A boat has a speed of 15ms-1 in still water. It is used to cross a river 500m wide along which there is a strong current of 0.9ms-1. The boat is being directed to the opposite bank, but is carried downstream.

Calculate:
a) The boat velocity relative to the bank
b) The distance the boat is carried downstream
c) The time taken to cross
d) The direction the boat must take, relative to the water if it is to cross directly
e) The time taken to cross directly, if it took the direction you calculated in part b.

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been there, done that

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a) The boat velocity relative to the bank:

The boat velocity relative to the bank can be calculated by finding the vector sum of the boat's velocity in still water and the velocity of the current. Since these velocities are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the boat's velocity relative to the bank.

The magnitude of the boat's velocity in still water is given as 15 m/s, and the magnitude of the current's velocity is given as 0.9 m/s. Therefore, the magnitude of the boat's velocity relative to the bank can be calculated as follows:

Vrelative = √(Vboat^2 + Vcurrent^2)
= √(15^2 + 0.9^2)
≈ √(225 + 0.81)
≈ √(225.81)
≈ 15.04 m/s

So, the boat's velocity relative to the bank is approximately 15.04 m/s.

b) The distance the boat is carried downstream:
To find the distance the boat is carried downstream, we need to calculate the time it takes for the boat to cross the river and then multiply it by the velocity of the current.

The time taken to cross the river can be calculated using the equation:
time = distance / velocity

The distance the boat needs to cross is 500 m. The boat's velocity in still water is given as 15 m/s, but it is being carried downstream by the current, so its effective velocity is the boat's velocity in still water minus the current's velocity:

Veffective = Vboat - Vcurrent
= 15 - 0.9
= 14.1 m/s

Using this effective velocity, we can calculate the time taken to cross the river:

time = distance / velocity
= 500 / 14.1
≈ 35.46 seconds

So, the boat is carried downstream for approximately 35.46 seconds.

c) The time taken to cross:
The time taken to cross the river is the same as the time calculated in part b, which is approximately 35.46 seconds.

d) The direction the boat must take, relative to the water if it is to cross directly:
To cross directly, the boat needs to have an angle of motion that compensates for the current's velocity. The boat should be aimed slightly upstream to counteract the effect of the downstream current.

To calculate this angle, we can use the following trigonometric relationship:

tan(θ) = Vcurrent / Vboat

where θ is the angle the boat must take relative to the water.

Plugging in the given values:

tan(θ) = 0.9 / 15
θ ≈ 3.64 degrees

So, the boat must take a direction approximately 3.64 degrees upstream relative to the water to cross directly.

e) The time taken to cross directly, if it took the direction calculated in part d:
Since the boat is traveling directly across the river under the angle calculated in part d, the effective velocity of the boat relative to the bank will be the same as the boat's velocity in still water:

Veffective = Vboat - Vcurrent
= 15 - 0.9
= 14.1 m/s

Using this effective velocity and the distance of 500 m, we can calculate the time taken to cross directly:

time = distance / velocity
= 500 / 14.1
≈ 35.46 seconds

So, the time taken to cross directly, if it took the direction calculated in part d, is approximately 35.46 seconds.