Kinda desperate here lol.
A submarine is heading due East at 10 m/s. The submarine is 5000 m E30degrees N of a destroyer, top speed 35 m/s. The destroyer immediately turns on a course to intercept the submarine. But the submarine turns to last as long as possible from being sunk and calls for air support. How long has the submarine got before being rammed?
My prof said the answer is 200, but may someone explain how?
A submarine is heading due East at 10 m/s. The submarine is 5000 m E30degrees N of a destroyer, top speed 35 m/s. The destroyer immediately turns on a course to intercept the submarine. But the submarine turns to last as long as possible from being sunk and calls for air support. How long has the submarine got before being rammed?
My prof said the answer is 200, but may someone explain how?
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The key to the problem is knowing that the course which maximize the time to closure is the submarine running directly away from the destroyer.
You know that because there are three limits to consider:
1) submarine heads directly towards ship -- minimizes time of closure.
2) submarine heads directly away from ship -- subs maximum velocity is directed away from the ship always.
3) submarine always moves orthogonal to the destroyer's trajectory -- the difference in subs component away from the ship is always smaller than in case (2)
if it cost time for the ship and the sub to turn, then the strategy might change. But for now, since turns are free (cost zero time) you can only maximize the vector component along the line of closure.
that said, (35-10) * dt = 5000 :: dt = 200.
Of course, if this were a tactical analysis problem rather than a math/physics problem, a better answer would be for the submarine to set dive planes, reduce speed to silent running after diving below the thermocline, then execute a silent, low speed turn to an escape trajectory determined by the bottom topography. Once out of range of the destroyer, then the sub can surface to radio the destroyer's location -- but even that would be unnecessary because satellite surveillance monitoring the area would already know where the destroyer was by its long-lived wake.
You know that because there are three limits to consider:
1) submarine heads directly towards ship -- minimizes time of closure.
2) submarine heads directly away from ship -- subs maximum velocity is directed away from the ship always.
3) submarine always moves orthogonal to the destroyer's trajectory -- the difference in subs component away from the ship is always smaller than in case (2)
if it cost time for the ship and the sub to turn, then the strategy might change. But for now, since turns are free (cost zero time) you can only maximize the vector component along the line of closure.
that said, (35-10) * dt = 5000 :: dt = 200.
Of course, if this were a tactical analysis problem rather than a math/physics problem, a better answer would be for the submarine to set dive planes, reduce speed to silent running after diving below the thermocline, then execute a silent, low speed turn to an escape trajectory determined by the bottom topography. Once out of range of the destroyer, then the sub can surface to radio the destroyer's location -- but even that would be unnecessary because satellite surveillance monitoring the area would already know where the destroyer was by its long-lived wake.