How to deal with vectors (Motion in two dimension)

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Homework Help Overview

The discussion revolves around a problem involving motion in two dimensions, specifically focusing on vector displacement, average speed, and average velocity. Participants are analyzing the components of motion along the x and y axes, considering directions and magnitudes of velocities.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of total vector displacement, questioning the use of time versus distance in the displacement formula. There are attempts to clarify the correct interpretation of components in vector notation, particularly regarding the distinction between time and displacement.

Discussion Status

Some participants have provided guidance on the need to focus on spatial displacement rather than time in the vector calculations. Others are exploring the implications of direction on the components of displacement and velocity. There is a recognition of the need for clarity in understanding the definitions of displacement and velocity vectors.

Contextual Notes

There are indications of confusion regarding the treatment of time in the calculations, with some participants questioning whether time differences should be treated as positive. Additionally, there is mention of the challenges faced by the original poster in self-studying the material.

RuthlessTB
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Homework Statement


2v02800.png


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)

Homework 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

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 jB) 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.
 
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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.
 
Do I need the angel in order to calculate the distance in these three points traveled?
 
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?
 
RuthlessTB said:

Homework Statement


2v02800.png


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)
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.

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)
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.

Homework Equations


A) Total Vector Displacement
Δr= (t3-t2-t1) i + (V3-V2-V1) j
Again, "displacement" is in space only- the first component is NOT time.

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

C) Average Velocity
Vav = Displacement/ΔTime

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 jB) 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.
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.
 
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, got to try to solve it later and post my solution.
 

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