Dynamics - Lesson 2: Rectilinear Motion Example Problem
HTML-код
- Опубликовано: 6 сен 2024
- 𝐌𝐲 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐍𝐨𝐭𝐞𝐛𝐨𝐨𝐤 for notes! Has graph paper, study tips, and Some Sudoku puzzles or downtime between classes! amzn.to/3pTSZXw
𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐬 𝐏𝐮𝐳𝐳𝐥𝐞 𝐁𝐨𝐨𝐤 Sodoku, Mazes, Word Search, Word Scramble - 400 puzzles and solutions - Perfect Gift! amzn.to/3KxtS3L
𝐂𝐚𝐩𝐬𝐭𝐨𝐧𝐞 𝐃𝐞𝐬𝐢𝐠𝐧 𝐉𝐨𝐮𝐫𝐧𝐚𝐥 For your senior design project documentation! 100 pages of numbered graph papers. amzn.to/3OtZJ6z
𝐅𝐄 𝐑𝐞𝐯𝐢𝐞𝐰 𝐉𝐨𝐮𝐫𝐧𝐚𝐥 For you notes and work while preparing for the FE Exam. 100 pages of graph paper! amzn.to/3ONRILn
𝐏𝐄 𝐉𝐨𝐮𝐫𝐧𝐚𝐥 𝐒𝐓𝐀𝐑 𝐋𝐨𝐠 Record your projects in preparation of your PE application or for your annual review at work...(Situation, Task, Analysis, Results) amzn.to/3rjyAMp
𝐇𝐨𝐰 𝐭𝐨 𝐀𝐜𝐞 𝐒𝐭𝐚𝐭𝐢𝐜𝐬 Great book to review or have on your shelf to make sure you never forget Statics! amzn.to/455QRKM
Top 15 Items Every Engineering Student Should Have!
1) TI 36X Pro Calculator amzn.to/2SRJWkQ
2) Circle/Angle Maker amzn.to/2SVIOwB
3) Engineering Paper amzn.to/2K7vYbP
4) Mechanical Pencils 40pk amzn.to/2GGRG4y
5) Desk/Bed Clamp Reading Light amzn.to/2YzN5qG
6) Stainless Steel Rulers amzn.to/31aqMcP
7) Small Stapler amzn.to/2ynDOrk
8) Post it Tabs for Page Markers amzn.to/2SSIP4q
9) Noise Canceling Wireless Headphones amzn.to/3aMLbdY
10) Wireless Mouse amzn.to/2ROkY5Y
11) Multicolor Pen/Pencil amzn.to/3331AXy
12) Portable Charger amzn.to/2RsitHG
13) Backpack Rain Cover amzn.to/2Gq4ATM
14) Amazon Dot amzn.to/2Gs42g4
15) Portable Bluetooth Speaker amzn.to/3aIMZV0
This video is NOT sponsored. Some product links are affiliate links which mean if you buy something we'll receive a small commission.
Follow me on Instagram / jeff_hanson_phd
Please consider supporting the channel by subscribing, commenting, and liking each video. This moves these videos up the search list and helps me greatly. Thank you for watching.
My Patreon: / jeff_hanson Please help keep the free videos coming by donating. Thank you for the many blessings.
This video is copyrighted by the Jeff Hanson for the private use of our audience. Any other use of this video or any pictures, descriptions, or accounts of the video without written consent from Jeff Hanson is strictly prohibited.
Sir, please carry on doing this! Your videos really are the best online, in my opinion!
Same
Thank you so much, Sir. Your videos are life saving for Mechanical engineering students.
Prof Hanson, I have recommended your online courses to my all of my classmates in statics, solids and dynamics because of your commitment to help engineering students. Your teaching style is second to none! Keep up the great work!
agreed! :)
Thanks master. I come here because of lockdown of covid-19. So, who else with me?
Covid was 4 years ago how... İts like yesterday
Thank you so much sir. You’re videos saved my bacon in statics last semester and are saving it aging this semester in dynamics.
If only I found this gentleman earlier. You are truly the greatest.
For anyone else wondering: If you graph the position function you'll see that the path of motion is longer before its first critical point (change of direction and v=0). Therefore the particle can take longer before that point and still have constant acceleration. Hope that helps!
I wondering about this should the time before and after the critical should be same?.Because if the acceleration is negatif and means the car in deceleration should not the time before the critical point be 1s and after it be 2s...and refer to the question that say the time that the car take only 3s..So why there are 4s for the total time..hope you can reply it..I really in confusing right now..
@@maturies139 at the critical point (peak) v=o so in the velocity equation v=6t-3t^2 sub v for 0 and solve for t
so it took 2 seconds to get to one point and 1 second to get back to the original spot?
The initial velocity is zero. The car has to accelerate and then decelerate for the velocity to also equal zero at t = 2 seconds. When it goes the other way, it doesn't decelerate at any point . It reaches the original position at a velocity much greater than zero.
Lmao I was just thinking this exactly, word-for-word, and here it is in the first comment. Thanks for asking this, you helped someone in the future.
Yes, don't be confused it's just a default thing
@@pointless7676
But velocity is not zero at 3s.
We need to find the distance covered by the car from 0 - 3 seconds; I think we can't assume that the position-time curve is symmetric about t=2 which is indeed a critical point. If we calculate the length of the curve from 0-3 (calculus), we will get distance = 8.81m showing that the car actually travels about 0.81m more. Can you clarify this please sir? 🙏
that's how professors are suppose to be
Professor Hanson , thank you for good start in Dynamics. This first example really increase my understanding of Rectilinear Motion.
The anti derivative of the velocity function yields a constant C. We have no limits of integration so you must account for the C, should you not?
The integral is definite, so there won't be a constant C to worry about.
should the average speed be 2 m/s ? I thought you traveled 8 m and it took you 2s forward and 2s back, therefore 8m/(4s) = 2m/s ?
If you graph the position function you'll see that the path of motion is longer before its first critical point (change of direction and v=0). Therefore the particle can take longer before that point and still have constant acceleration. Hope that helps!
you're videos are a great help you helped me get an A in statics now I'm taking dynamics and solids in the fall
Can someone explain why the 5th formula doesn't work for this? is it not constant acceleration because 'a' is reliant on 't'?
2:33 "PICK UP THE PHONE HONEYY!" 5:43 -when i see my calculus I marks
Hello Dr. Jeff, to calculate the position at t=3, I tried using the ads = vdv formula and integrating v from 0 to -9m/s and a remains a constant at -12m/s^2. But I didn’t get the same answer. Shouldn’t both methods yield the same answer?
Sir i have a question, why didn't we took constant while performing the integration for part (a)?
because he did a definite integral from t=0 to t
SIR I LOVE YOUR VIDS.YOUR VIDS SAVED MY LIFE BEFORE EXM.PLS BE MY FATHER
Thanks so much!! I had to take an incomplete in this class after getting hit by a car on my bike :-( This is going to be a LIFE SAVER for my catching up and final exam!!
....but theoretically... don't we need to include the constant for the position at any time??
Thanks sir! Helping me a lot.. I'm in week 4 and I don't understand a single thing.. tq sir!
Thank you so much Prof Hansen for your videos. They are a life saver. Do you have to have any videos on Calculus specifically Integrations?
Question what is instantaneous velocity and acceleration?
Thank you so much for the series, please carry on doing these Dynamics videos please! and should the velocity be 2m/s
If you graph the position function you'll see that the path of motion is longer before its first critical point (change of direction and v=0). Therefore the particle can take longer before that point and still have constant acceleration. Hope that helps!
How do you integrate to get 6*t2/2-3*t3/3 @4:30
please keep going! i really need your videos to pass the course
Maaaaaan you are the beeeest god damn it , wish u all the best sir
if s=0 doesn't that mean at t=3 the car has travelled 0m because it went back to initial position?
God will bless you in plenty folds, this has really helped me
my savior, good job prof... keep up the good videos
Very helpful videos thank you. I have watched only two videos, but I think I will check the rest too.
Love u doctor❤❤
Very well teaching. Thanks sir! I finally understood!
I feel like I need to pay this man instead of my school. Can I made a donation? (for real)
I have a Patreon if you would seriously like to consider making a donation. It helps me out a lot! The link is in the description.
wow! this is on weekend!! Great work Dr.
Better than the professor i had. He said just go look at the book LOL
Isn't dv/dt used for instantaneous acceleration?If the question was asking for instantaneous acceleration shouldn't it use the term be at t=3 instead of when t=3?(for part a) wouldn't the latter imply average acceleration?
will there be more dynamic videos loading soon, I just started my dynamics class today. I need your help. Thanks :)
Daniel Wong there will be more. Starting again next week.
Yes thank you!!
Great explanations,professor
how we can be sure it also traveled 4m while coming back? I mean in general .. while the speed isn't constant !
great explanations Professor
Saved my life, thanks
To find position at t=3s . Can we use the S=So +Vot +1/2 at2 ?
I don’t think the acceleration is constant that’s why
Sir, you are a treasure. thank you so much.
Sir, I have a question. How were you able to determine that @2 seconds the position of the particle is exactly at the middle of the parabolic line?
Wow, already the second video?
You rock, sir! :D
thank you so much . that was really good .
Hi can i know which book did you used as a reference please?
You are the Master!
Is that a continuous motion?..
why is it v=0 when it is at 2 sec? i don't get the graph
Why blur your logo top right corner?
THANK YOU!
thanks for the vids!
You are a LEGEND
Thanks for such a dandy series!!
So it took you 2 seconds to get to 4m, and took you only 1 sec for another 4m? if this was 'constant' acceleration, I have doubts sir and needs clearance. please respond
If you graph the position function you'll see that the path of motion is longer before its first critical point (change of direction and v=0). Therefore the particle can take longer before that point and still have constant acceleration. Hope that helps!
LUKE OSTER that helps a lot sir! thank you
Then why assume that after 3 sec, the car travelled 8m, when the graph is not symmetrical
you are alife saver i would love for you to be my lecturer
thank you so much
how many dynamics lessons are u planning to upload
Also, when do you plan on uploading more
i have some dynamics videos ;)
please share the link, Thank you
You are my hero
That is just pure awesomnesssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss
keep it up sir!
6:00 Skrrrt
I just had to keep hitting your 6:00 minute mark over and over. Funny stuff.
Sir please send the solution of the problem
When a particle moves on a rough cycloid whose axis is vertical and vertex downwards , the time taken by the particle to reach a point is independent to the starting point
PLEASE I AM BEGGING UPLOAD MORE
With all respect, the diagram shouldn't supposed to be drawn that way. The turning point is not in the middle. The right side direction should be from 0s to 2s and the left should be from 2s to 3s. And the total distance should be calculated separately for 0s-2s and 2s-3s and adding them (ignoring the negative sign for the 2-3s).
It took me forever to figure out how V = 6t-3t^2 got to be 3t(2-t). This is why dynamics is so frustrating, why would anyone think of solving such a simple problem this way?
I've solved a lot of problems in calc and differential equations by pulling out factors. It's kind of the first step in determining when the velocity is equal to 0.
Steven Spielberg is teaching us
i dont get d integration part pls get back to meee
If you plug in the units in the formula, the car did not travel 4m but 4seconds 😂
If Jeff had taught at my university maybe I wouldn't have dropped out.
Dr Hanson I would do ANYTHING for an A. 🥵🥵🥵😒💦