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Chapter 3: Problem 77

A boy is attempting to swim directly across a river; he is able to swim at a speed of \(0.500 \mathrm{m} / \mathrm{s}\) relative to the water. The river is \(25.0 \mathrm{m}\) wide and the boy ends up at \(50.0 \mathrm{m}\) downstream from his starting point. (a) How fast is the current flowing in the river? (b) What is the speed of the boy relative to a friend standing on the riverbank?

Short Answer

Expert verified
Answer: The speed of the river's current is 1.00 m/s, and the boy's speed relative to the riverbank is 1.12 m/s.

Step by step solution

01

(Step 1: Set up the given information)

First, let's represent the horizontal motion with vector Vx, and the vertical motion with vector Vy. The boy's swimming speed across the river is Vy, which is 0.500 m/s. He ends up across the river (25.0 m) and downstream (50.0 m) from his starting point.
02

(Step 2: Calculate the time it takes for the boy to cross the river)

The time it takes for the boy to cross the river is the same in both directions. We can find this time by using the equation for distance: Distance = Speed × Time Rearranging for time, we get: Time = Distance / Speed We know the distance and speed of the boy in the vertical direction, so we'll plug them in to find the time: Time = 25.0 m / 0.500 m/s = 50.0 s
03

(Step 3: Calculate the speed of the river's current)

Now we'll use the horizontal distance the boy swam to find the speed of the river's current. Let Vx be the speed of the current, then: Horizontal Distance = Vx × Time Rearranging for Vx, we get: Vx = Horizontal Distance / Time Plugging in the values, we have: Vx = 50.0 m / 50.0 s = 1.00 m/s The speed of the river's current is 1.00 m/s.
04

(Step 4: Calculate the boy's speed relative to the riverbank)

To find the boy's speed relative to the riverbank, we need to combine the horizontal speed (Vx) and the vertical speed (Vy) as vectors. To do this, we'll use the Pythagorean theorem: VB_boy = √(Vx^2 + Vy^2) Plugging in the values, we get: VB_boy = √((1.00 m/s)^2 + (0.500 m/s)^2) = √(1.00 + 0.250) = 1.12 m/s (rounded to 2 decimal places) The boy's speed relative to the riverbank is 1.12 m/s.

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Most popular questions from this chapter

Sheena can row a boat at \(3.00 \mathrm{mi} / \mathrm{h}\) in still water. She needs to cross a river that is 1.20 mi wide with a current flowing at $1.60 \mathrm{mi} / \mathrm{h} .$ Not having her calculator ready, she guesses that to go straight across, she should head \(60.0^{\circ}\) upstream. (a) What is her speed with respect to the starting point on the bank? (b) How long does it take her to cross the river? (c) How far upstream or downstream from her starting point will she reach the opposite bank? (d) In order to go straight across, what angle upstream should she have headed?
Vector \(\overrightarrow{\mathbf{A}}\) is directed along the positive \(x\) -axis and has magnitude 1.73 units. Vector \(\overrightarrow{\mathbf{B}}\) is directed along the negative \(x\) -axis and has magnitude 1.00 unit. (a) What are the magnitude and direction of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}} ?\) (b) What are the magnitude and direction of \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}} ?\) (c) What are the magnitude and direction of \(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}} ?\)
A car travels east at \(96 \mathrm{km} / \mathrm{h}\) for \(1.0 \mathrm{h}\). It then travels \(30.0^{\circ}\) east of north at \(128 \mathrm{km} / \mathrm{h}\) for \(1.0 \mathrm{h} .\) (a) What is the average speed for the trip? (b) What is the average velocity for the trip?
A marble is rolled so that it is projected horizontally off the top landing of a staircase. The initial speed of the marble is $3.0 \mathrm{m} / \mathrm{s} .\( Each step is \)0.18 \mathrm{m}\( high and \)0.30 \mathrm{m}$ wide. Which step does the marble strike first?
The citizens of Paris were terrified during World War I when they were suddenly bombarded with shells fired from a long-range gun known as Big Bertha. The barrel of the gun was \(36.6 \mathrm{m}\) long and it had a muzzle speed of \(1.46 \mathrm{km} / \mathrm{s} .\) When the gun's angle of elevation was set to \(55^{\circ},\) what would be the range? For the purposes of solving this problem, neglect air resistance. (The actual range at this elevation was \(121 \mathrm{km} ;\) air resistance cannot be ignored for the high muzzle speed of the shells.)
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