# Math Help - Particle Motion...

1. ## Particle Motion...

1. A particle starts at time $t = 0$ and moves along the x-axis so that its position at any time $t \geq 0$ is given by $x(t)=(t-1)^3(2t-3)$.

EDIT: Find total distance traveled when t = 0 and when t = 5

I've seen this kind of questions in a different format, but I'm not sure how I should approach this one.. Help would be greatly appreciated!

2. Originally Posted by kdl00
1. A particle starts at time $t = 0$ and moves along the x-axis so that its position at any time $t \geq 0$ is given by $x(t)=(t-1)^3(2t-3)$.

I've seen this kind of questions in a different format, but I'm not sure how I should approach this one.. Help would be greatly appreciated!
What problem? You are given a statement about a particle but there is no question about it nor are you told to do anything with it.

What do you want to know? What are you trying to do?

3. Originally Posted by HallsofIvy
What problem? You are given a statement about a particle but there is no question about it nor are you told to do anything with it.

What do you want to know? What are you trying to do?
Hm... that is the question on the worksheet i was given. But I know that by looking at the previous questions, it wants me to find the total distance when t=0 and t=5

4. Okay.

The "net" distance would be x(5)- x(0). but "total" distance might be different- the object might have gone past x(5) then back.

What I would do is this: differentiate the x(t) function to fine the "velocity" function and determine where that is equal to 0. Points where the derivative is 0 show where x itself is a local max or min- in other words where you might have turned around and back tracked.

Once you have found, say, $t_0$, $t_1$, [tex]t_2[tex], etc. as points where the derivative is 0, Start at t= 0 and determine the first "turn around" time. If it is less than t=5, calculate the distance from x(0) to that "turn around" point. Find the next "turn around time" if there is one. If it is less than 5, calculate the distance between those two "turn around points"- the difference is x values may be negative but remember that distance is always positive- take the absolute value.

5. Originally Posted by HallsofIvy
Okay.

The "net" distance would be x(5)- x(0). but "total" distance might be different- the object might have gone past x(5) then back.

What I would do is this: differentiate the x(t) function to fine the "velocity" function and determine where that is equal to 0. Points where the derivative is 0 show where x itself is a local max or min- in other words where you might have turned around and back tracked.

Once you have found, say, $t_0$, $t_1$, [tex]t_2[tex], etc. as points where the derivative is 0, Start at t= 0 and determine the first "turn around" time. If it is less than t=5, calculate the distance from x(0) to that "turn around" point. Find the next "turn around time" if there is one. If it is less than 5, calculate the distance between those two "turn around points"- the difference is x values may be negative but remember that distance is always positive- take the absolute value.

Ahh ok... my teacher didn't stress on some of the part you mentioned. Thank you, this helped a lot. I understand how to do this problem now.