Find distance traveled using acceleration graph

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SUMMARY

The discussion focuses on calculating the distance traveled using an acceleration graph over the first 20 seconds. The user attempted to integrate the acceleration function twice to derive the position but initially arrived at incorrect values of 100m and 137.5m. It was identified that the user neglected to account for the distance traveled at uniform velocity between 10 to 15 seconds, which is crucial for accurate distance calculation.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with kinematic equations related to motion.
  • Knowledge of interpreting acceleration graphs.
  • Ability to analyze piecewise functions for uniform motion.
NEXT STEPS
  • Review the principles of definite integrals in calculus.
  • Study kinematic equations for motion under constant acceleration.
  • Learn how to analyze acceleration graphs to determine velocity and position.
  • Explore examples of piecewise functions in physics to understand uniform motion segments.
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and integration, as well as educators looking for examples of motion analysis using graphs.

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Homework Statement


Here is a link to the graph.

http://www.webassign.net/pse/pse6_p2-12.gif

Find the distance traveled in the first 20 seconds.


Homework Equations



Integral once = velocity
Integral twice = position

I attempted to use the following formula:

a = [ 2 ( x - x(sub-i) ) ] / 2

The Attempt at a Solution



When I integrated twice i came up with, 100m however this is incorrect
Then I somehow came up with 137.5, this is incorrect.
I attempted to use the following formula:

a = [ 2 ( x - x(sub-i) ) ] / 2

unfortunately I came up empty handed.

Please help! this problem is killing me.
Thank you for your time.
 
Physics news on Phys.org
While coming up with 137.5 m, I think, you have left out the distance traveled with uniform velocity during 10 to 15 seconds.
 

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