A river has a steady speed of vs. A student swims upstream a distanced and back to the starting point. (a) If the student can swim at a speed of v in still water, how much time t.n does it take the student to swim upstream a distance d? Express your answer in d, v, and vs.
A river has a steady speed of vs. A student swims upstream a distanced and back to the starting point. (a) If the student can swim at a speed of v in still water, how much time t.n does it take the student to swim upstream a distance d? Express your answer in d, v, and vs.
College Physics
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Swimming Against the Current: A Study in Physics**
A river has a steady speed of \( v_s \). A student swims upstream a distance \( d \) and back to the starting point.
**(a)** If the student can swim at a speed of \( v \) in still water, how much time \( t_{\text{up}} \) does it take the student to swim upstream a distance \( d \)? Express your answer in terms of \( d \), \( v \), and \( v_s \).
\[ t_{\text{up}} = \underline{\hspace{3cm}} \]
**(b)** Using the same variables, how much time \( t_{\text{down}} \) does it take to swim back downstream to the starting point?
\[ t_{\text{down}} = \underline{\hspace{3cm}} \]
**(c)** Sum the answers found in parts (a) and (b) and show that the time \( t_a \) required for the whole trip can be written as
\[ t_a = \frac{2d/v}{1 - v_s^2/v^2} \]
(Do this on paper. Your instructor may ask you to turn in this work.)
**(d)** How much time \( t_b \) does the trip take in still water?
\[ t_b = \underline{\hspace{3cm}} \]
**(e)** Which is larger, \( t_a \) or \( t_b \)?
- \( t_b \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb5f65923-b5e6-4be3-9be2-727b112c89ca%2F0f3541d1-4e4b-4e56-a34a-c4762ec83a96%2Fwxdas0q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Swimming Against the Current: A Study in Physics**
A river has a steady speed of \( v_s \). A student swims upstream a distance \( d \) and back to the starting point.
**(a)** If the student can swim at a speed of \( v \) in still water, how much time \( t_{\text{up}} \) does it take the student to swim upstream a distance \( d \)? Express your answer in terms of \( d \), \( v \), and \( v_s \).
\[ t_{\text{up}} = \underline{\hspace{3cm}} \]
**(b)** Using the same variables, how much time \( t_{\text{down}} \) does it take to swim back downstream to the starting point?
\[ t_{\text{down}} = \underline{\hspace{3cm}} \]
**(c)** Sum the answers found in parts (a) and (b) and show that the time \( t_a \) required for the whole trip can be written as
\[ t_a = \frac{2d/v}{1 - v_s^2/v^2} \]
(Do this on paper. Your instructor may ask you to turn in this work.)
**(d)** How much time \( t_b \) does the trip take in still water?
\[ t_b = \underline{\hspace{3cm}} \]
**(e)** Which is larger, \( t_a \) or \( t_b \)?
- \( t_b \)
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