Calculating Relative Velocity on a Moving Ship: Magnitude and Direction

[Clari]

A man walks with a velocity of 2m/s due North on a ship, which moves due East at 3m/s. Find the relative velocity of the man to the Earth, in magnitude and direction.

I do it this way:
rel. vel. of man to ship = Vm -Vs
Vm = rel. vel. of man to ship + Vs
....

But the answer is 3.6 m/s in direction N56(degrees)E, so I have no idea how it gets that...Help me please...Thanks!

P.S. Vm is relative velocity of man to earth

 

[Leong]

[tex]\vec{v_m}=[/tex] velocity of the man relative to earth
[tex]\vec{v_s}[/tex] =velocity of the ship relative to Earth = 3i
[tex]\vec{v_{mc}}= velocity\ of\ the\ man\ relative\ to\ the\ ship=
\vec{v_m}-\vec{v_s}= 2j[/tex]
[tex]\vec{v_m}=2j + 3i[/tex]. The rest will be left to you. j is the unit vector points to north. i is the unit vector points to east.

 

[Clari]

Hello Leong!
your help is much appreciated~ ^v^ But I don't understand why the rel. vel. of man to ship is 2j.

 

[faust9]

The North South vector is the y-axis is the j axis.

The East west vector is the x-axis is the i axis.

So, if the ship is traveling east then its velocity vector is 3i

If the man is traveling north then his velocity vector is 2j

adding the two vectors together yields [itex]\vec{v}=3i+2j[/itex]

 

[Chi Meson]

But you teacher/texbook isprobably looking for a so-called "resultant" velocity and direction. THink of faust's answer as "3 horizontal and 2 vertical" ( know it's actually east and north, but...)

draw a horizontal line 3 cm long. Now starting at the end of that line (the right side) draw a vertical line up two cm. Draw a third line from the start of the first line to the end of the second line, and measure it. You'll notice it is 3.6 cm.

Using math, you can see that it is the hypotenuse of the triangle you just drew. Use the pythagorean theorem to get the mathematical answer. Use trigonometry to find the angle.

 

[Leong]

The relative velocity of man to ship is 2j because the man is walking on the ship which is as well moving. if you are on the ship, you will see the man walking north; that is the velocity of the man relative to you or the ship is 2j. if you are outside the ship or we can say on the earth, then you will see the man walking with the velocity given by the answer.

 

[Clari]

Hello Chi Meson!
I would like to know when I should draw the diagram with all the vectors starting from one point and when to draw by the method as you described...I am confused now..When my teacher talk about change of velocity, she said I should draw in the former method.

 

[Fredrik]

I would like to know when I should draw the diagram with all the vectors starting from one point and when to draw by the method as you described.

If you try both methods, you'll see that the result is the same. Chi meson's triangle will look like a mirror reflection of yours, but that's not relevant if all you're looking for is the length of the hypothenuse.

The difference between the two methods is that the one Chi meson suggested will work even when the two vectors aren't perpendicular to each other.

 

[Chi Meson]

The method I described is commonly called the "tail to tip" method. No matter how individula vectors are "given" to you, you can pick them up and put them down wherever you want; as long as they are the same length, and pointing in the same orientation, they are the same vector.

The tail to tip method is useful to get a quick idea of what your matematical answer should be (especially if there are more than two vectors to add together). The rule is simple, put the tail of the next vector at the tip of the previous vector until all the vectors are strung together. THe resultant will be the vector drawn from the tail of the first vector to the tip of the last vector.

But when you simply have two vectors that are perpendicular, go straight to the math, because the resultant will always be the hypotenuse of a right triangle where the original vectiors are the two sides.