How to find Instantaneous acceleration with points from a graph

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SUMMARY

This discussion focuses on calculating instantaneous acceleration using given velocity and time values. The final velocity is 4 m/s, the initial velocity is 1 m/s, the final time is 5 seconds, and the initial time is 2 seconds. The method involves using the equation v = u + at to solve for acceleration, as well as determining the slope of the velocity vs. time graph between the specified points. For straight-line motion under constant acceleration, the average and instantaneous accelerations are equal, and the slope method can also be applied to curves for better estimates.

PREREQUISITES
  • Understanding of kinematic equations, specifically v = u + at
  • Basic knowledge of graph interpretation, particularly velocity vs. time graphs
  • Familiarity with the concept of slope in mathematics
  • Concept of instantaneous vs. average acceleration
NEXT STEPS
  • Study the derivation and application of kinematic equations in physics
  • Learn how to calculate the slope of a line on a graph
  • Explore the concept of tangents and their use in finding instantaneous rates of change
  • Investigate the differences between average and instantaneous acceleration in various motion scenarios
USEFUL FOR

Students studying physics, particularly those focusing on kinematics, as well as educators and anyone looking to deepen their understanding of motion analysis and acceleration calculations.

adamgriffis
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Homework Statement



How to find instantaneous acceleration when the velocity final is 4 m/s, the velocity initial is 1 m/s, the time final is 5 seconds, and the time initial is 2 seconds?

Homework Equations





The Attempt at a Solution


I have no idea how to do the second derivative for the equation


 
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adamgriffis said:
How to find instantaneous acceleration when the velocity final is 4 m/s, the velocity initial is 1 m/s, the time final is 5 seconds, and the time initial is 2 seconds?
Hi adamgriffis! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Since you haven't included a graph, it's reasonable to assume this is a question involving straight-line motion under constant acceleration. For such motion, there is a set of equations with which you must become adept because you will rely on them for further study.

Have you encountered an equation, or one similar to: v = u + a·t https://www.physicsforums.com/images/icons/icon5.gif
 
Last edited by a moderator:
Though I can't be 100% sure, I'm going to assume that the graph you have mentioned is a v vs. t graph of uniform acceleration. There are a number of ways to determine the instantaneous acceleration between the points you listed, which would be (2, 1) and (5, 4). The first would be NascentOxygen's excellent suggestion to use v = u + at (final velocity = initial velocity + acceleration * time), and solve for the acceleration.

The second way is useful in graphs. This only works with straight-line graphs.* Assuming that the slope between the two points you listed is constant and unchanging, solving for the slope (slope = change in y divided by change in x) will yield the instantaneous acceleration at any time between those two points. If you want to know why, when an object moves at constant velocity/acceleration, its average and instantaneous velocities/accelerations are equal at all times.

Hope this helped.

*Note: If you have a curve plotted on a graph, it is possible to use this slope method again. Since instantaneous acceleration is defined as the change in velocity as the limit of change in time approaches zero, getting two points as close as possible to the point of acceleration you are trying to find will yield a good estimate. A more exact result would come from plotting a line tangent to the point of acceleration.
 

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