<p>The average velocity in the interval 8:05 to 8:36 is 331.2. So, use $v_1=$ 331.2 at $t_1=$ 08:20:30 under the assumption that acceleration is constant.</p>

<p>The average velocity between 8:36 and 9:27 is $v_2=$ 306.5, so use $t_2=$ 9:01:30</p>

<p>Acceleration is $(306.5-331.3)/(41\times60)=-0.01\ $km/s$^2$.</p>

<p>The same calculation gives me $0.029\ $km/s$^2$ and $-0.038\  $km/s$^2$.</p>

<p>Is this just arithmetic error on the first value (using 51m instead of 41m), a small rounding issue on the second and the third I agree with.</p>

<p>Other than that, then some unstated alternative interpolation scheme has been used.</p>

CC BY-SA 3.0
2016-08-02T08:06:30.780Z
2016-08-25T23:28:04.727Z
2016-08-25T23:28:04.727Z
ProfRob
<p>I am looking at data used in a project to get the velocity and acceleration of a CME (Coronal Mass Ejection). I understand how the velocity was calculated, and for the acceleration, I get the v2-v1 part of the formula for acceleration that was used (ex. 306.5-331.2), but what time intervals (t2-t1) were used to calculate the accelerations?</p>

<p><a href="https://i.sstatic.net/cmifs.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/cmifs.png" alt="enter image description here"></a></p>

<velocity><coronal-mass-ejection>
How was the acceleration calculated for this CME (Coronal Mass Ejection)?
John Michaels
# How was the acceleration calculated for this CME (Coronal Mass Ejection)?

(Asked by: username_0 on 2016-08-02)

I am looking at data used in a project to get the velocity and acceleration of a CME (Coronal Mass Ejection). I understand how the velocity was calculated, and for the acceleration, I get the v2-v1 part of the formula for acceleration that was used (ex. 306.5-331.2), but what time intervals (t2-t1) were used to calculate the accelerations?

[![enter image description here](https://i.sstatic.net/cmifs.png)](https://i.sstatic.net/cmifs.png)

**Tags:** velocity, coronal-mass-ejection


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## 1 answers

### Answer by username_1 on 2016-08-02. Score: 2

The average velocity in the interval 8:05 to 8:36 is 331.2. So, use $v\_1=$ 331.2 at $t\_1=$ 08:20:30 under the assumption that acceleration is constant.

The average velocity between 8:36 and 9:27 is $v\_2=$ 306.5, so use $t\_2=$ 9:01:30

Acceleration is $(306.5-331.3)/(41\times60)=-0.01\ $km/s$^2$.

The same calculation gives me $0.029\ $km/s$^2$ and $-0.038\ $km/s$^2$.

Is this just arithmetic error on the first value (using 51m instead of 41m), a small rounding issue on the second and the third I agree with.

Other than that, then some unstated alternative interpolation scheme has been used.


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