Acceleration as a function of displacement (1 Viewer)

[mrpotatoed]

If given the acceleration as a function of displacement, (d^2x/dt^2), do you integrate normally to find the velocity as a function of time, and use the (v^2)/2 thingy to find velocity also as a function of displacement?

 

[kawaiipotato]

If given the acceleration as a function of displacement, (d^2x/dt^2), do you integrate normally to find the velocity as a function of time, and use the (v^2)/2 thingy to find velocity also as a function of displacement?

Yes, use dv/dt to find velocity as a function of time and d(1/2 v^2)/dx to find velocity as a function of displacement. In 4 unit, it's more common to use vdv/dx to find velocity as a function of displacement.

 

[mrpotatoed]

Thanks, I actually posted this in the wrong forum, should be for 3u haha, but guess its still relevant. I was getting confused with how the whole derivative notations linked up so thanks for confirming this, helps explain a lot lol.

 

[Ekman]

Yes, use dv/dt to find velocity as a function of time and d(1/2 v^2)/dx to find velocity as a function of displacement. In 4 unit, it's more common to use vdv/dx to find velocity as a function of displacement.

This is only useful when acceleration is given in terms of velocity. However if acceleration is given in terms of displacement, it is more suitable to use d((1/2 v^2) / dx =a, as you can directly integrate the displacement.

 

[kawaiipotato]

This is only useful when acceleration is given in terms of velocity. However if acceleration is given in terms of displacement, it is more suitable to use d((1/2 v^2) / dx =a, as you can directly integrate the displacement.

Why is it more suitable? In both cases, you're differentiating with respect to displacement. Integrating vdv = v^2 / 2

 

[Ekman]

Why is it more suitable? In both cases, you're differentiating with respect to displacement. Integrating vdv = v^2 / 2

One could argue that using dv/dt is suitable as well, since you would be using related rates to integrate dv/dt which would lead you to integrate vdv/dx. My point is that using d(1/2v^2)/dx is quicker as you only have to worry about one side of the equation, as opposed to integrating vdv then the other side of the equation. I guess it just comes down to preference.

 

[mrpotatoed]

If the acceleration has x in it though, you cant integrate normally can you, to find velocity in terms of t? You have to use the 1/2v^2 to integrate and get v in terms of x, then reciprocate it to get dt/dx, then integrate again and rearrange to get displacement in terms of time, and then finally differentiate that. Is there an easier way I don't know of?

 

[InteGrand]

If the acceleration has x in it though, you cant integrate normally can you, to find velocity in terms of t? You have to use the 1/2v^2 to integrate and get v in terms of x, then reciprocate it to get dt/dx, then integrate again and rearrange to get displacement in terms of time, and then finally differentiate that. Is there an easier way I don't know of?

No, it's pretty much this (I think it's explained in the Pender (Cambridge) 3U Year 12 textbook). Something to note is that you need to decide whether to take a positive square root or a negative square root when you "get v in terms of x" before integrating.