Asked by Lor

a ferry is crossing a river. If the ferry is headed due north with a speed of 2.5 meters per second relative to the water and the river's velocity is 3 meters per second to the east, what will the boat's velocity be relative to earth?
Answered by Henry
X = hor. = 3m/s.
Y = ver. = 2.5m/s.

tanA = Y / X = 2.5 / 3 = 0.8333,
A = 39.8 deg.

Vb = X / cosA = 3 / cos39.8 = 3.9m/s.
@ 39.8 deg.
Answered by DEE
Vfw=√(Vfr)²+(Vwr)²
Vfw=√(2.5)²+(3)²
=15.25 m/s
θ=tan^-1(2.5/3)=40°
Vfw=15.25 @ 40°
Answered by DEE
Vfw=√(Vfr)²+(Vwr)²
Vfw=√(2.5)²+(3)²
=3.9m/s
θ=tan^-1(2.5/3)=40°
Vfw=13.9 @ 40°
Answered by DEE
Vfw=√(Vfr)²+(Vwr)²
Vfw=√(2.5)²+(3)²
=3.9m/s
θ=tan^-1(2.5/3)=40°
Vfw=3.9 @ 40°
Answered by Anonymus pro
V= 7.9>67= 8549°
9489054°= PI 3.14x8^6= 968766-56= 98^x*09^y
=98493894898989^xy
Answered by Physics teacher (Jhon)
V= S= T= D=
9837295872398759283x9847885748^98989= 39874938274337710/399384
= 0950960940.0450490549^4989584^x^y= 485792875982375983
934938943894/3904893849389= 4985894.983798578943789983784

No one from my class discovered this procedure, just one student and got a A+ (10) for his grade, it's a dificult excercise but just try this way and you would got a A+
Answered by Phillip
84975237893578932758973289572398573987589237598798327832975923^39827589729875983275893728953729^xy= 3948782374983279547239842,902384293004 e
Answered by balls
773489342847384732874839274326523636290327936932 is the final answer.
Answered by john cena
10 =1
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