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 \(v_r = 1.6\) mi/h (south - direction) and the velocity of the swimmer \(v_s=4\) mi/h. We want the resultant velocity to be due east. Using the right - triangle relationship for velocity vectors (since the river flow and the swimmer's intended direction of crossing are perpendicular in the non - resultant sense). Let \(\theta\) be the angle (E of N) that the swimmer heads. The vertical (north - south) component of the swimmer's velocity must cancel the river's velocity.
The vertical component of the swimmer's velocity \(v_{s,y}=v_s\sin\theta\). We know that \(v_{s,y}=v_r\) (to cancel the river's south - ward flow). So, \(\sin\theta=\frac{v_r}{v_s}\).

Step2: Calculate the angle \(\theta\)

Substitute \(v_r = 1.6\) mi/h and \(v_s = 4\) mi/h into the formula \(\sin\theta=\frac{v_r}{v_s}\). Then \(\sin\theta=\frac{1.6}{4}=0.4\).
To find \(\theta\), we use the inverse - sine function \(\theta=\sin^{- 1}(0.4)\). Using a calculator, \(\theta\approx23.6^{\circ}\)

Answer:

\(23.6^{\circ}\)