How Do You Calculate Instantaneous Acceleration in Physics Problems?

AI Thread Summary
Instantaneous acceleration can be calculated as the slope of the velocity-time curve at a specific point, which is effectively a derivative. For the first part of the problem, the correct instantaneous acceleration at 4 seconds is 1.33 m/s², as the time interval is only 3 seconds. The second part confirms that the instantaneous acceleration at 7 seconds is zero, attributed to a zero slope rather than zero velocity. Understanding that instantaneous acceleration can be derived from average acceleration is crucial. This approach clarifies the calculations needed for solving such physics problems effectively.
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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.
 
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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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