Question

a river flows due south at 1.6 mi/h, and a swimmer attempts to cross the river from the west side to the east side. in what direction should the swimmer head (in degrees e of n), at a velocity of 4 mi/h, in order to arrive at a landing point due east of the starting point? (round your answer to one decimal place.)

Explanation:

Step1: Analyze the velocity vectors

Let the velocity of the river be \(v_{r}=1.6\) mi/h (south - direction) and the velocity of the swimmer relative to the water be \(v_{s}=4\) mi/h. We want the resultant velocity \(v\) to be in the east - direction.
If we consider the angle \(\theta\) (measured from the north towards the east), then the vertical (north - south) component of the swimmer's velocity relative to water must cancel the river's velocity.
The vertical component of the swimmer's velocity relative to water is \(v_{s}\sin\theta\). Since the river is flowing south (\(v_{r} = 1.6\) mi/h) and we want no net vertical displacement, we have \(v_{s}\sin\theta=v_{r}\).

Step2: Solve for \(\theta\)

We know that \(v_{s} = 4\) mi/h and \(v_{r}=1.6\) mi/h. Substituting into the equation \(v_{s}\sin\theta=v_{r}\), we get \(\sin\theta=\frac{v_{r}}{v_{s}}\).
Substitute \(v_{r} = 1.6\) and \(v_{s}=4\) into the formula: \(\sin\theta=\frac{1.6}{4}=0.4\).
Then \(\theta=\sin^{- 1}(0.4)\).
Using a calculator, \(\theta\approx23.6^{\circ}\)

Answer:

\(23.6\)