A heavy, 6 m long uniform plank has a mass of 30 kg. It is positioned so that 4 m is supported on the deck of a ship and 2 m sti
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Consider the travel of the speedboat from A to B and back.
Refer to the first figure.
The boat travels at 5 m/s relative to the water, and the downstream speed of the water is 0.5 m/s.
Therefore,
V₁=5 m/s, u = 0.5 m/s, sinθ = 0.5/5 = 0.1 => θ = arcsin(0.1) = 5.74°.
The boat should travel upstream at 5 m/s, at an angle of 5.74°.
Similarly, return speed from B to A is 5 m/s, at 5.74° upstream.
The horizontal component of velocity from A to B (or vice versa) is
5cos(5.74°) = 4.975 m/s.
The time required to travel 1000 m is
1000/4.975 = 202.02 = 3.367 min.
The time for the return trip is
t₁ = 2*3.367 = 6.734 min.
Consider travel from C to D and back, as shown in the second figure.
The resultant velocity upstream from C to D is
V₁ = 5 - 0.5 = 4.5 m/s
The time required to travel 1000 m fromC to D is
1000/4.5 = 222.22 s
The resultant velocity downstream from D to C is
V₂ = 5 + 0.5 = 5.5 m/s
The time required to travel fro D to C is
1000/5.5 = 181.82 s
Total time for the return trip between C and D is
t₂ = 222.22 + 181.82 = 404.04 s = 6.734 min
Answer:
The first boat should travel upstream at an angle of 5.74°. The time for the return trip between A and B is t₁ = 6.734 min or 404.04 s.
The second boat travels upstream against the current, and downstream with the current. The time for the return trip between C and D is t₂ = 6.734 min or 404.04 s.
Refer to the first figure.
The boat travels at 5 m/s relative to the water, and the downstream speed of the water is 0.5 m/s.
Therefore,
V₁=5 m/s, u = 0.5 m/s, sinθ = 0.5/5 = 0.1 => θ = arcsin(0.1) = 5.74°.
The boat should travel upstream at 5 m/s, at an angle of 5.74°.
Similarly, return speed from B to A is 5 m/s, at 5.74° upstream.
The horizontal component of velocity from A to B (or vice versa) is
5cos(5.74°) = 4.975 m/s.
The time required to travel 1000 m is
1000/4.975 = 202.02 = 3.367 min.
The time for the return trip is
t₁ = 2*3.367 = 6.734 min.
Consider travel from C to D and back, as shown in the second figure.
The resultant velocity upstream from C to D is
V₁ = 5 - 0.5 = 4.5 m/s
The time required to travel 1000 m fromC to D is
1000/4.5 = 222.22 s
The resultant velocity downstream from D to C is
V₂ = 5 + 0.5 = 5.5 m/s
The time required to travel fro D to C is
1000/5.5 = 181.82 s
Total time for the return trip between C and D is
t₂ = 222.22 + 181.82 = 404.04 s = 6.734 min
Answer:
The first boat should travel upstream at an angle of 5.74°. The time for the return trip between A and B is t₁ = 6.734 min or 404.04 s.
The second boat travels upstream against the current, and downstream with the current. The time for the return trip between C and D is t₂ = 6.734 min or 404.04 s.