how do I do this?

Question

A boat capable of making 9.0 km/h in still water is used to cross a river flowing at a speed of 4.0 km/h. (a) At what angle must the boat be directed so that its motion will be straight across the river? Express the angle of the boat in the upstream direction from the direct path across the river. Also what is its resultant speed relative to the shore

Answer 1

Angle that the boat must be directed is 26.34 degrees, pointing upriver, or into the current.

Sin (Angle) = Opposite side/Hypoteneuse
Angle = Inv Sin (4/9)

For speed relative to bank either:
speed = sqrt(9^2-4^2) = 8.062

Cos(26.39) * 9 = Adjacent side = 8.062

Answer 2

Ok, first you must remember that velocity is a vector since it has magnitude and directon, so you can represent it as straight lines.

So you write down a diagram with the velocity vector of the boat supplied by the river and the velocity vector of the boat sopplied by it's engine. The resultant vector you get after adding the other two is going to be the horizontal vector across the river, so you get a rectangular triangle.

In order to get the angle you have tu use the function sin. Remember:

Sin θ = Opposite side/ hypotenuse.

in this case the opposite side is going to be the river velocity and the hypotenuse is going to be the boat velocity given by it's engine, so you solve for θ:

θ= sin-1(os/h)

and you get

θ=sin-1(os/h) = sin-1 (4/9)=26.4º

now you can get the resultant velocity with two methods, using the cos function or using pitagoras.

cos function:

cos θ = adjascent side/hypotenuse

as = h*cos θ

as = 9 km/h * cos 26.4º = 8.06 km/h

Pitagoras:

h^2=os^2+as^2

so

as = √(h^2 - os^2)

as = √((9km/h)^2 - (4km/h)^2)

as = 8.06 km/h

hope this helps.

Answer 3

.y
.l..\
.l...\
.l....\
.l.....\
.l......\.....
-x......(o,o)
.x=4km/h
y= target
what they said expains the rest

the hypotenuese is 9km/h (o,o) to (x,y)length

solve for y and do the trig y is the speed relative to shore

(ignore the dots used to make graph)