Can you solve for delta d in jerk motion using this formula?

  • Thread starter Thread starter RobotGymnast
  • Start date Start date
  • Tags Tags
    Jerk Motion
AI Thread Summary
To solve for delta d in jerk motion, the formula incorporates time, initial velocity, initial acceleration, and constant jerk, expressed as delta d = vt + (1/2)at^2 + (1/6)Jt^3. The discussion emphasizes that jerk is the derivative of acceleration, and if mass is constant, acceleration can be derived from jerk using the relationship J = ma. There is a mention of potential confusion when substituting different physics equations, leading to varied results. Integrating the equation for jerk motion twice aligns with the correct formula for delta d. Understanding the context of impulse is also highlighted as relevant to the discussion.
RobotGymnast
Messages
4
Reaction score
0
What formula would one use to solve for delta d with regards to t, v1, a1, and constant J?
 
Physics news on Phys.org
RobotGymnast said:
What formula would one use to solve for delta d with regards to t, v1, a1, and constant J?

Please define your symbols, and provide the context of your question. Are you familiar with the concept of an impulse?

http://en.wikipedia.org/wiki/Impulse

.
 
Kinematics. As in,
Code:
delta d = vt + (1/2)at[SUP]2[/SUP]

except assuming that jerk motion is constant, rather than acceleration.
 
Last edited:
Assuming that the mass is constant, if the force of the jerk J is constant, considering that F = ma so J = ma so a = J/m since J and m are constant A is constant.

So if your implying there is an acceleration in place before the jerk, you would use vector addition to figure out the resultant acceleration. And use your formula to solve for d.
 
*jerk motion (the derivative of acceleration). whoops.
 
Well, going from the equation using velocity s=v t, to the equation using acceleration v t + \frac{1}{2}a t^2 I would assume that the equation using jerk is s=v t + \frac{1}{2}a t^2 + \frac{1}{6}j t^3.

There's my guess.
 
That's what I got. What confuses me was that I also got a different answer by substituting physics equations into one another. I ended up getting ¼ Jt^3 rather than 1/6 Jt^3. But integrating the equation for jerk motion twice gives the answer you gave, which I think is correct.
 
Back
Top