Particle Problem

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legacyofpiracy

Junior Member
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Oct 20, 2005
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A particle moves along the x-axis in such a way that its position at time t for t is greater than or equal to 0 is given by

x= (1/3)t^3-3t^2+8t

I need to show that at time t=0 the particle is moving to the right, find all the values in which it moves left, the position at t=3, and at t=3, the total distance traveled.

Any help would be greatly appreciated
 
The derivative x’(t)=t<SUP>2</SUP>−6t+8 is the velocity function.
The velocity is zero at t=2 & t=4. What does that mean for ‘movement’?
If the velocity is positive what is the direction of the particle?

BTW: I usually will not respond to any post that includes “ASAP”.
 
I am really sorry about the ASAP in the heading, it was extreemly rude of me and i really appreciate your response.

You see the trouble is I am not quite sure what they are asking because I know how to find the velocity and how to read the graph of it, but the questions do not specifically ask for the velocity. Rather they ask for me to show when the particle is moving left and right and the position of the particle at a given time.
 
Define the right direction to be positive. The particle is either moving in the positive direction or the negative direction. If the particle is moving in the positive direction then its velocity will be positive.

Alternatively, you can think of x vs t being like y vs x, in that time is on the x-axis, and x is on the y-axis. When the curve to the cubic is moving up, the particle is moving to the right; when the curve is going down, the particle is moving to the left.
 
If the velocity is positive, the derivative is positive, the distance is increasing! WHY?
If negative, the distance is decreasing. What does that mean about right/left?
In this problem x(0)=8, so the particle starts at 8. How does it move?
 
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