I Kinematics cases with non-constant acceleration

[fog37]

TL;DR

Kinematics cases with non-constant acceleration

Hello,

I understand that, for 1D kinematic problems where the acceleration function $a_x$ is initially given along with the initial conditions, we can use calculus (differentiation and integration) to get the position $x(t)$ and velocity $v_x (t)$ of the moving object.

• When the acceleration is a function of position only, the velocity will also be a function of position and not depend on time, i.e. the object will always have the same velocity when it is found at a particular position.
• When the acceleration is a function of time $t$ only, regardless of where the object's position, the object's velocity will have specific values at instants of time after motion starts (Ex: a rocket accelerating moving upward has a time-dependent acceleration...or is it an example of position dependent acceleration?)

If the acceleration depended instead on two independent variables, for ex: $a_x (v,t)= 3t^2 +v^3$, we would need to perform double integration to get position and velocity function as a function of time. Is it that simple or am I missing some subtle points? Not all function can be analytically integrated or differentiated...

Thanks!

 

[anuttarasammyak]

For your example it means a differential equation of
\frac{d^2x}{dt^2}-3t^2-(\frac{dx}{dt})^3=0
We have to solve this equation to know x(t) which also tells us v(t) and a(t). Usually it is very difficult to get analytical solution but we may get numerical solution by computer.

 

[fog37]

Thank you.

At the end of the day, the equation to solve remains $$\frac {d^2 x} {dt^2}=...$$ with whatever is found on the right-had side, which is generally derivatives of the variable $x$ of different order...

 

[kuruman]

Thank you.

At the end of the day, the equation to solve remains $$\frac {d^2 x} {dt^2}=...$$ with whatever is found on the right-had side, which is generally derivatives of the variable $x$ of different order...

Not necessarily. Depending on the differential equation you have, it might be easier to solve $\dfrac{dv}{dt}=\dots~$ then integrate to find $x(t)$.
In your example, you will have to solve $\dfrac{dv_x}{dt}=3t^2+v_x^3$ which is still not separable but at least it's a first-order differential equation.

 

[anuttarasammyak]

Computation example
https://www.wolframalpha.com/input/?i=dv/dt=3t^2+v^3,+v(0)=0+&lang=en