Velocity & Acceleration Question

Click For Summary

Homework Help Overview

The discussion revolves around a scenario where Dave walks in one direction and then turns around to walk in the opposite direction. The focus is on understanding the concepts of velocity and acceleration at the moment of turning around.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definitions of velocity and acceleration, questioning how they apply at the moment of direction change. There are attempts to clarify the relationship between displacement, velocity, and acceleration, particularly at the instant of turning around.

Discussion Status

Participants are actively engaging with the definitions and implications of velocity and acceleration. Some have provided concrete examples to illustrate their points, while others are seeking clarification on the relationships between the variables involved. There is a recognition of the need to differentiate between displacement direction and the direction of motion.

Contextual Notes

There appears to be some confusion regarding the signs of displacement and velocity, as well as the interpretation of instantaneous changes in motion. Participants are encouraged to think critically about these aspects without reaching a definitive conclusion.

whitehorsey
Messages
188
Reaction score
0

Homework Statement



Dave walks to the left and then turns around and walks to the right without pausing. At the exact moment that he turns around is
a. velocity & acceleration zero
b. velocity zero & acceleration non -zero
c. velocity non-zero, acceleration zero
d. velocity and acceleration non-zero
e. all of the above depending on the situation

Homework Equations


The Attempt at a Solution



I think his velocity is non-zero because he's still walking while his acceleration is zero because his velocity is still constant?

Also, if a particle has a constant position it has a non-zero velocity? (because it has to be changing) I believe you have a non-zero velocity when the acceleration slope is zero and when an object falls near the surface of the Earth.
 
Last edited by a moderator:
Physics news on Phys.org
Take this back to the definition of velocity and acceleration ...

displacement is a change in position;
velocity is rate of change of displacement;
acceleration is rate of change of velocity;
all three are vectors - which affects what counts as a "change".
... you need to think what is involved in instantaneous changes.

So - at the exact instant Dave turns around, is his displacement changing? Is his velocity changing?
Put it another way - is there any way to change direction without accelerating?
 
Simon Bridge said:
Take this back to the definition of velocity and acceleration ...

displacement is a change in position;
velocity is rate of change of displacement;
acceleration is rate of change of velocity;
all three are vectors - which affects what counts as a "change".
... you need to think what is involved in instantaneous changes.

So - at the exact instant Dave turns around, is his displacement changing? Is his velocity changing?
Put it another way - is there any way to change direction without accelerating?

Ah is displacement changing because if we look at it on the x-axis he would have been going in the negative direction to the positive direction. I think. His velocity would be zero in the left direction when he changes displacement. So he has to accelerate since he is changing velocity. Therefore, velocity and acceleration are both non-zero?
 
Hmmm... let's make it concrete.
We first see Dave traveling right to left, to the right of the origin.

So his initial displacement will be +x (from the origin), his speed will be -v (changing displacement in the -x direction), and let's give him a constant speed for this part of the journey to keep things simple, so the acceleration is zero.

He reaches the origin at some time T - where he turns around and heads back the other way at velocity +v
At time T, before he turns around,
1. what is his displacement from the origin?
2. what is his velocity?
At time T, after he turns around,
3. what is his displacement from the origin?
4. what is his velocity?

5. what is the change in displacement?
6. what is the change in velocity?
 
Simon Bridge said:
Hmmm... let's make it concrete.
We first see Dave traveling right to left, to the right of the origin.

So his initial displacement will be +x (from the origin), his speed will be -v (changing displacement in the -x direction), and let's give him a constant speed for this part of the journey to keep things simple, so the acceleration is zero.

He reaches the origin at some time T - where he turns around and heads back the other way at velocity +v
At time T, before he turns around,
1. what is his displacement from the origin?
2. what is his velocity?
At time T, after he turns around,
3. what is his displacement from the origin?
4. what is his velocity?

5. what is the change in displacement?
6. what is the change in velocity?


Why is his speed -v if he is going +x from the origin?

1. +x
2. -v
3. -x
4. +v
5. -x -x = -2x
6. final velocity - initial velocity = 2v
 
OK - you can handle velocities just fine.
Displacements need a bit of work.

1. This tells me that when Dave is at the origin, his displacement from the origin is some distance x to the right of the origin?
3. Now you are saying that when he is at the origin, his displacement from the origin is some distance x to the left of the origin?

I think you are confusing the direction of a displacement with the direction Dave is facing.
Displacement is always from some position. When the displacement is from the origin, then it is the same as the position vector... so it may be easier for you to think of it in terms of position vectors instead.

Considering your answer to #6 - what can you say about the acceleration?
 
Simon Bridge said:
OK - you can handle velocities just fine.
Displacements need a bit of work.

1. This tells me that when Dave is at the origin, his displacement from the origin is some distance x to the right of the origin?
3. Now you are saying that when he is at the origin, his displacement from the origin is some distance x to the left of the origin?

I think you are confusing the direction of a displacement with the direction Dave is facing.
Displacement is always from some position. When the displacement is from the origin, then it is the same as the position vector... so it may be easier for you to think of it in terms of position vectors instead.

Considering your answer to #6 - what can you say about the acceleration?

Thank You for your help! I got it! :]
 
Well done - you'll find a lot of science is just asking the right questions.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
12
Views
2K
Replies
4
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 29 ·
Replies
29
Views
4K
Replies
67
Views
5K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
2
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 38 ·
2
Replies
38
Views
4K