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Chapter 3: Problem 22

A boat travels at a speed of \(v_{\mathrm{BW}}\) relative to the water in a river of width \(D .\) The speed at which the water is flowing is \(v_{\mathrm{W}}\) a) Prove that the time required to cross the river to a point exactly opposite the starting point and then to return is \(T_{1}=2 D / \sqrt{v_{B W}^{2}-v_{W}^{2}}\) b) Also prove that the time for the boat to travel a distance \(D\) downstream and then return is \(T_{1}=2 D v_{\mathrm{B}} /\left(v_{\mathrm{BW}}^{2}-v_{\mathrm{w}}^{2}\right)\)

Short Answer

Expert verified
Question: Calculate the time it takes for a boat to cross a river of width D and return to its starting point in two different scenarios: a) The boat aims to reach exactly the opposite point on the shore, and b) the boat travels a distance D downstream and then returns to the starting point. The boat's velocity relative to water is \(v_{\mathrm{BW}}\) and the river's current velocity is \(v_{\mathrm{W}}\). Answer: a) \(T_1 = 2\frac{D}{\sqrt{v_{\mathrm{BW}}^2 - v_{\mathrm{W}}^2}}\), b) \(T_2 = \frac{2Dv_{\mathrm{BW}}}{v_{\mathrm{BW}}^2 - v_{\mathrm{W}}^2}\).

Step by step solution

01

Determine the boat's actual velocity component perpendicular to the river#

First, let's determine the boat's velocity component perpendicular to the river, which we can call \(v_{\mathrm{B}}\). Since the boat has to overcome the river's current, we can determine \(v_{\mathrm{B}}\) using the Pythagorean theorem: \(v_{\mathrm{B}}^2 = v_{\mathrm{BW}}^2 - v_{\mathrm{W}}^2\). Here, \(v_{BW}\) is the boat's velocity relative to water, and \(v_W\) is the river's current velocity.
02

Calculate the time required to cross the river#

Now that we have the boat's velocity perpendicular to the river (\(v_{\mathrm{B}}\)), we can calculate the time it takes for the boat to cross the river using the formula \(t = \frac{d}{v}\). Since the river's width is given as \(D\), the time required to cross the river is \(t_{\mathrm{cross}} = \frac{D}{v_{\mathrm{B}}} = \frac{D}{\sqrt{v_{\mathrm{BW}}^2 - v_{\mathrm{W}}^2}}\).
03

Double the time to account for the return trip#

As the boat turns back and returns exactly to the starting point, the time required for the return trip is the same as the time it took to cross the river initially. Therefore, the total time required (\(T_1\)) is twice the time calculated in Step 2: \(T_1 = 2t_{\mathrm{cross}} = 2 \frac{D}{\sqrt{v_{\mathrm{BW}}^2 - v_{\mathrm{W}}^2}}\). b) Time required to travel a distance \(D\) downstream and return to the starting point
04

Determine the boat's velocity components downstream and upstream#

We will first find the boat's velocity components downstream (\(v_{\mathrm{DS}}\)) and upstream (\(v_{\mathrm{US}}\)). When the boat is moving downstream, we will add the river's current velocity to the boat's speed relative to water: \(v_{\mathrm{DS}} = v_{\mathrm{BW}} + v_{\mathrm{W}}\). Similarly, when going upstream, we will subtract the river's current velocity from the boat's speed relative to water: \(v_{\mathrm{US}} = v_{\mathrm{BW}} - v_{\mathrm{W}}\).
05

Calculate the time required to travel downstream and upstream#

We can use the formula \(t = \frac{d}{v}\) again to calculate the time taken when traveling downstream (\(t_{\mathrm{DS}}\)) and upstream (\(t_{\mathrm{US}}\)). Since both distances are given as \(D\), we have \(t_{\mathrm{DS}} = \frac{D}{v_{\mathrm{DS}}} = \frac{D}{v_{\mathrm{BW}} + v_{\mathrm{W}}}\) and \(t_{\mathrm{US}} = \frac{D}{v_{\mathrm{US}}} = \frac{D}{v_{\mathrm{BW}} - v_{\mathrm{W}}}\).
06

Calculate the total time required for both parts of the trip#

Finally, we will sum up the time taken for both parts of the trip to get the total time \(T_2 = t_{\mathrm{DS}} + t_{\mathrm{US}} = \frac{D}{v_{\mathrm{BW}} + v_{\mathrm{W}}} + \frac{D}{v_{\mathrm{BW}} - v_{\mathrm{W}}}\). This can be simplified by finding the common denominator and combining the fractions: \(T_2 = \frac{2Dv_{\mathrm{BW}}}{v_{\mathrm{BW}}^2 - v_{\mathrm{W}}^2}\).

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