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3 years ago

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At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 22 knots and ship B is sailing north at 23 kn

ots. How fast (in knots) is the distance between the ships changing at 3 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

1 answer:

Arturiano [62]3 years ago

5 0

Answer:

30.66 knots

Explanation:

Distance of ship A from B at noon = 50 NM

$\frac{da}{dt}$ = Velocity of ship A = 22 knots = 22 NM/h

$\frac{db}{dt}$ = Velocity of ship B = 23 knots = 23 NM/h

Distance travelled by ship A from noon to 3 PM = 22×3 = 66 NM

a = Total distance travelled by ship A = 50+66 = 116 NM

b = Total distance travelled by ship B till 3 PM = 23×3 = 69 NM

c = Distance between Ship A and B at 3 PM = √(116²+69²) = 134.97 NM

a²+b² = c² (Pythagoras theorem)

Now differentiating with respect to time

$2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\\\Rightarrow a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\\\Rightarrow \frac{116\times 22+69\times 23}{134.97}=\frac{dc}{dt}\\\Rightarrow \frac{dc}{dt}=30.66\ NM/h$

∴ The velocity with which the distance is changing at 3 PM (3 hours later) is 30.66 knots

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now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

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