Non-constant acceleration find the equations of motion when a=2-0.1t.
Автор: Zak's Lab
Загружено: 2024-08-24
Просмотров: 517
Описание:
When the acceleration of an object is changing, we have to go back to first principles to find the equations of motion. So, we start with the derivative relationships between position, velocity and acceleration: v(t)=x'(t), a(t)=v'(t).
🧠 Access full flipped physics courses with video lectures and examples at https://www.zakslabphysics.com/
Given a non-constant acceleration a=2-0.1t, we guess the antiderivative to find the velocity as a function of time, and that's v=2t-0.05t^2+C, where C is an arbitrary constant. We were given the initial conditions for non-uniform acceleration problem: v(0)=0 and x(0)=0, so we plug in v=0 at t=0 and find that C is zero. This allows us to write down our first equation of motion for an object with decreasing acceleration: v(t)=2t-0.05*t^2.
Now we guess the antiderivative of the velocity function in order to find the position function: x(t)=t^2-0.05/3*t^3+C. Again, we apply the initial conditions, this time using the fact that x=0 when t=0 to find that C is zero.
So we finally have the position function when acceleration is changing as a=2-0.1t: x(t)=t^2-0.0167*t^3.
Повторяем попытку...
Доступные форматы для скачивания:
Скачать видео
-
Информация по загрузке:
![RMS values for current and voltage - full calculus derivation [AC circuit physics]](https://imager.clipsaver.ru/6HiiPLn-vAI/max.jpg)
![Почему работает теория шести рукопожатий? [Veritasium]](https://imager.clipsaver.ru/ggI1xKzoANs/max.jpg)
![Пожалуй, главное заблуждение об электричестве [Veritasium]](https://imager.clipsaver.ru/6Hv2GLtnf2c/max.jpg)
