Solving Vector Problem B: Find Magnitude & Direction

[cowgiljl]

A ship sails due east across lake at 19 mi/hr. The ship encounters a current flowing due northat 3.00 mi/hr

A calculate the resultant velocity ( be sure to calculate the magnitude and direction)

I drew a picture

R^2 = 19^2 + 3^2
R = about 19 mi/hr



B) In what direction should the ship head in order to offswet the effect of the current
tan-1 (3/19) = 8.97 degrees n of east

Just need some reassurance I am doing these right

 

[himanshu121]

First of all write vector eqn in terms of i,j,k components

then apply
$$V_a_b = V_a - V_b$$

rel velocity of a wrt b= vel of a - vel of b

 

[himanshu121]

Always try to write vector eqn if you are not able understand graphically and derive the results accordingly

 

[HallsofIvy]

Two points:

1) The problem said "calculate the resultant velocity " NOT "speed".
Technically, velocity is a vector quantity.
2) The problem specifically said " be sure to calculate the magnitude and direction"

You have only given the resultant speed (magnitude of the velocity).

You should either find the angle (arctan(3/19) north of east) or
(much simpler) write the two given velocities as vectors and add:
Using the standard convention that <1, 0> is east, <0, 1> north,
19 m/h n is <0, 19> and 3 m/h e is <3, 0>

The resultant velocity is <3, 19>. That was easy!