Calculating Displacement in Relative Motion

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Homework Help Overview

The discussion revolves around calculating the total displacement of a passenger walking on a bus that is moving in a specific direction. The problem involves relative motion, specifically how the passenger's movement relates to the bus's velocity and direction.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore whether drawing vectors is necessary for solving the problem. Some question the completeness of the problem statement and the information provided, particularly regarding the passenger's velocity relative to the ground.
  • There is a discussion about the definitions of speed and velocity, with some participants expressing confusion over how direction factors into these concepts.
  • Several participants attempt to clarify the relationship between speed, direction, and velocity, while others question the relevance of the bus's speed in determining the passenger's velocity.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's interpretations. There is no explicit consensus, but various lines of reasoning are being explored regarding the definitions of velocity and the implications of the problem statement.

Contextual Notes

Some participants note that the problem may lack sufficient information to calculate the displacement accurately, and there is an emphasis on the importance of providing a complete problem statement in homework submissions.

Balsam
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Homework Statement


A person on a bus walks to the front of the bus at 3.0 km/h relative to the bus, while the bus travels south at 15km/h. What is the passenger's total displacement?

Homework Equations


displacement= final position-initial position
total displacement=displacement1+displacement2

The Attempt at a Solution


Do you draw vectors to do this? I have no clue what to do.
 
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Is that the exact problem statement word for word? I ask because it's not possible to calculate the displacement without more information.

Meanwhile try working out how fast the passenger is traveling (with respect to the ground).
 
CWatters said:
Is that the exact problem statement word for word? I ask because it's not possible to calculate the displacement without more information.

Meanwhile try working out how fast the passenger is traveling (with respect to the ground).

The question
CWatters said:
Is that the exact problem statement word for word? I ask because it's not possible to calculate the displacement without more information.

Meanwhile try working out how fast the passenger is traveling (with respect to the ground).
The question asks for the passesnger's velocity relative to the road, but I was unsure about how to find displacement because you need displacement to calculate velocity.
 
Balsam said:
The question

The question asks for the passesnger's velocity relative to the road, but I was unsure about how to find displacement because you need displacement to calculate velocity.
No, you don't. Velocity is a vector quantity. Vector quantities can be added and subtracted vectorially.
 
SteamKing said:
No, you don't. Velocity is a vector quantity. Vector quantities can be added and subtracted vectorially.

So how would you calculate the velocity, because the question only gives you 2 speed values?
 
Balsam said:
So how would you calculate the velocity, because the question only gives you 2 speed values?
Read the problem statement carefully again. It doesn't give you just speed; a direction is also specified for the travel of the bus and the person walking in the aisle.

Speed + Direction = Velocity.
 
SteamKing said:
Read the problem statement carefully again. It doesn't give you just speed; a direction is also specified for the travel of the bus and the person walking in the aisle.

Speed + Direction = Velocity.
I thought velocity was the rate at which an object is displaced while speed is the rate at which an object changes position-- doesn't that mean that you can't add a direction to speed and turn it into a velocity value, because they're not the same? Ex. An object that takes two seconds to move from a starting position forwards 2m and back 2m has a speed of 2m/s, but it has a velocity of 0m/s because it's not being displaced- I don't know if that's true, but that's what I think.
 
Balsam said:
I thought velocity was the rate at which an object is displaced while speed is the rate at which an object changes position-- doesn't that mean that you can't add a direction to speed and turn it into a velocity value, because they're not the same? Ex. An object that takes two seconds to move from a starting position forwards 2m and back 2m has a speed of 2m/s, but it has a velocity of 0m/s because it's not being displaced- I don't know if that's true, but that's what I think.
Displacement is a change in position. It is a vector.
Distance is a change in position. It is a scalar.

http://www.physicsclassroom.com/class/1DKin/Lesson-1/Distance-and-Displacement

Nothing is said in the problem about the bus going backwards and forwards. As far as you know, the bus is traveling south at 15 km/hr till the end of time. The guy in the aisle is walking forward at 3 km/hr relative to the bus. If you were standing on the side of the road as this bus and passenger passed you, what would be the speed of the passenger relative to you?
 
SteamKing said:
Read the problem statement carefully again. It doesn't give you just speed; a direction is also specified for the travel of the bus and the person walking in the aisle.

Speed + Direction = Velocity.

Balsam said:
I thought velocity was the rate at which an object is displaced while speed is the rate at which an object changes position-- doesn't that mean that you can't add a direction to speed and turn it into a velocity value, because they're not the same? Ex. An object that takes two seconds to move from a starting position forwards 2m and back 2m has a speed of 2m/s, but it has a velocity of 0m/s because it's not being displaced- I don't know if that's true, but that's what I think.

What you are saying is true. (Average) speed = sum of the absolute values of displacement, divided by the total trip time. The summation is over the separate "legs" of the journey. In your example, the speed on the forward portion is 2 m/s, so the displacement is 2t (m) (where t = forward trip time in seconds). The displacement on the backward portion is 2t as well; t must be the same because the distances and speeds are the same on both legs of the trip. So, the total absolute displacement = 2t + 2t = 4t, while the total trip time = 2t; thus, speed = 4t/2t = 2 (m/s). And, of course, velocity = 0, just as you said.
 
  • #10
Thats not relevant to this problem.

Velocity has two components, speed and direction.

The problem statement contains all the info needed to work out both components on their own. No need to add the two components.

Some examples of velocity...

100mph heading North.

60 meters per second vertically upwards.

4kts heading west along the equator.
 
  • #11
CWatters said:
Thats not relevant to this problem.

Velocity has two components, speed and direction.

The problem statement contains all the info needed to work out both components on their own. No need to add the two components.

Some examples of velocity...

100mph heading North.

60 meters per second vertically upwards.

4kts heading west along the equator.
So wouldn't the person's velocity just be 3km/h[N]? If so, what was the point of the bus' speed being given?
 
  • #12
Balsam said:
So wouldn't the person's velocity just be 3km/h[N]? If so, what was the point of the bus' speed being given?
According to you, the problem asks for the passenger's velocity relative to the road. This is why Item No. 1 in the HW template asks posters to provide a complete problem statement.

CWatters said:
Thats not relevant to this problem.

Velocity has two components, speed and direction.

The problem statement contains all the info needed to work out both components on their own. No need to add the two components.
Emphasis added.
Except, according to Post #3, the problem asks for the passenger's velocity relative to the road.
 
  • #13
SteamKing said:
Except, according to Post #3, the problem asks for the passenger's velocity relative to the road.

I agree. I think the OP was confused by..

SteamKing said:
Speed + Direction = Velocity.

eg It's not regular addition.
 

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