Question

How am I supposed to solve this?

A swimmer wants to cross a river, from point A to point B, as shown in the figure. The distance d1 (from A to C) is 159 m, the distance d2 (from C to B) is 121 m, and the speed vr of the current in the river is 5 km/hr. Suppose that the swimmer's velocity relative to the water makes an angle of θ = 45 degrees with the line from A to C, as indicated in the figure.

To swim directly from A to B, what speed us, relative to the water, should the swimmer have?

Express the swimmer's speed in kilometers per hour.

Answer 1

Answer:

4.02 km/hr

Explanation:

5 km/hr = 1.39 m/s

The swimmer's speed relative to the ground must have the same direction as line AC.

The vertical component of the velocity is:

uᵧ = us cos 45

uᵧ = √2/2 us

The horizontal component of the velocity is:

uₓ = 1.39 − us sin 45

uₓ = 1.39 − √2/2 us

Writing a proportion:

uₓ / uᵧ = 121 / 159

(1.39 − √2/2 us) / (√2/2 us) = 121 / 159

Cross multiply and solve:

159 (1.39 − √2/2 us) = 121 (√2/2 us)

220.8 − 79.5√2 us = 60.5√2 us

220.8 = 140√2 us

us = 1.115

The swimmer's speed is 1.115 m/s, or 4.02 km/hr.

Answer 2

We are to find the speed Us relative to the water, should the swimmer have expressed in kilometer per hour

Therefore, the speed, Us of the swimmer relative to the water is

Us = 4.02km/hour

The speed Us of the swimmer relative to the water can be resolved into its vertical and horizontal component.

Therefore, the horizontal component of speed, Us is Usx = -(Us × Cos45°).

The negative sign above is due to the position of the speed, Us on the negative x-axis.

And, the vertical component of speed, Us is Usy = Us × Sin45°

Also, the horizontal component of speed Vr of the river is,

Vrx = Vr × Cos0° = 5km/hour

And, the vertical component of speed Vr of the river is,

Vry = Vr × Sin0° = 0km/hour

This is so because, the speed Vr makes an angle 0° with the horizontal.

Therefore, total speed in the horizontal and vertical direction, V'x and V'y are (5-UsCos45), and (UsSin45) respectively.

Therefore, the time Ty required to move from point A to C is,

Ty = d1/(UsSin45).

Also, the time Tx required to move from point C to B is,

Tx = d2/(5 - UsCos45)

However, Tx = Ty.

Therefore,

d1/(UsSin45) = d2/(5 - UsCos45).

i.e 159 / (0.7071Us) = 121 / (5 - 0.7071Us).

Cross product

85.56Us = 795 - 112.43Us

85.56Us + 112.43Us = 795

197.99Us = 795

The speed, Us = 795 / 197.99

The speed, Us of the swimmer relative to the water, is,

Us = 4.02km/hour.

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