# How to deal with vectors (Motion in two dimension)

1. Oct 13, 2012

### RuthlessTB

1. The problem statement, all variables and given/known data

X-axis
t1= 3 min
t2= 2 min (Since it's west, it is -2)
t3= 1 min (Since it's northwest, it is -1)

Y-axis
V1y= 20 m/s (Since it's south, it is -20)
V2y= 25 m/s (It's -25)
V3y= 30 m/s (Stays positive since it's in the northwest)

2. Relevant equations
A) Total Vector Displacement
Δr= (t3-t2-t1) i + (V3-V2-V1) j

B) Average Speed
Not really sure, I guess it is
Average Speed = Distance/TotalTime

C) Average Velocity
Vav = Displacement/ΔTime

3. The attempt at a solution
A) Total Vector Displacement
Δr= (-1 i + 30 j) - (-2 i + -25 j) - (3 i + -20 j)
Δrx= ((-1)-(-2)-3) i = -2 i
Δry= ((30)-(-25)-(-20)) j = 75 j
Δr= -2 i + 75 j

B) Average Speed
Total time is 6 minutes = 360 seconds
but I don't know should I use the velocity as a distance or I need to calculate it too?

C) Average Velocity
To calculate the displacement, I take the square root of x^2 + y^2
Displacement = Square root of (-2)^2 + (75)^2 = 75.03

Δt, -2 minutes = -120 seconds
My instructor said time should be positive, but I don't know if Δtime should be positive too or it's okay to be negative.

Vav= 75.03/-120 = -0.63 m/s

I hope anyone could tell me if what I did is right and what I need to do in order to calculate the average speed.

2. Oct 13, 2012

### lewando

Your RE for total vector displacement is incorrect and is at the root of your problems. i j notation is fine, but your i component needs to be the sum of the x-axis components of displacement (distance), not time. Similarly, your j component needs to be the sum of y-axis displacements, not velocities.

3. Oct 13, 2012

### RuthlessTB

Do I need the angel in order to calculate the distance in these three points traveled?

4. Oct 13, 2012

### lewando

For each leg, you need to be angle-aware (direction-aware) in order to determine the i (x-axis, East +, West -) component and the j (y-axis, North +, South -) component of displacement.

Take one leg at a time, beginning with the first, what do you get for i, j displacement?

5. Oct 13, 2012

### HallsofIvy

Staff Emeritus
This makes no sense. Since this is two-dimensional movement, you want the x- axis to be east-west movement, not time. And, assuming that time moves forward for this person like it does for the rest of us, time difference is always positive.

Taking his initial position to be (0, 0), positive x east, positive y north, as usual, his first leg is from (0, 0) to (0, -40). His second leg will be from (0, -40) to (-50, -40). On his third leg, since he drives at 30 m/s for 1 second, he will have gone a distance of 30 m. Since he is driving "northwest" that will be $-30\sqrt{2}{2}= -15\sqrt{2}$ west and $-30\sqrt{2}{2}= 15\sqrt{2}$ north. He will have driven from (-50, 40) to $(-50- 15\sqrt{2}, -40+ 15\sqrt{2})$.
His total vector displacement will be from (0, 0) to ($(-50- 15\sqrt{2}, -40+ 15\sqrt{2})$.

The "displacement" is the length of that vector.

Again, "displacement" is in space only- the first component is NOT time.

Frankly, you did nothing right. You don't seem to know what "displacement vectors" and "velocity vectors" are. Your errors are so fundamental that I strongly recommend that you talk about this with your teacher.

6. Oct 13, 2012

### RuthlessTB

Actually my instructor is horrible, he just gives the laws or equations with some simple information and moving on. I am self-studying this that is why I am facing difficulties.
Anyway, gotta try to solve it later and post my solution.