A boat crosses a river from port A to port B, which are just on the opposite side. The speed of the water is Vw and that of boat is VB relative to water. Assume VB = 2 VW. What is the time taken by the boat, if it has to cross the river directly on the AB line?a)b)c)d)Correct answer is option 'A'. Can you explain this answer?

Question

Answer

Let the velocity of the boat, if it has to cross the river directly on water the line AB be v and the angle between and vB be θ. Then, from the figure sinθ=vwvB Given, vB=2vw ∴ sinθ=vw2vw=12 ⇒ θ=30o Now, the time is taken by the boat to cross the river directly from A to B t=DvA=DvBcosθ =DvBxcos30o or t=2DvB√ 3

Answer

Answer

To cross the river directly from A to B, the boat’s velocity v_B relative to water at angle θ to AB must cancel the river’s velocity v_W. Thus, we have:

v_B sin θ = v_W

Given v_B = 2v_W, we substitute this into the equation:

sin θ = v_W / (2v_W) = 1/2

Therefore, θ = 30°.

The velocity component across the river is:

v_B cos θ = v_B × √3/2

The time to cross distance D is:

t = D / (v_B cos θ) = D / (v_B × √3/2) = 2D / (v_B √3)

Answer: A.

Answer

To cross the river directly from A to B, the boat’s velocity v_B relative to water at angle θ to AB must cancel the river’s velocity v_W. Thus, we have:

v_B sin θ = v_W

Given v_B = 2v_W, we substitute this into the equation:

sin θ = v_W / (2v_W) = 1/2

Therefore, θ = 30°.

The velocity component across the river is:

v_B cos θ = v_B × √3/2

The time to cross distance D is:

t = D / (v_B cos θ) = D / (v_B × √3/2) = 2D / (v_B √3)

Answer: A.

Answer

To cross the river directly from A to B, the boat’s velocity v_B relative to water at angle θ to AB must cancel the river’s velocity v_W. Thus, we have:

v_B sin θ = v_W

Given v_B = 2v_W, we substitute this into the equation:

sin θ = v_W / (2v_W) = 1/2

Therefore, θ = 30°.

The velocity component across the river is:

v_B cos θ = v_B × √3/2

The time to cross distance D is:

t = D / (v_B cos θ) = D / (v_B × √3/2) = 2D / (v_B √3)

Answer: A.

Answer

To cross the river directly from A to B, the boat’s velocity v_B relative to water at angle θ to AB must cancel the river’s velocity v_W. Thus, we have:

v_B sin θ = v_W

Given v_B = 2v_W, we substitute this into the equation:

sin θ = v_W / (2v_W) = 1/2

Therefore, θ = 30°.

The velocity component across the river is:

v_B cos θ = v_B × √3/2

The time to cross distance D is:

t = D / (v_B cos θ) = D / (v_B × √3/2) = 2D / (v_B √3)

Answer: A.

Answer

To cross the river directly from A to B, the boat’s velocity v_B relative to water at angle θ to AB must cancel the river’s velocity v_W. Thus, we have:

v_B sin θ = v_W

Given v_B = 2v_W, we substitute this into the equation:

sin θ = v_W / (2v_W) = 1/2

Therefore, θ = 30°.

The velocity component across the river is:

v_B cos θ = v_B × √3/2

The time to cross distance D is:

t = D / (v_B cos θ) = D / (v_B × √3/2) = 2D / (v_B √3)

Answer: A.

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