How to find the total distance travelled in a kinematics integration question? 
I completed question a which is $\int _0^3\:t^2-6t+5 = -3$ or more specifically $3m$ to the left
Now isnt distance the absolute value of the displacement? $\left|\int _{t1}^{t2}\:v\left(t\right)\right|=\left|x1\:-x2\right|$
So for question b i did this:
$\left|\int _0^3\:t^2-6t+5\right| = 3$
and apparently this is incorrect. I am a bit confused what I did wrong, everything seems logical to me. So if you can try question b) that will be great and please point out what I am missing. Thnkas
 A: In physics questions, displacement is the final distance from the starting point, whereas distance is the total traveled distance.
So, in such questions, you can find the displacement by integrating the function, like you did.
However, to find the distance traveled, you have to take the integral of the absolute value of the function, not the absolute value of the integral of the function, which is what you did.
If you were to take the absolute value of the function, then you would notice that there is a point on your graph in the interval given where the function touches the x-axis and then bounces back upwards.
So, you would have to (if integrating by hand), split the integral up into two parts, and then adding them up. Since, in the interval (0,3), the function you gave has a root at x=1, the integral will become:
$$\int_0^3v(t)\,dt=\int_0^1|v(t)|\,dt+\int_1^3|v(t)|\,dt$$
That's how you'll be able to find the correct answer.
A: 
Velocity $ v = (t-1)(t-5) = t^2-6t +5 $ at time instants  $ t=1$ and $t=5$ vanish as seen in graph.
Acceleration is integral of velocity. We can take arbitrarily distance marker zero at goalpost $ t=0, x=0 $
$$ x= \dfrac{t^3}{3} -3 t^2 + 5 t + 0  $$
Now just read off from the graph of cubical parabola that represents distance logged
Red graph shows distances$ x$ ( you can find more accurate values in a hand calculator). We have  at times
$ t= (1,2,3,5,6.8,7)$ seconds  $ x= (2.5,0.8,-3, -8.3,0,2.4 );$
Positive $x$ means forward distance logged, negative means gone back.
At 3 seconds 3 meters gone back. The distance you got is correct. If sign is not considered the displacement is 3 meters in question 2 is okay. If sign is considered we know its relation to a desired direction.
Until 6.8 seconds still no headway made, only thereafter travel has forward displacements.
By differentiation we find that maximum and minimum distances occur at times $t=1$ and $t= 5$ seconds.
A: A very important point in calculus mechanics is looking at where the graph's roots lie. A second VERY important point is that displacement is often different to distance. Your answer to part $b$ was correct, but only for displacement, not distance, which is what the question asked for.
The reason these points are important is demonstrated in your question. The question asks for change in position. Consider your graph:
$$v(t)=t^2-6t+5=(t-5)(t-1)$$
From the facftorsied form we see that the graph has a root WITHIN the interval the question has asked for (at $(0,1)$ ). This means any integration between $0$ and $3$ will be comprised partly of a positive value equal to the integral between $0$ and $1$ but ALSO a negative value equal to the integral between $1$ and $3$, resulting in the displacement but NOT the distance. The distance can be found by adding the positive integral between $0$ and $1$ to the absolute value, ie the positive value of the second integral between $1$ and $3$. I hope that helped!
Those are important points; so don't hesitate to ask for any clarification. I'm also studying in High School in England so I know where you're coming from in that question :)
A: I'm going to give an answer that does nothing to deal with the problem above, but is meant to clear up some conceptual misunderstanding. There's a key thing you're missing here - the difference between displacement and distance. Displacement is the integral of velocity, whereas distance is the integral of speed. Let's say at time $t=t_0$ we are at a position $x=x_0$, and then at time $t=t_1$ we are at a position $x=x_1$. Now regardless of how I got from $x_0$ to $x_1$, the displacement is the same. Whereas the distance covered could be anything that is at least $x_1-x_0$. I might have traversed the gap only once, or maybe I did it twice, or maybe I stopped halfway in the middle and turned around but changed my mind and went back... and so on. Below is a picture to illustrate this.

While all of these curves illustrate the same displacement between the start and end point, they don't all illustrate the same distance.
A: Some definitions:

*

*$$\text{average speed} = \frac{\text{distance travelled}}{\text{time
   elapsed}}$$

*$$\text{speed at time }t=\left| v(t)\right|$$

*$$\text{average value of } f(t) \text{ over }
   [t_1, t_2]=\frac{\int_{t_1}^{t_2}f(t)\,\mathrm{d}t}{t_2-t_1}$$
Therefore
\begin{align}\text{distance travelled over }[t_1, t_2] &= \text{average speed}\times\text{time elapsed}\\
&=\frac{\int_{t_1}^{t_2}\left| v(t)\right|\,\mathrm{d}t}{t_2-t_1} \times \left(t_2-t_1\right)\\
&=\boxed{\int_{t_1}^{t_2}\left| v(t)\right|\,\mathrm{d}t}\\
&\neq \left| {\int_{t_1}^{t_2} v(t)\,\mathrm{d}t } \right|\\
&=\text{magnitude of displacement over }[t_1, t_2].\end{align}
The magnitude of displacement is the shortest distance between Positions $1$ and $2$, so can be smaller than the distance travelled from Position $1$ to $2$ (what an odometer measures).
