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How to find Instantaneous acceleration with points from a graph

  • #1

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

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
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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 [Broken]

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 [Broken]
 
Last edited by a moderator:
  • #3
5
0
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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