Explanation Given data: The river flows at 0.52 m s . Still water velocity is 1.78 m s . Formula used: Write the expression for velocity. v = Δ x Δ t (1) Here, Δ x is the displacement, and Δ t is the time interval. Explanation: Find the value of velocity in downstream. v downstream = 1.78 m s + 0.52 m s = 2.3 m s Modify equation (1) for downstream. v downstream = Δ x downstream Δ t downstream Δ t downstream = Δ x downstream v downstream Substitute 2.3 m s for v downstream , Δ t downstream = Δ x downstream 2.3 m s = Δ x downstream 2.3 s Find the value of velocity in upstream. v upstream = 1.78 m s − 0.52 m s = 1.26 m s Modify equation (1) for v upstream . v upstream = Δ x upstream Δ t upstream Δ t upstream = Δ x upstream v upstream Substitute 1.26 m s for v upstream , Δ t upstream = Δ x upstream 1.26 m s = Δ x upstream 1

16th Edition

ISBN: 9781323515303

Author: Knight

Publisher: PEARSON C

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Question

Chapter 1, Problem 59GP

To determine

The distance in percentage, that the swimmer can takes to swim in downstream and come back to upstream compared with same trip in still water.

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Chapter 1, Problem 58GP

Chapter 1, Problem 60GP

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Chapter 1 Solutions

COLLEGE PHY2053 W/MODIFIED ACCESS>BI<

Ch. 1 - A softball player slides into second base. Use the...Ch. 1 - A car travels to the left at a steady speed for a...Ch. 1 - A ball is dropped from the roof of a tall building...Ch. 1 - Prob. 5CQ Ch. 1 - Give an example of a trip you might take in your...Ch. 1 - Write a sentence or two describing the difference...Ch. 1 - The motion of a skateboard along a horizontal axis...Ch. 1 - You are standing on a straight stretch of road and...Ch. 1 - Two friends watch a jogger complete a 400 m lap...Ch. 1 - A softball player hits the ball and starts running...

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