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## Curvilinear Motion

1. The problem statement, all variables and given/known data
A particle moves along the curve $y(x)=x^2-4$ with constant speed of 5 m/s. Determine the point on the curve where the max magnitude of acceleration occurs and compute its value.

2. Relevant equations radius of curvature
$$|a|=\sqrt{a_n^2+a_t^2}$$

3. The attempt at a solution
I have assumed that 'constant speed' means $$v_t=5 \Rightarrow a_t=0$$

So now I have to find where a_n is maximum. Any hints to get me going here?

Thanks,
Casey
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 Recognitions: Gold Member So I got the right answer. But I don't know how to spell out a proper procedure for future problems. I found y'(x)=0; at x=0. So x=0 is a critical point. Then I just plugged that into $$a_n=\frac{[1+\frac{dy}{dx}^2]^{3/2}}{\frac{d^2y}{dx^2}}$$ and that worked. But do I know that x=0 is a MAXIMUM, if I do the second derivative test, I get positive 2, so I would have thought it was a minimum.
 Recognitions: Gold Member Can anyone point out how I would know it was a maximum? Shouldn't the 2nd derivative test work here?

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## Curvilinear Motion

 Quote by Saladsamurai Can anyone point out how I would know it was a maximum? Shouldn't the 2nd derivative test work here?
But really. Do it. Do it. (Dodgeball reference)
 Blog Entries: 1 Recognitions: Gold Member Science Advisor Staff Emeritus You should note here that x = x(t), that is, position is a function of time. And that the maximum acceleration would occur when x'''(t)=0 and not when y''(x)=0.

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 Quote by Hootenanny You should note here that x = x(t), that is, position is a function of time. And that the maximum acceleration would occur when x''(t)=0 and not when y''(x)=0.
I see. So how would I determine, if and where a maximum would be if I am given y(x)?

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 Quote by Hootenanny You should note here that x = x(t), that is, position is a function of time. And that the maximum acceleration would occur when x'''(t)=0 and not when y''(x)=0.
 Quote by Saladsamurai I see. So how would I determine, if and where a maximum would be if I am given y(x)?
Any takers? 'cause I don't know...
 Have you studied parametric equations? I'm guessing that's exactly what you're doing right now, from the looks of this problem. If you can restate the problem by writing independent eqations for y(t) and x(t) (i.e. parametrically), then you'll be set. The key is that you know how x and y are related, and you know that the magnitude of the velocity (i.e. the speed) is constant. Do you know how to express the velocity vector for parametric equations?
 Recognitions: Gold Member I have actually never studied them. This is for a dynamics course. I have probably studied them, but never heard them worded as "parametric". Give me an example, if you would.
 Oh, then maybe that's not the way to go. It's where you have separate equations for x(t) and y(t) and then you use the chain rule a lot to express dy/dx in terms of dx/dt and dy/dt. If you haven't seen it, then you probably have to stick with expressing the normal acceleration as a function of the curvature. Do you know how to do that? I'm not sure I remember, myself ...
 Recognitions: Gold Member No...you're right, it does have to do with the chain rule. I am just confused by it....that, is I do not have a nice procedure as to where to start or how to set it up. I have y as a function of x. Now what?
 Can I jump in? First of all constant velocity means $u^2=u_x^2+u_y^2=C$. Thus first take the derivative of $y(t)=x(t)^2-4$ to find the relation bewteen $u_x,\,u_y$ and plug it in $u^2$

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 Quote by Rainbow Child Can I jump in? First of all constant velocity means $u^2=u_x^2+u_y^2=C$. Thus first take the derivative of $y(t)=x(t)^2-4$ to find the relation bewteen $u_x,\,u_y$ and plug it in $u^2$
You lost me. I am given y as a function of x... not time. i'll look it over some more, though.
 The $y,\,x$ are actually $$y(t),\,x(t)[/itex]. Can you take the derivative of $y(t)=x(t)^2-4$ with respect to $t$? Hint. $u_x=\dot{x}(t),u_y=\dot{y}(t)$  Recognitions: Gold Member I am sorry, I cannot follow any of your notations. I have y(x) not y(t). . . where are you getting t? What is u? Where am I?  Ok! The partile is moving in the plane x-y. Thus x,y varies with time, i.e. $x(t),y(t)$ and the velocity $\vec{u}$ has two compontents, i.e. $u_x=\frac{d\,x(t)}{d\,t},\,u_y=\frac{d\,y(t)}{d\,t}$ and the magnitude of the velocity is [tex]u^2=u_x^2+u_y^2$$ Do you follow up to now?

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 Quote by Rainbow Child Ok! The partile is moving in the plane x-y. Thus x,y varies with time Do you follow up to now?
Not this part. I get all the maths after that, but why do i automatically assume that it varies with time? It says nothing about time.