# Position of a particle moving along the $x$-axis question

I am in need of help solving this question.

The position of a particle moving along the $x$-axis is given by the function $s(t) = t^3 -6t^2 +9t$ where $s$ is in meters and $t$ is in seconds.

When is the particle moving to the right?

What does the diagram to represent the motion of the particle look like?

When is it going to the right? When to the left?

What is the total distance traveled during the first 5 seconds?

What is the acceleration at time t and after 4 seconds?

What does the graph of the position, velocity, and acceleration functions for $0 <t < 5$ look like?

When is the particle speeding up ? When is it slowing down?

Thanks so much!

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@user72714: Welcome to MSE! It really helps readability if you format your questions using MathJax. Have a got your equation correct? Also, what have you tried and where are you confused? If this is HW, it should be tagged as such. Thanks – Amzoti Feb 17 '13 at 21:47
Where are you stuck? Each question is one liner if use definition. – Kaster Feb 17 '13 at 21:52

Lots of questions! We have $s(t)=t^3-6t^2+9t$. So if the velocity is denoted by $v(t)$, we have $$v(t)=s'(t)=3t^2-12t+9=3(t-1)(t-3).$$ The particle is moving to the right when the velocity is positive, and to the left when the velocity is negative.

Looking at $3(t-1)(t-3)$, we note that it is positive when $t\gt 3$, also when $t\lt 1$. So in the time interval $(-\infty,1)$ and in the time interval $(3,\infty)$, to the degree this makes physical sense, we have motion to the right. In the time interval $(1,3)$ we have motion to the left.

The acceleration $a(t)$ is the derivative of velocity. So $a(t)=6t-12$.

There is some possible ambiguity (or trick) in the question about speeding up. The velocity is increasing when the acceleration is positive, that is, when $t\gt 2$. The velocity is decreasing when $t\lt 2$.

You should be able to do the rest of the parts. But "total distance travelled in the first $5$ seconds" is tricky, so we do some detail.

The net change in displacement is easy, it is $s(5)-s(0)$. But for total distance travelled, we need to take account of the fact that we are travelling to the right when $t$ is between $0$ and $1$, also when $t$ is between $3$ and $5$, while between $1$ and $3$ we are travelling to the left. So while $s(1)-s(0)$ and $s(5)-s(3)$ are positive, the number $s(3)-s(1)$ is negative. Thus the total distance travelled in the first $5$ seconds is $$|f(1)-f(0)|+|f(3)-f(1)|+|f(5)-f(3)|.$$

Maybe Don't Read: Velocity is not the same thing as speed. The speed at time $t$ is the absolute value of velocity, so it is $3|(t-1)(t-3)|$. We may want to know when speed is increasing. That's a different question than asking when velocity is increasing.

To find out you where speed is increasing, you can find out where the derivative of $(3(t-1)(t-3))^2$ is positive. This derivative is $9(t-1)(t-3)(2t-4)$. It is not hard to find out where this is positive: for $t\gt 3$ and for $1\lt t\lt 2$.

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You seem to have ignored the question about the total distance traveled in the first 5 seconds. You need the speed (integrate $ds$ for that and make sure $s$ is always nonnegative). – Barbara Osofsky Feb 18 '13 at 4:12
@BarbaraOsofsky: Thanks, that may cause the OP some problems, so I will add something. – André Nicolas Feb 18 '13 at 4:25