Doubt Calculating the Total Displacement for this Person Walking

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Homework Help Overview

The discussion revolves around calculating total displacement for a person walking, specifically focusing on the application of the displacement formula and the treatment of directional signs in the context of kinematics.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the correct application of the displacement formula, questioning how to handle positive and negative signs for different directional movements. There is a discussion about whether to treat displacements as vectors or scalars and how to sum them appropriately.

Discussion Status

Participants are actively engaging with the problem, with some clarifying their understanding of displacement versus distance. There is a recognition of the need to consider direction when calculating displacement, but no consensus has been reached on the best approach to take.

Contextual Notes

Some participants express confusion regarding the definitions of distance and displacement, as well as the implications of directionality in their calculations. There is an indication that the problem may involve specific homework rules regarding how to present answers.

Remle
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Homework Statement
You start walking home from school. After walking 1.3 km North, you get a phone call on your cell from your mom asking if you can meet her at the mall. You will have to turn around and walk 2.5 km South. Determine your distance and displacement to get to the mall.
Relevant Equations
d=df-di
I having a little bit of problem with ##\Delta d = d_f - d_i##. When substituting fo ##d_f## and ##d_i##, should I follow the signs rule (positive or negative)?
For example,
The problem shows that the displacement is ##1.2~\rm{km}, south## by solving ##\Delta d = -2.5 + 1.3## and I get that, but if I use the formula above the equation would appear like this ##\Delta d = -2.5 - 1.3## which gives me ##-3.8~\rm{km}## or ##3.8~\rm{km}, south##.

What am I missing?

source: http://www.studyphysics.ca/2007/20/01_kinematics/08_velocity.pdf
 
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There are two displacements, one is 1.3 km North and the other is 2.5 km South. The overall displacement is the sum of the two. Note that the displacements have opposite directions so you are adding a positive and a negative number.
 
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kuruman said:
There are two displacements.
THIS is what I needed. So first displacement is ##\Delta d = 1.3 - 0## and the second is ##\Delta d = 0 - 2.5## so to speak.
 
Remle said:
THIS is what I needed. So first displacement is ##\Delta d = 1.3 - 0## and the second is ##\Delta d = 0 - 2.5## so to speak.
If you are given two points (a start point and an end point), then the dispalcement is the position vector of the end point minus the position vector of the start point.

But, in this case you are given the displacements, so there is no need for any subtraction:
$$\Delta d_1 = 1.3km, \ \Delta d_2 = - 2.5km$$
 
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@Remle -- what did you get for the distance answer?
 
berkeman said:
@Remle -- what did you get for the distance answer?
Sorry for the late response. For the distance since is a scalar just had to add all the numbers. ##d = 1.3 + 2.5 = 3.8~\rm{km}##.
 
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Remle said:
Sorry for the late response. For the distance since is a scalar just had to add all the numbers. ##d = 1.3 + 2.5 = 3.8~\rm{km}##.
No, distance is not a scalar, it is a magnitude. Scalars have sign.
Taking North as positive you can find displacement using scalars: 1.3km+(-2.5)km N = -1.2km N, or 1.2km S.
For distances you add the magnitudes |1.3|+|-2.5|=3.8.
 

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