Acceleration vs Time Find Velocity

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SUMMARY

The discussion focuses on calculating the velocity of a particle at t=4 seconds given its acceleration versus time graph. The initial velocity is established as 8 m/s at t=0. Participants emphasize that integrating the acceleration function yields the velocity, and they suggest using the equation v = u + at for linear graphs. The area under the acceleration-time graph is also discussed as representing the change in velocity.

PREREQUISITES
  • Understanding of kinematic equations, specifically v = u + at
  • Knowledge of integration in calculus
  • Familiarity with interpreting graphs in physics
  • Basic concepts of velocity and acceleration
NEXT STEPS
  • Study the relationship between acceleration and velocity through integration
  • Explore geometric interpretations of area under curves in physics
  • Review kinematic equations and their applications in linear motion
  • Practice problems involving acceleration-time graphs to reinforce concepts
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and graph interpretation, as well as educators looking for effective teaching strategies in motion analysis.

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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