Recent content by FelidaeTrick

All integrals boundaries are (0,D) and relating to x. v is a function of x. T=∫1/v ∫v'=U --> ∫[v'-U/D]=0 Adding Lagrange Multiplier: T=∫[1/v-λ(v'-U/D)] That functional needs be maximised. Using EL: v''=-1/(λv^2) Solution to this isn't simple, and moreover seems not to be correct.

Yes, I tried. but I couldn't derive something valueable

Guess it must involve variational calculus. I'm desperate... :-(

Slowest That means that it took the object the longest period of time to reach the end. In other words, the time that took for the object to get to the end is to be maximised.

Homework Statement I need to find the acceleration, a(t), for which a particle moves on a straight line of a length D. V(0)=0 V(final, that is at x = D) = U a is always positive, but NON-INCREASING! What is a(t) for which the particle reaches the end of the line the slowest as possible?Homework...