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?

V= 7.9>67= 8549°

9489054°= PI 3.14x8^6= 968766-56= 98^x*09^y
=98493894898989^xy

84975237893578932758973289572398573987589237598798327832975923^39827589729875983275893728953729^xy= 3948782374983279547239842,902384293004 e

773489342847384732874839274326523636290327936932 is the final answer.

10 =1

To find the boat's velocity relative to the Earth, we need to use vector addition to combine its velocity relative to the water with the velocity of the river.

First, let's break down the problem into a coordinate system. We'll consider the north direction as the y-axis and the east direction as the x-axis. The ferry's velocity relative to the water is 2.5 meters per second north, and the river's velocity is 3 meters per second east.

Now, we can use vector addition to find the boat's velocity relative to the Earth. We add the two velocities by adding their corresponding components:

The boat's velocity in the x-direction (east) is the sum of the ferry's velocity in the x-direction (zero, since it's heading north) and the river's velocity in the x-direction (3 meters per second east).

The boat's velocity in the y-direction (north) is the sum of the ferry's velocity in the y-direction (2.5 meters per second north) and the river's velocity in the y-direction (zero, since it's heading east).

So, the boat's velocity relative to the Earth is (3 meters per second east, 2.5 meters per second north).

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°

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°

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+

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°

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.