How do find the displacement when it is =ing to distance

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Homework Help Overview

The discussion revolves around understanding the concepts of displacement and distance in the context of a ball moving in a straight line with uniform velocity, hitting a wall, and returning to its starting point. The problem involves drawing a velocity-time graph and calculating total distance traveled and displacement.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate displacement using the formula displacement = velocity × time but struggles with the concept due to the ball returning to its starting point. Some participants question the relationship between velocity direction and displacement.

Discussion Status

Participants have engaged in clarifying the concepts of velocity and displacement, with one suggesting that the negative velocity during the return trip contributes to the overall displacement calculation. There is a sense of progress as the original poster expresses understanding after considering the area under the velocity-time graph.

Contextual Notes

The original poster is new to the forum and is seeking help with fundamental concepts. There is an emphasis on understanding the distinction between distance and displacement, particularly in scenarios involving changes in direction.

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How do find the displacement when it is =ing to distance!

I am new here ,so please bear with me by helping in this simple problem:confused::cry:
Problem:A ball moes on ta smooth floor in a straight line with uniform velocity 10meters/seconds for 6seconds.At t=6seconds, the ball hits a wall and comes back along the same line to the starting point with same speed.Draw the velocity -time graph and use it to find the total distance traveled by the ball and its displacement.
My try:
I had tried to make rough graph but the problem is that it is not coming displacement as 0
Because if velocity =10m/s & time=12(total time after returning)
.: displacement=v*t=10*12=120
But as it comes back to its original place ,I don't understand , How does it happen ?&
How do I represent it?:frown:
.Plz can anyone clear the consepts ?!:frown:
 
Last edited:
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Hint: velocity, apart from a magnitude, also has a direction.
 


THANKS A LOT CompuChipip!:smile::smile::smile:
I think I got the answer, as there is a negetive velocity due to moving back of the ball.:
It appears like this
|V
|
| __________
| .
| .
| .
|0_________ 6_________12______ (i don't know but the image is not coming correct)
| . . T
| . .
| . .
| .__________.
|V'
So,total area=(10*6)+(-10*6)=60-60=0=displacement!I am sooo happy!:smile:
Thanks million for ur help!Keep helping!:approve::smile::smile:
 


Code:
|V
|
| __________                              
|           .                
|           . 
|           .
|0_________ 6_________12______       
|           .          .           T
|           .          .
|           .          .
|           .__________.
|V'

Yep, exactly right! :)
 

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