(Figure 1)A swimmer wants to cross a river, from poi... Step 1: This velocity can be dissected into two segments: one that is perpendicular to the current (aiding the swimmer in moving directly across the river) and one that aligns with the current (being carried by its flow). Step 2: Distance are d₁ = 200 m and d₂ = 150 m Velocity of the current in the river is V₁ = 5 Km/hr The swimmer's velocity relative to the water makes an angle of 0 =45° Break the Us component in two direction in vertical and in horizontal direction As we know that the us cose is responsible for clearing d1 distance and 5 - us sine is responsible for d₂ distance. velocity = v = t us cose t = = d₁ Us cose distance time d₁ t (1) The time which we have find out above is taken to cover the d² distance 5 - us sine t = d2 5-us sine = d2 t ...(2) The time taken to cover distance d₂ is equal to the time taken by distance d₁ Compare eq (1) and eq (2), we have d₂ = d₁ 5-us sine Us cose

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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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why do we subtract 5 from ussintheta for the velocity? and why do we set d1=d2, why do we assume that the time taken to cover d2 is equal to the time taken by distance d1?

(Figure 1)A swimmer wants to cross a river, from poi...
Step 1:
This velocity can be dissected into two segments: one that is perpendicular to the current (aiding the
swimmer in moving directly across the river) and one that aligns with the current (being carried by its flow).
Step 2:
Distance are d₁ = 200 m and d₂ = 150 m
Velocity of the current in the river is V₁ = 5 Km/hr
The swimmer's velocity relative to the water makes an angle of 0 =45°
Break the Us component in two direction in vertical and in horizontal direction
As we know that the us cose is responsible for clearing d1 distance and
5 - us sine is responsible for d₂ distance.
velocity =
v =
t
us cose
t
=
=
d₁
Us cose
distance
time
d₁
t
(1)
The time which we have find out above is taken to cover the d² distance
5 - us sine
t
=
d2
5-us sine
=
d2
t
...(2)
The time taken to cover distance d₂ is equal to the time taken by distance d₁
Compare eq (1) and eq (2), we have
d₂
=
d₁
5-us sine Us cose
Transcribed Image Text:(Figure 1)A swimmer wants to cross a river, from poi... Step 1: This velocity can be dissected into two segments: one that is perpendicular to the current (aiding the swimmer in moving directly across the river) and one that aligns with the current (being carried by its flow). Step 2: Distance are d₁ = 200 m and d₂ = 150 m Velocity of the current in the river is V₁ = 5 Km/hr The swimmer's velocity relative to the water makes an angle of 0 =45° Break the Us component in two direction in vertical and in horizontal direction As we know that the us cose is responsible for clearing d1 distance and 5 - us sine is responsible for d₂ distance. velocity = v = t us cose t = = d₁ Us cose distance time d₁ t (1) The time which we have find out above is taken to cover the d² distance 5 - us sine t = d2 5-us sine = d2 t ...(2) The time taken to cover distance d₂ is equal to the time taken by distance d₁ Compare eq (1) and eq (2), we have d₂ = d₁ 5-us sine Us cose
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