Acceleration vs Time Find Velocity

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Homework Help Overview

The problem involves analyzing a graph of acceleration versus time for a particle moving along the x-axis, with an initial velocity provided. The goal is to determine the particle's velocity at a specific time based on the graph's information.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the significance of the area under the acceleration-time graph and its relation to velocity. Some consider integrating the graph to find velocity, while others question the appropriateness of specific integration methods and constants. There is also mention of using the equation v = u + at in the context of a linear graph.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants are considering geometric methods to find the area under the graph, while others are reflecting on the implications of initial conditions and integration constants. No explicit consensus has been reached yet.

Contextual Notes

Participants are navigating the constraints of the problem, including the need to interpret a graph without complete information on the function's specifics. There is also a focus on the initial velocity and how it interacts with the area under the curve.

circuscircus
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Homework Statement


The graph shows acceleration versus time of a particle moving along the x axis. Its initial velocity is 8m/s at t=0. What is the particle's velopcity at t=4s?

http://img259.imageshack.us/img259/6684/graphcc9.jpg

Homework Equations





The Attempt at a Solution



vi=8
ti=0
vf=??
tf=4

I was thinking of integrating y=-x+4 but I don't think I'm suppose to do that and also I don't know how to find +c.
 
Last edited by a moderator:
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What does the area under an a-t graph represent?
 
learningphysics said:
What does the area under an a-t graph represent?

Well integrating the acceleration gets you velocity, does it not?
 
circuscircus said:
I was thinking of integrating y=-x+4 but I don't think I'm suppose to do that and also I don't know how to find +c.

You don't really want to integrate "y=-x+4"... with those variables.
When you integrate the function with the appropriate physical variables, what is the physical interpretation of your "c" constant of integration?

You don't need to actually do the integral [with calculus]... you can do it geometrically using learningphysics' suggestion.
 
Last edited:
Well the area under the chart is 8 at 4 seconds but that's the vi so how doe sthat come into play?
 
circuscircus said:
Well the area under the chart is 8 at 4 seconds but that's the vi so how doe sthat come into play?

Try your integration method, as well.
Now, rethink your answer.
 
correct me if I'm wrong, but i believe the equation v=u+at can be used, since this is a linear graph??

you know, u(initial velocity), and the product (at) is 0. the initial velocity is equal to the final velocity.
 

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