(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity straight north relative to the
Earth, assuming its speed relative to the water remains 7.00 m/s ? (b) What would its speed be relative to the Earth?
1 answer:
Answer:
Direction of ship: 9.45° West of North
Ship's relative speed: 7.87m/s
Explanation:
A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0
Vx=0
Therefore, -VsSin∅+VcCos∅40°
Sin∅ = Vc/Vs × Cos 40°
Sin∅ = 1.5/7 ×Cos40°
Sin∅= 0.164
∅= Sin-¹ (0.164)
∅= 9.45° W of N
B. Ship's relative speed:
Vy= VsCos∅ + Vcsin40°
= 7Cos9.45° + 1.5sin40°
= 7×0.986 + 1.5×0.642
= 7.865
= 7.87m/s
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