Velocity-time & displacement time graphs

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SUMMARY

The discussion centers on calculating average velocity and acceleration from velocity-time and position-time graphs. The average velocity from the given velocity-time graph, which starts at (0,6) and ends at (8,12) with a slope of 0.75, is confirmed to be 12, as per the answer sheet. To find acceleration from a position-time graph, the second derivative of velocity with respect to time is utilized, specifically using the formula: d²x/dt² = (x(t+Δt) - 2x(t) + x(t-Δt)) / (Δt)².

PREREQUISITES
  • Understanding of velocity-time graphs
  • Knowledge of position-time graphs
  • Familiarity with derivatives in calculus
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the concept of derivatives in calculus
  • Learn how to construct and interpret velocity-time graphs
  • Explore numerical methods for calculating derivatives
  • Investigate the relationship between position, velocity, and acceleration
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Students in physics, educators teaching kinematics, and anyone looking to deepen their understanding of graph analysis in motion.

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I am starting to panic...

How do you find the average velocity from a Velocity-time graph. The graph has a slope of 0.75 and starts at (0,6) and ends at (8,12)

I calculated V.avg = 9. (( 6+ 12)*0.5) but the answer sheet says 12? How?

And, in general how do you find the acceleration from a position- time graph

edit: I was looking at the wrong answer sheet >.< for the first part it is 12!
but I still don't know how to find the acceleration from a position - time graph, easily.I know I am supposed to find the instantaneous velocity at several points and then construct a velocity - time graph. But is there an easier way?
 
Last edited:
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sakonpure6 said:
I am starting to panic...

How do you find the average velocity from a Velocity-time graph. The graph has a slope of 0.75 and starts at (0,6) and ends at (8,12)

I calculated V.avg = 9. (( 6+ 12)*0.5) but the answer sheet says 12? How?

And, in general how do you find the acceleration from a position- time graph

edit: I was looking at the wrong answer sheet >.< for the first part it is 12!
but I still don't know how to find the acceleration from a position - time graph, easily.I know I am supposed to find the instantaneous velocity at several points and then construct a velocity - time graph. But is there an easier way?

Yes, there is a more direct way. The acceleration is the second derivative of the velocity with respect to time. To get the acceleration directly, you need to evaluate the second derivative with respect to time numerically. A formula you can use for this is:

[tex]\frac{d^2x}{dt^2}=\frac{x(t+\Delta t)-2x(t)+x(t-\Delta t)}{(\Delta t)^2}[/tex]

This is the acceleration at time t.
 

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