Relative Velocity Question

1. Oct 18, 2003

mcintyre_ie

Id be very grateful for any help with this relative velocity question:

To a man walking North at 4m/s, a constant wind appears to have a speed f 3m/s. The man then changes direction, and when he is walking West at 3m/s the wind appears to have a speed of 4 m/s. There are two possible wind velocities.

If I and j are unit vectors in the east and north directions, respectively:
1) Show that -3i + 4j is one possible wind velocity
2) Find the second possible wind velocity.

Ok, so this is what I¡¦ve got so far:
Vm (Velocity of man) = 4m/s = 4j m/s. (When wind appears to have Vw (Velocity of wind) of 3m/s)
Vm (Velocity of man) =3m/s = 3i m/s (wind appears to have Vw (Velocity of wind) of 4m/s)

Vm = 4j
Vwm = xi
Vw = ai + bj

Vwm = Vw ¡V Vm
Xi = ai + bj ¡V 4j
Xi = ai + bj +0i -4j
„Ã xi = ai
„Ã x = a

„Ã 0 = b ¡V 4
„Ã b=4

Vm = -3i
Vwm = yi ¡V yj
Vw = ai + bj

Vwm = Vw ¡V Vm
Yi ¡V yj = (ai + bj) + 3i
„Ã yi = ai +3i
„Ã -y = b
„Ã y = -4

Y = a ¡V 3
-4 = a ¡V 3
1 = a

This gives me a velocity of 1i + 4j.

Have i gone totally wrong, or have i just found the answer to part two instead of part one?
If so, what do i do next to find the second velocity, or, if its all just totally wrong, how should i actually go about doing the question?

2. Oct 19, 2003

mcintyre_ie

Any help is appreciated

3. Oct 20, 2003

mcintyre_ie

Anybody at all?

4. Oct 20, 2003

HallsofIvy

Staff Emeritus
Okay, I saw this the other day, meant to respond and didn't get around to it!

The man is, originally walking n so, setting up a coordinate system with i pointing east, j pointing north, his velocity vector is, as you say, 4j.
I'm not at all sure what you mean by
"Vwm = Vw ¡V Vm
Xi = ai + bj ¡V 4j
Xi = ai + bj +0i -4j"

You defined Vm and Vw but you didn't say what Vwm was except "xi" which isn't a definition since you hadn't define xi! What is "iV"?

Here's how I would do it: writing the wind's velocity vector as ai+ bj (your "Vw"), the velocity of the wind relative to the man is (ai+ bj)- 4j= ai+ (b-4)j. The speed of the wind relative to the man is &radic;(a2+ (b-4)2). We are told that this 3.

When the man is walking west at 3 m/s, his velocity vector is -3i (you have 3i: that would be walking east) and the velocity of the wind relative to him is (a+ 3)i+ b. The speed of the wind relative to the man is &radic;((a+ 3)2+ b2). We are told that that is 4.

Squaring both sides of &radic;(a2+ (b-4)2)= 3, we get (a2+ (b-4)2= 9.
Squaring both sides of &radic;((a-3)2+ b2) we get (a-3)2+ b2= 16.
That gives 2 equations to solve for a and b.

Actually, we don't need to solve to answer the first part. It just asks that we show that -3i+ 4j (that is, a=-3, b= 4) satisfies the conditions. Putting a= -3, b= 4 in the first equation, we have
(-3)2+ (4-4)2= (-3)2+ 02= 9 which is correct.
Puting a=-3, b= 4 in the second equation, we have
(-3+ 3)2+ (4)2= 16 which is, again, correct.
Yes, a wind speed of -3i+ 4j satisfies both conditions.

Going back to the equations that must be satisfied,
(a2+ (b-4)2= 9 and
(a+3)2+ b2= 16 we can multiply the squares and get:
a2+ b2- 8b+ 16= 9 and
a2+ 6a+ 9+ b2 = 16.
Subtracting the second equation from the first, the "a2" and "b2" terms cancel and we have
-8b- 6a= -14 or, dividing by -2, 4b+ 3a= 7.
(Notice that b= 4, a= -3 satisfy this.)

putting a= 7/3- (4/3)b into the fist equation, we have
(7/3-(4/3)b)2+ (b-4)2= 9 or
49/9- (56/9)b+ (16/9)b2+ b2-8b+ 16= 9 which is the quadratic equation (25/9)b2- (128/9)b+ 112/9= 0.

That is the same as 25b2- 128 b+ 112= 0.

Knowing that b= 4 is one solution, it shouldn't be hard to factor and find the other possible value for b and then find a.