Is the slope of the curve the acceleration?

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SUMMARY

The discussion centers on the relationship between speed and acceleration in the context of projectile motion, specifically a football kicked into the air. Participants confirm that the slope of a speed versus time graph represents the magnitude of acceleration, denoted as dv/dt. It is clarified that while the acceleration remains constant, the slope changes sign at the peak of the trajectory when plotting speed instead of velocity. This distinction is crucial for accurately interpreting the graph.

PREREQUISITES
  • Understanding of basic physics concepts, particularly projectile motion.
  • Familiarity with graphing speed versus time.
  • Knowledge of the difference between speed and velocity.
  • Basic calculus, specifically the concept of derivatives (dv/dt).
NEXT STEPS
  • Study the principles of projectile motion in physics.
  • Learn about the differences between speed and velocity in motion analysis.
  • Explore the concept of derivatives in calculus, focusing on their application in physics.
  • Investigate how air resistance affects projectile motion and graph interpretation.
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Students of physics, educators teaching mechanics, and anyone interested in understanding the dynamics of projectile motion and graph interpretation.

Omid
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A football is kicked into the air at some angle theta and lands down field. Ignoring air friction, draw a rough plot of its speed versus time. Is the slope of the curve the acceleration? Explain.

I think the slope of the curve is the absolute value or the magnitude of acceleration, isn't it?
 
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In graphic of Speed versus time the slope will be instant acceleration or dv/dt
 
Cyclovenom is correct, of course. The only addition I would make is that, if you're plotting speed instead of velocity, the slope is going to change sign as you pass through the peak. The acceleration will, of course, be constant, so by taking the slope, you'll be getting the value with possibly the wrong sign. I'm sure this is what you meant by the "magnitude" of the acceleration, so basically: yes, you're right.
 

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