How Do You Find Instantaneous Velocity at 1s?

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SUMMARY

The discussion focuses on calculating the instantaneous velocity of an object at 1 second using its position versus time graph. The user initially derived the equation of the line as y = (7/2)x - 2 based on points (0, -2) and (2, 7). However, they realized that the instantaneous velocity at 1 second should be derived from the slope of the tangent line at that specific point, not the average slope over the interval. The correct approach involves verifying the slope calculation to ensure accurate results.

PREREQUISITES
  • Understanding of basic calculus concepts, specifically derivatives.
  • Familiarity with linear equations and slope calculations.
  • Knowledge of position versus time graphs in physics.
  • Ability to interpret instantaneous velocity from graphical data.
NEXT STEPS
  • Study the concept of derivatives in calculus to understand instantaneous rates of change.
  • Learn how to find the slope of a tangent line on a curve using calculus techniques.
  • Explore the relationship between position, velocity, and acceleration in kinematics.
  • Practice with various position versus time graphs to calculate instantaneous velocity accurately.
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Students studying physics, educators teaching kinematics, and anyone interested in understanding the principles of motion and calculus applications.

hackett5
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The position versus time for a certain object moving along the x-axis is shown. The object’s initial position is −2 m. Find the instantaneous velocity at 1s.
http://img23.imageshack.us/img23/5285/phy1w.jpg

Using points (0,-2) and (2,7) I found the equation of the line to be =7/2x-2. Since this is a straight line I thought the tangent line would have the same slope. So, at 1s the instantaneous velocity would be 7/2, but this is not the correct answer. Where am I going wrong?
 
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Slope of the line = (y2 - y1)/(x2 - x1).
From 0 to 2 seconds this will be the velocity of the particle.
 
Thanks for pointing out my now obvious mistake! I didn't even think to check to see if I had gotten the slope correct, I just assumed my understanding of instantaneous velocity was wrong.
 

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