Using calculus to find stopping distance of object towed from a vessel
I studied this thirty years ago but it is all very rusty. The problem is to find out how far a towed object will travel if the boat that is pulling it suddenly stops. (We will be working in ice, and so might possibly stop suddenly ... we don't want to towed object to get caught in the propellers).
The towed object is called a "steamer", a plastic-coated oil-filled cylinder with a 50 mm diameter and usually about 5 km long. It is filled with transducers (hydrophones) and is used in conducting seismic surveys.
Assuming 5 knots vessel speed and a 5 km Sercel streamer.
Initial velocity : v(t=0) = 5 kts = 5 x 1852 m / 3600 sec = 2.57 m/sec
Mass of one filled 150 m section is 320 kg, so total mass of streamer = (5000/150) x 320 kg = 10,667 kg
Tension at shooting speed is around 1000 kg so this is the drag force at 5 knots (around 10,000 Newtons)
Now ... force = mass x acceleration
So ... acceleration = force / mass = 10,000 / 10,667 = 0.9375 m/sec/sec ... actually deceleration in this case, of course.
Now we can draw a diagram of velocity against time. Velocity is the "y" axis, time is the "x" axis.
The streamer velocity will cut the "y" axis at 2.57 m/sec and cut the "x" axis at 2.74 seconds.
The distance travelled is the area under the line = 2.57 x 2.74 / 2 = 3.521 metres
However this is wrong as the drag force is not constant, it is proportional to velocity through the water squared (I think) so calculus must be used.
Could somebody solve this one for me ... please.