Calculus/Calculus

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Question
A PARTICLE MOVES on the x-axis in such a way that its position at time t is given by x(t)=(2t-1)(t-1)^2

at what times t is the particle at rest
during what interval of time is the particle moving to the left.

Thank you very much.

Answer
The particle is at rest when x'(t)=0 , & the particle is moving to
the left when x'(t)<0. So 1st let's derive :
x'(t)=2(t-1)^2+(2t-1)*2*(t-1)=2[t^2-2t+1]+2[2t^2-3t+1]=2[4t^2-5t+2].
x'(t)=0 -> 4t^2-5t+2=0 NO Solutions !! which means
x'(t) is always positive non zero !
The particle always moving right & never stops

Alon.

Calculus

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Alon Mandes

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Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .

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1. I'm a team member of mathnerds (math site for answering questions) 2. I'm a team member in the Student's Union of the Technion, helping students who have problems in mathematics. 3. 2 years of experience as a math teacher in college. 4. I give free homework help for high school students in Mathematics & Physics. 5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" , "Complex Functions".

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Hi-Tech company : GSM4VOIP ; job possition : Algorythm developer.

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M.A in Mathematics & Bs.c in Electronics.

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