Need clarification about x-t and v-t graph

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Homework Help Overview

The discussion revolves around understanding the implications of negative slopes on x-t (position-time) and v-t (velocity-time) graphs in the context of motion. Participants are exploring the meanings of these slopes and their relationship to velocity and acceleration.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants describe the interpretations of positive and negative slopes on both x-t and v-t graphs, questioning whether a negative slope indicates slowing down or speeding up. They also discuss the relationship between average velocity and displacement.

Discussion Status

The discussion is active, with participants providing insights into the meanings of slopes and engaging in clarifying questions. Some guidance has been offered regarding the relationship between velocity and acceleration signs, but no consensus has been reached on the implications of negative slopes.

Contextual Notes

Participants are navigating through definitions and relationships in kinematics, with some expressing confusion about the physical meanings behind graph slopes. There is a mention of homework constraints regarding the interpretation of average velocity and displacement.

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Homework Statement


I'm confused with what a negative slope means on an x-t and v-t graph.

On an x-t graph, I understand that a positive constant slope means uniform velocity and when the slope is 0, then the object is stationary. On a v-t graph, constant slope means constant acceleration and a zero slope means constant speed.

What does the negative slope mean? is the object slowing down?

Can someone please help me get a clearer understanding of this? THANKS


Homework Equations





The Attempt at a Solution

 
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On an x vs. t graph:

If the slope is positive, then the velocity is constant and positive.
If the slope is zero, then velocity is zero.
If the slope is negative, then velocity is constant and negative.

On a v vs. t graph:

If the slope is positive, then the acceleration is constant and positive.
If the slope is zero, then the acceleration is zero.
If the slope is negative, then the acceleration is constant and negative.

Generally speaking, acceleration is the derivative of the equation for velocity with respect to time (a = dv/dt), and velocity is the derivative of the equation for position (v = dx/dt).

Hope that helps.
 
Linear negative slope on a x-t graph means uniform negative velocity. The object is moving in the opposite direction.

Linear negative slope on a v-t graph means constant deceleration. Or said in another way, constant negative acceleration.
 
mplayer said:
Linear negative slope on a x-t graph means uniform negative velocity. The object is moving in the opposite direction.

Linear negative slope on a v-t graph means constant deceleration. Or said in another way, constant negative acceleration.

but does it mean that the object is slowing down?
 
mizzy said:
but does it mean that the object is slowing down?

Depends. If acceleration and velocity are both the same sign (negative vs. positive), then it's speeding up. If the signs are different, it's slowing down. I think that's correct.
 
ok. so basically the negative or positive sign just tells us the direction and not if is speeds or slows down. correct??

i came across this question? If the average velocity of an object at a certain time interval is zero, what is its displacement? What's happening to the object?

If the average velocity is zero, the object is stationary. is that right?
 
mizzy said:
ok. so basically the negative or positive sign just tells us the direction and not if is speeds or slows down. correct??

i came across this question? If the average velocity of an object at a certain time interval is zero, what is its displacement? What's happening to the object?

If the average velocity is zero, the object is stationary. is that right?

Correct on the signs. Well, sort of - opposing signs mean slowing down when you're talking velocity and accel.

Correct on the displacement. Δx = Δv / Δt ; if Δv = 0, then Δx = 0/t.
 
Wow, that seems very cumbersome if you're just trying to understand what is being represented on a graph. It's almost like memorization if you are trying to attach the concept of slope to its actual physical meaning. I've got a question for all of you - is there really such a thing as "instantaneous velocity", in the real physical world? If you can answer that, then you understand motion graphs.
 

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