3.) [15 points] A physics teacher is on the west side of a small lake and wants to swim across and up at a point directly across from his starting point. He notices that there is a current in the lake and
that a leaf floating by him travels 4.2m [S] In 5.0s. He is able to swim 1.9 m/s in calm water,
(a) What direction will he have to swim in order to arrive at a point directly across from his position?

Given that the there is a water current across the lake, and the physics teacher intends to swim directly across the lake, the direction the physics teacher will have to swim is found as follows;

The speed of the water current is given by the speed of the floating leaf traveling with the water current

Distance traveled by the leaf = 4.2 m South

Time of travel of the leaf = 5.0 s

Speed of leaf = 4.2/5 = 0.84 m/s = Speed of the water current

Swimming peed of the teacher, v = 1.9 m/s

To swim directly across the lake, the teacher has to swim slightly in the opposite direction of the water current, the y-component of the teacher's swimming speed should be equal to and opposite that of the speed of the water current.

Y-component of v = v×sin(θ), where θ is the angle of the direction, the teacher should swim

Therefore;

1.9 × sin(θ) = 0.84

sin(θ) = 0.84/1.9 = 0.44

θ = 26.24°

That is the teacher should swim in a direction 29.24° North of East.

ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-03-30T10:06:42+02:00

To cross the lake the teacher has to swim in a direction 29.24° North of the East

Finding the direction of speed required:

The speed of the water current can be derived from the speed of the floating leaf :

The distance traveled by the leaf L = 4.2 m South

Time taken T = 5s

So, the speed of the leaf is:

u = 4.2/5

u = 0.84 m/s South

So, the speed of the current is 0.84 m/s South

Now, it is given that the speed of the teacher is, v = 1.9 m/s East

To cross the lake the speed of the teacher must be in a Northeast direction so that the North component of the speed of the teacher cancels out the speed of the current which is directed towards the South.

Let, the speed of the teacher makes an angle of θ from the EAST.

So, the North component is given by:

v(north) = vsinθ

it must be equal to the speed of the current:

vsinθ = u

1.9 × sinθ = 0.84

sinθ = 0.84/1.9

sinθ = 0.44

θ = 26.24°

The teacher should swim in a direction 29.24° North of East.

Learn more about vector components:

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