a swimmer can swim in still water at a speed of 9.50 m/s. he intends to swim directly across the river that has a downstream current of 3.75 m/s. what is his velocity relative to the bank?
Answers
Still-water speed = 9.5 m/s
River speed = 3.75 m/s down stream.
The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.
The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s
The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°
Answer: 10.2 m/s at 21.5° downstream.
The relative velocity of swimmer to the bank is [tex]\boxed{10.21\text{ m/s}}[/tex] and [tex]\boxed{21.54^\circ}[/tex] downstream.
Further Explanation:
The swimmer swims across the river from one end to the other. The river is flowing downstream.
The swimmer will cross the river with some relative velocity.
Given:
The velocity of swimmer is [tex]9.50\text{ m/s}[/tex].
The velocity of river is [tex]3.75\text{ m/s}[/tex] downstream.
Concept:
Consider the direction of velocity of swimmer in positive x-direction.
The velocity of swimmer in vector form:
[tex]V_s=9.50\hat i[/tex]
Consider the direction of downstream current of river in negative y-direction.
The velocity of downstream current in vector form:
[tex]V_r=- 3.75\hat j[/tex]
The relative velocity of swimmer to the bank is the resultant of two vectors in x and y direction.
Magnitude of relative velocity:
[tex]{V_{sr}}=\sqrt{V_r^2+V_s^2}[/tex]
Substitute [tex]9.50\text{ m/s}[/tex] for [tex]V_s[/tex] and [tex]3.75\text{ m/s}[/tex] for [tex]V_r[/tex] in above equation.
[tex]\begin{aligned}{V_{sr}}&=\sqrt{3.75^2+9.50^2}\text{ m/s}\\&=10.21\text{ m/s}\end{aligned}[/tex]
The magnitude of relative velocity of swimmer is [tex]10.21\text{ m/s}[/tex] .
The direction of relative velocity:
[tex]\theta={\tan^{-1}}\left({\dfrac{{{V_r}}}{{{V_s}}}}\right)[/tex]
Substitute [tex]9.50\text{ m/s}[/tex] for [tex]V_s[/tex] and [tex]-3.75\text{ m/s}[/tex] for [tex]V_r[/tex] in above equation.
[tex]\begin{aligned}\theta&={\tan^{-1}}\left({\dfrac{{{-3.75}}}{{{9.50}}}}\right)\\&=21.54^\circ\end{aligned}[/tex]
The direction of relative velocity is [tex]21.54^\circ[/tex] downstream.
Thus, the relative velocity of swimmer to the bank is [tex]\boxed{10.21\text{ m/s}}[/tex] and [tex]\boxed{21.54^\circ}[/tex] downstream.
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Answer Details:
Grade: Middle School
Subject: Physics
Chapter: Scalars and vectors
Keywords:
Swimmer, still, water, speed, 9.50 m/s, intends, directly, river, downstream, current, 3.75 m/s, velocity, relative, bank, vector, direction, 21.5degree and 10.21 m/s.
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Specifically, the object on Planet A, which has a stronger force of gravity, will weigh more than the object on Planet B, which has a weaker gravitational force.
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