MHB What Is the Total Distance Traveled by the Particle from t = 0 to t = 3?

[karush]

$\tiny{207 \quad DOY}$
A particle moves along the x-axis. The velocity of the particle at time t is $6t - t^2$.
What is the total distance traveled by the particle from time $t = 0$ to $t = 3$ ? $(A)\,3 \quad (B)\,6 \quad (C)\,9 \quad (D)\,18\quad (E) \, 27$

Spoiler

$\displaystyle\int_0^3 6t - t2 \: dt = 3t^2-\dfrac{1}{3}t^3\Bigg|_0^3=27-9=18\quad (D)$

ok think this is correct...

the reason I am posting these is bc I am preparing a list of about 30+ AP exam problems which will be sent via a cell phone about twice a week to HS students who will eventually take the AP calculus Exam.
if you have ever looked up solutions to these, they range from a quick one line to a major production.
I want the simplest and shortest as possible and still maintain integrity of the topic
Most AP problems do not require heavy calculation, many can done just by observation.

MHB has a been a great source to get to the main core of these problems and would be a help tremendously to those who will need to take the test.

 

[skeeter]

Understand there is a difference between "total distance" and "displacement". In this particular problem, they are the same since $v(t) \ge 0$ over the time interval $0 \le t \le 3$.

Strictly speaking ...

displacement (change in position) is a vector quantity, $$\Delta x = \int_{t_1}^{t_2} v(t) \, dt$$

distance traveled is a scalar quantity, $$d = \int_{t_1}^{t_2} |v(t)| \, dt$$

Had the time interval been $0 \le t \le 8$, then the two values would not be the same since the object changes direction at $t=6$

$$\Delta x = \int_0^8 6t-t^2 \, dt = \dfrac{64}{3}$$

... the object ends up 64/3 units in the positive direction from its initial position.

$$d = \int_0^8 |6t-t^2| \, dt =\int_0^6 6t-t^2 \, dt - \int_6^8 6t-t^2 \, dt = \dfrac{152}{3}$$

... the object travels a total distance of 152/3 units, ending up at the relative position mentioned above.

 

[topsquark]

I just wanted to make sure the point is made here. (What skeeter said is right on, but I want to re-inforce the point.)

[math]\overline{s(t)} = \int_a^b \overline{v(t)} ~ dt[/math] ( I'm using [math]\overline{v(t)}[/math] as a vector quantity. I don't like the LaTeX arrow.)

If you are looking for the distance traveled and v(t) is negative over a domain then you have to take the absolute value of it in the integration. That gives you the distance traveled. [math]s(t) = \int_a^b | \overline{v(t)} | ~ dt[/math].

If you are looking for the displacement then you don't need to worry about whether the vector points in a negative direction. So [math]\overline{s(t)} = \int_a^b \overline{v(t)} ~ dt[/math].

-Dan

 

[karush]

well, I was wondering why the integral was showing up in some examples and just a division in others

 

[HallsofIvy]

If you are given a function, v(t), that tells the velocity of an object at each time, t, then $\int |v(t)|dt$ gives the total distance traveled. If instead you are given an average velocity,v, over time t, then vt gives the distance traveled.

Neither of those involves a division. That would be "going the other way"- if you are given a function, x(t), that tells the distance traveled at each time, t, then $\frac{dx}{dt}$ give the velocity at each time, t. If, instead, you are given a total distance traveled, d, in time t, then $\frac{d}{t}$ gives the average velocity.