Acceleration of a particle on a parabola

[Jenny Physics]

Homework Statement

A particle moves along a parabola on the x-y plane with equation $y^{2}=2px$ with constant speed $1000m/s$.What is the magnitude of its acceleration?

Homework Equations

Parametric equations $\vec{r}=(b^{2}t^{2}/(2p),bt)$.

The Attempt at a Solution

$\vec{v}=\frac{d\vec{r}}{dt}=(b^{2}t/p,b)\implies |\vec{v}|=b\sqrt{b^{2}t^{2}/p^{2}+1}=1000$

$\vec{a}=\frac{d\vec{v}}{dt}=(b^{2}/p,0)\implies |\vec{a}|=b^{2}/p$

Not sure were to go from here.

 

[Mark44]

Homework Statement

A particle moves along a parabola on the x-y plane with equation $y^{2}=2px$ with constant speed $1000m/s$.What is the magnitude of its acceleration?

Homework Equations

Parametric equations $\vec{r}=(b^{2}t^{2}/(2p),bt)$.

I think this is more complicated than it needs to be, with the added parameter.
You could use $\vec{r}=(\frac{t^2}{2p}, t)$. This gives the proper relationship between the x and y components. It also makes for a simpler expression for $\vec v$ and the magnitude of the acceleration depends only on p.

The Attempt at a Solution

$\vec{v}=\frac{d\vec{r}}{dt}=(b^{2}t/p,b)\implies |\vec{v}|=b\sqrt{b^{2}t^{2}/p^{2}+1}=1000$

$\vec{a}=\frac{d\vec{v}}{dt}=(b^{2}/p,0)\implies |\vec{a}|=b^{2}/p$

Not sure were to go from here.

BTW, very nice job in formatting your work!

 

[LCKurtz]

It looks to me like the $b$ must depend on $t$ for the constant speed to happen. And consequently the acceleration must depend on $t$.

 

[LCKurtz]

After thinking a bit more about it, I see no reason to assume a form for $y$. So I would start by assuming a parameterization like $$\vec r(t) =\left \langle \frac{y^2(t)}{2p},y(t) \right \rangle$$Then see what $|\vec r'(t)| = C$ gets you. I don't think it will be as simple as $y=bt$.

 

[Jenny Physics]

After thinking a bit more about it, I see no reason to assume a form for $y$. So I would start by assuming a parameterization like $$\vec r(t) =\left \langle \frac{y^2(t)}{2p},y(t) \right \rangle$$Then see what $|\vec r'(t)| = C$ gets you. I don't think it will be as simple as $y=bt$.

$|\vec{r}'|=\frac{dy}{dt}\sqrt{\left(\frac{y^{2}}{p^{2}}+1\right)}=C$.

 

[LCKurtz]

Well, I haven't worked it all out, but my thought is that you want to keep the variable $t$ if you are going to use the parametric form.

 

[Jenny Physics]

Well, I haven't worked it all out, but my thought is that you want to keep the variable $t$ if you are going to use the parametric form.

I fixed an issue in post #5. It seems I end up with a differential equation.

 

[LCKurtz]

Yes, I think it gives a DE and it is about to get very messy. It is making me curious what level course you are taking where this problem came from. Have you had DE's? This problem is looking to me like it is non-trivial unless there is some clever insight that I am missing. The DE I came up with will only give $y(t)$ implicitly and the second derivative looks like it isn't going to be simple to work with.

 

[Jenny Physics]

Yes, I think it gives a DE and it is about to get very messy. It is making me curious what level course you are taking where this problem came from. Have you had DE's? This problem is looking to me like it is non-trivial unless there is some clever insight that I am missing. The DE I came up with will only give $y(t)$ implicitly and the second derivative looks like it isn't going to be simple to work with.

Yes I have taken a DE class (this is from a physics class that assumes knowledge of DE). The equation you end up would have to be solved numerically?

 

[ehild]

The acceleration along an in-plane curve has a tangential and a centripetal component.
https://www.real-world-physics-problems.com/curvilinear-motion.html
As the speed is constant, the tangential acceleration is zero. The centripetal acceleration is equal to V²/R where R is the radius of curvature and V is given. It will depend on the position along the curve.