Acceleration as a function of position example
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- čas přidán 9. 09. 2024
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Dynamics Tutorial: Acceleration as a function of position example
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Thank you so much for this amazing video! You have explained it so well!
Thanks for the feedback!!🍓
i'm wondering why is the integral of vdv equals to v^2/2. It makes sense if v is a variable, but isn't v a function (v(x)) ???
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I actually watched this video hoping to see how the integration would happen, how would you do that last step?
www.integral-calculator.com/ is a pretty good pace to start, you can put in definite integrals and there is an option to show steps. Alternatively if you have a wolfram pro student account they also show the steps but it's not free =/
Symbolab is also a good website to check out
So what is final equation after taking derivative? And s=2 when substituting in equation?
Hey for the final time I got 1.14 sec after plugging in the numbers. did you approximate or do the whole integral?
once I have the V(s), how can i turn that into V(t) if given initial conditions
Thanks!
whatup homie. Thanks a lot. I got a 60% final tomorrow, if I pass, i will comment this comment. Thanks Homie.
sup dowg, how'd u do?
what is the function of x respect to t?can we found it?
If initial velocity is not given, should I assume that it is 0?
Depends on the wording of the entire problem. If they say starts at a rest, then it's definitely zero. Could be tricky wording through so watch out
@@Engineer4Free Alright. Thank you so much! 😊
the very last peice (ds/(1+2s^2)) why wasnt the limiti there 0-2 instead of 0-s?
Good catch. It should be written as the integral from 0-2. The approximation of that integral from 0-2 is correct as being 1.25
Hey, how would this work if we have speed in the opposite direction? Since we can't have a negative inside a square root.
I mean, if we had acceleration that was increasing the opposite direction, starting from a positive value
Hey I just glanced at it, depending on the numbers, it's possible that even if a(s) is negative, the number inside the brackets of the square root could still be positive, which is really what is important. I haven't made a dynamics video in a really long time but do hope to get back to working on the series soon, so I'll take a closer look at this problem when I do that.
@@Engineer4Free wow, thanks for the reply. Yeah, I realized that just before my quiz. Thank you for the content man, you rock.
where does that -1/2 come at 2:09 ?
Hey thanks for the good question! We are taking the definite integral. So to go from the left hand side of the arrow to the right hand side of the arrow at that part of the video, we plug the value of the upper bound (which is v) into the variable of the expression (v^2)/2 which just gives us the same expression "(v^2)/2". Then we plug in the value of the lower bound (which is 1), which gives us "(1^2)/2" = "1/2". We then subtract the expression with the lower bound in it from the one with the upper bound in it to solve the definite integral. Notice that we also do this for the integral containing the S, but because the lower bound is zero and the whole expression is a single term, we technically end up with S^2 - 0, but I don't bother writing the "- 0" term. If you need to brush up on definite integrals, I made a video on them you can check it out here: www.engineer4free.com/4/using-definite-integrals-to-find-the-area-under-a-curve
Huge thanks for the swift reply! You're teaching me more than my lecturer does :)
No worries, glad to be of help!!
Where did v*dv come from?
velocity is equal to the rate of change of distance. the rate of change of distance is written as ds/dt. so V=ds/dt. In the expression (dV/ds)*(ds/dt) I just replaced ds/dt with V, because V=ds/dt. now the expression becomes (dV/ds)*V, or (V*dV)/ds. Because (V*dV)/ds = 2s, we can multiply both sides by 2s to get V*dV = 2s*ds. that's where the V*dV comes from.
the last integration process is horrible haha
lol there's a reason I skipped the work in this video :p
That's actually the part I'm confused about. How would you integrate that?
@@evaliu6359 By closing your eyes and hope that the answer magically appears