Question

Please help... The velocity vector of a boat has an x-component of 3.40 m/sec and a y-component of (-2.20) m/sec.

What is the magnitude and direction of the velocity vector?

Answer 1

for a vector\[\vec v=\langle x,y\rangle\]this can be written as a right triangle

Answer 2

from trig we know that\[\tan\theta=\frac yx\implies \theta=\tan^{-1}\frac yx\]and the length of the hypotenuse (which is the magnitude of the vector) is \[\|\vec v\|=\sqrt{x^2+y^2}\]

Answer 3

for all we know..magnitude=\[\sqrt{x ^{2}+y ^{2}}\]...if x and y be the components in respective direction...and the angle is tan^-1(y/x)

Answer 4

thats wit x axis

Answer 5

So, I still don't know what the answer would be

Answer 6

let shift the Vy parllel to it as vector doesn't change on shifting parllel to itself so
so in it using pythogoras theorem OB=\[\sqrt{OA^2+AB^2}\]= 7.48m/s
and angle is tan^-1(AB/OA)=-32.90525approx
:) which can u see from final fig