 - y)^2 + y^2) (dz/dt)
Now, we can solve this equation to find dz/dt, which will give us the rate at which the other ship moves.
Note: One minute is equal to 1/60 hours, so we will use this value for t.
To solve the equation, we can rearrange it as follows:
dz/dt = (600 - 2y) / [2 sqrt((200 - y)^2 + y^2)]
We need to find the value of y when t = 1/60 hours. To do this, we'll need to plug in the values and solve for y:
dz/dt = (600 - 2y) / [2 sqrt((200 - y)^2 + y^2)]
t = 1/60 hours
Once we find the value of y, we can substitute it back into the equation dz/dt to determine how quickly the other ship is moving after one minute.