Negative acceleration and instantaneous velocity on an object.

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SUMMARY

The discussion focuses on analyzing negative acceleration and calculating instantaneous velocity for an object moving through a straight tunnel. It is established that the object experiences negative acceleration from approximately 27 seconds to 45 seconds, as its velocity decreases during this interval. The correct instantaneous velocity at t=20s is calculated as 0.3 m/s, while the instantaneous velocity at t=45s should be approximately -0.43 m/s, indicating that the object is moving in the negative direction. The distinction between average and instantaneous velocity is emphasized, with the latter requiring the slope of the tangent to the graph.

PREREQUISITES
  • Understanding of kinematics, specifically acceleration and velocity.
  • Familiarity with graph interpretation and tangent line concepts.
  • Knowledge of instantaneous versus average velocity calculations.
  • Ability to analyze motion in one dimension.
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  • Study the concept of instantaneous velocity in-depth, focusing on tangent lines to position-time graphs.
  • Learn about negative acceleration and its implications in kinematics.
  • Review the differences between average velocity and instantaneous velocity calculations.
  • Explore graphical analysis techniques for motion, including inflection points and slope interpretation.
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Students studying physics, particularly those focusing on kinematics, as well as educators looking to enhance their teaching methods in motion analysis.

CaptFormal
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Homework Statement



The position of an object along a straight tunnel as a function of time is plotted below:

graphxt5.jpg


1. Are there any time intervals in which the object has a negative acceleration? If so, list them and explain how you know.

2. Calculate the instantaneous velocity at t=20s and t-45s.

Homework Equations



N/A

The Attempt at a Solution



So, for the first question I stated that at 30s to 37s the object's velocity is decreasing, thus the acceleration is negative. However, I was told that this is only part of the interval. What am I missing?

For the second question I did the following calculations:

at 20s = (6 m)/(20s) = 0.3 m/s
at 45s = (16m)/(45s) = 0.36 m/s

I did get the instantaneous velocity at 20s correct but not the instantaneous velocity at 45 s. Where am I going wrong?

Thanks for your assistance.
 
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I would say that the velocity is decreasing from about 27 seconds (at the inflection point) to about 46s (at the other inflection point). Draw tangent lines to the graph, and note the parts where those slopes are getting smaller and smaller (the slope of the slopes).

I think you must be close for the one at 45 seconds. It should be around what you got, I'd say more around .43m/s though it's an estimate when you are using a graph rather than the function.
 
CaptFormal said:

Homework Statement


So, for the first question I stated that at 30s to 37s the object's velocity is decreasing, thus the acceleration is negative. However, I was told that this is only part of the interval. What am I missing?

The velocity is decreasing up until 45 s. Remember that velocity can be negative, so if velocity goes from -1 m/s to -2 m/s, it's decreasing.
For the second question I did the following calculations:

at 20s = (6 m)/(20s) = 0.3 m/s
at 45s = (16m)/(45s) = 0.36 m/s

I did get the instantaneous velocity at 20s correct but not the instantaneous velocity at 45 s. Where am I going wrong?

The velocity should be negative, not positive. Also, you're calculating average velocity. The question is looking for instantaneous velocity, which is the slope of the tangent of the graph.
 

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