Re: [Math4u] Particle path problem

I don't have time to do this problem. What you want to do is either accelerate towards x(t) with acceleration a from t(0) to t(1) and then deaccelerate with acceleration -a from t(1) to t(2)

or initially accelerate away and then accelerate towards x(t)

such that x(2) =x(t) and v(2)=v(t)

You have 2 equations and 2 unknowns

First equation for velocity

v(2)=v(t)=a*(t1-0)+(-a(t2-t1)=a*(2*t1-t2)

Second equation for distance

x=integral of v which is the integral of t

which may look something like

-(1/2)*a*(t2)^2+2*a*(t1)*(t2)-a*(t1)^2 (I am probably wrong)

You have 2 unknowns, a and t1. Generate an expression for "a" or t1 using the first equation and substitute it into the second equation (after confirming and correcting the second equation). Then you will have one equation and one unknown.

Regards,

Brian

 

 

----- Original Message -----

From: Mojtaba Fathi

To: Math4u@yahoogroups.com ; moji

Sent: Tuesday, October 09, 2007 7:13 AM

Subject: [Math4u] Particle path problem

Hi all

Please help me to solve this problem. I faced this when designing a cam and follower mechanism :

Consider a particle moving along x-axis. at the beginning, the particle is in origin and hes no velocity, ie x(0)=0, v(0)=0.
Now, wa want to move particle from it's initial position to a new position so that x(t2)=x2,v(t2)=v2. (t2 > 0 , x2 > 0 and v2 > 0 are given).
Find the equation of motion x=x(t) such that
1. x(t) is continuous within range 0 <= t <= t2
2. v(t)=dx(t)/dt is continuous within range 0 <= t <= t2
3. maximum value of absolute of particle acceleration becomes minimum within range 0 <= t <= t2, i.e. L=max(abs(d2x(t)/dt2) becomes as minimum as possible.

Thanks

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