Calculating acceleration from 2 tangent lines

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SUMMARY

The discussion focuses on calculating acceleration from two tangent lines on a position vs. time graph. The slopes of the tangent lines are identified as velocities, specifically 295 cm/s and 575 cm/s. To find the average acceleration, the formula used is (final velocity - initial velocity) / (final time - initial time), but it is emphasized that the time interval does not need to be the total duration. If acceleration is constant, it can be calculated over any time interval, making the process straightforward.

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives
  • Familiarity with kinematic equations
  • Knowledge of graph interpretation, particularly position vs. time graphs
  • Basic understanding of velocity and acceleration definitions
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  • Study the concept of derivatives in calculus to better understand tangent lines
  • Learn about kinematic equations for uniformly accelerated motion
  • Explore graphical analysis techniques for interpreting motion graphs
  • Investigate the relationship between velocity and acceleration in physics
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Students studying physics, particularly those focusing on motion analysis, as well as educators teaching kinematics and calculus concepts.

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I have a position vs. time graph which is slightly curved. I found the slope of 2 tangent lines which I know are the velocity. My question is how do I get the acceleration using these 2 slopes. One slope is 295cm/s and the other is 575cm/s. I know that avg acceleration is final velocity - initial velocity/ final time - initial time but because these points aren't at the initial and final times, I don't think I can use that.
 
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Then don't use final time in the acceleration equation! If acceleration is a constant, you can calculate it over any time interval.
 

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