2+x-(3-z); use x=6, and z=1
2 answers:
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Answer:
a) b.) -12knots, 8 knots c) No e)
Step-by-step explanation:
We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.
a)
We know that at time t , the ship A has moved and ship B has moved
. We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.
Using Pythagorean theorem, we can write the distance s as:
b)
We want to find for t=0 and t=1
c)
We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.
Ships have seen each other =
Since function is quadratic, concave up and has no real roots, we know that
for every t. So, the ships haven't seen each other.
d)
Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.
e)
Function ds/dt has a horizontal asympote in the first quadrant if
So, lets check this limit:
Notice that:
=√(speed of ship A² + speed of ship B²)
