Question

A physics teacher is on the west side of a lake and wants to swim across and end up at a point directly east from his starting point. He notices that there is a current in the lake and that a leaf floating by him travels 2.2 m [S] in 5.0 s. In calm water, he is able to swim 1.8 m/s. (b) [5 marks] Calculate his velocity relative to the shore. (c) [5 marks] If the lake is 545 m wide, how long will it take him to cross the lake?

Answer 1

The teacher's velocity relative to the shore is approximately 2.84 m/s. It will take him about 191.55 seconds to cross the lake.

To calculate the teacher's velocity relative to the shore, we need to consider his swimming speed and the velocity of the current.

Since he is swimming perpendicular to the current, we can use the Pythagorean theorem to find the resultant velocity.

The magnitude of his velocity relative to the shore is the square root of the sum of the squares of his swimming speed and the velocity of the current. So we have:

Velocity relative to the shore = √(swimming speed^2 + current velocity^2)

Plugging in the values:

Velocity relative to the shore = √(1.8^2 + 2.2^2) = √(3.24 + 4.84) = √8.08 ≈ 2.84 m/s

To calculate the time it will take him to cross the lake, we can use the formula:

Time = Distance / Velocity relative to the shore

Plugging in the values:

Time = 545 m / 2.84 m/s ≈ 191.55 s

Answer 2

Step-by-step explanation:

Current speed = 2.2 m / 5 s = .44 m/s

the horizontal component of his swimming must match this to go directly across the 'lake' ....

1.8 cos Φ = .44 shows the angle to be Φ = 75.85

Now the vertical distance across the 'lake' is 545 m

sin ( 75.85) = 545 / (distance he must swim)

distance he must swim = 562 meters

at 1.8 m/s this will take 562 m / 1.8 m/s = 312 seconds

Velocity relative to shore

1.8 sin 75.85 = 1.75 m/s