How Do You Calculate Instantaneous Acceleration in Physics Problems?

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SUMMARY

The discussion focuses on calculating instantaneous acceleration in physics problems, specifically at 4 seconds and 7 seconds. The correct instantaneous acceleration at 4 seconds is determined to be 1.33 m/s², calculated using the formula for average acceleration, where the time interval is 3 seconds. For 7 seconds, the instantaneous acceleration is confirmed to be zero, attributed to a zero slope on the velocity-time curve. The key takeaway is that instantaneous acceleration is derived from the slope of the velocity-time graph, reinforcing the concept that it is a derivative.

PREREQUISITES
  • Understanding of basic kinematics
  • Familiarity with the concept of derivatives
  • Knowledge of velocity-time graphs
  • Ability to apply the average acceleration formula
NEXT STEPS
  • Study the relationship between velocity and acceleration in kinematics
  • Learn how to calculate derivatives in physics contexts
  • Explore the use of velocity-time graphs for motion analysis
  • Investigate advanced topics in calculus related to motion
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Students studying physics, educators teaching kinematics, and anyone interested in understanding motion dynamics and acceleration calculations.

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Homework Statement


2je4m06.png

(X-Axis = Time)

1. Find the instantaneous acceleration at 4 s.
Answer in units of m/s2.

2. Find the instantaneous acceleration at 7 s.
Answer in units of m/s2.

Homework Equations


Unknown, I'm assuming you'll use:
Acceleration Average = Delta V(final)-V(initial)/Delta T

The Attempt at a Solution


I've used the equation above ^ and I've gotten for
1.) 3.1 - (-4)/4 = 1.775
2.) I'm assuming Zero considering the object is not moving.

Any help would be great, I'm not quite sure how to do these correctly, so guidance or reassurance would be great. Thanks.
 
Last edited:
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You have the correct idea. Though for part 'a' you made a small mistake the time that elapses is only 3s so the acceleration=(4/3)=1.33m/s^2

The idea is that instantaneous acceleration is equal to the slope of the velocity-time curve at any point. Thus it is a derivative...but since the slope for part 'a' is a constant that means that the instantaneous acceleration over that whole portion is also a constant that can be calculated using the equation for average acceleration.

Likewise part 'b' is zero as you predicted, but not because the velocity is zero...it is because the slope is zero.
 
kjohnson said:
You have the correct idea. Though for part 'a' you made a small mistake the time that elapses is only 3s so the acceleration=(4/3)=1.33m/s^2

The idea is that instantaneous acceleration is equal to the slope of the velocity-time curve at any point. Thus it is a derivative...but since the slope for part 'a' is a constant that means that the instantaneous acceleration over that whole portion is also a constant that can be calculated using the equation for average acceleration.

Likewise part 'b' is zero as you predicted, but not because the velocity is zero...it is because the slope is zero.

Thank you, very much appreciated.
 

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