# displacement and distance

• Dec 22nd 2008, 02:06 PM
vassago
displacement and distance
if i had a velocity function how would i find the displacement and distance traveled? the distance i'm guessing is an arclength but how do i find displacement? say $\displaystyle v(t) = 2t^2+3t+1$ and interval [-3,8].
• Dec 22nd 2008, 02:13 PM
Ziaris
$\displaystyle \int_{a}^b v(t)dt$ gives you total displacement. $\displaystyle \int_{a}^{b}\left|v(t)\right|dt$ gives you the total distance.
• Dec 22nd 2008, 02:20 PM
skeeter
Quote:

Originally Posted by vassago
if i had a velocity function how would i find the displacement and distance traveled? the distance i'm guessing is an arclength but how do i find displacement? say $\displaystyle v(t) = 2x^2+3x+1$ and interval [-3,8].

you have velocity as a function of position (x) ... is that what you meant?

or should velocity be a function of time, i.e $\displaystyle v(t) = 2t^2 + 3t + 1$ ?
• Dec 22nd 2008, 02:20 PM
vassago
I just made that equation up so I hope it works. but that means that I'd find the antiderivative of v(t), say V(t) is the antideriv, then find V(b)-V(a) and |V(b)-V(a)|. I'm really dumb when it comes to integrals.
and V(t) would be $\displaystyle 2/3t^3+3/2t^2+t$? Isn't there a constant, c, I would then be missing?
• Dec 22nd 2008, 02:21 PM
vassago
Quote:

Originally Posted by skeeter
you have velocity as a function of position (x) ... is that what you meant?

or should velocity be a function of time, i.e $\displaystyle v(t) = 2t^2 + 3t + 1$ ?

yes, velocity should have been a function of time, i fixed it. x is a bad habit.
• Dec 22nd 2008, 02:34 PM
skeeter
$\displaystyle v(t) = 2t^2 + 3t + 1 = 0$

$\displaystyle (2t + 1)(t + 1) = 0$

$\displaystyle t = -\frac{1}{2}$ , $\displaystyle t = -1$

total distance ...

$\displaystyle d = \int_{-3}^{-1} v(t) \, dt - \int_{-1}^{-\frac{1}{2}} v(t) \, dt + \int_{-\frac{1}{2}}^{8} v(t) \, dt$

the antiderivative of $\displaystyle v(t)$ is $\displaystyle x(t) = \frac{2t^3}{3} + \frac{3t^2}{2} + t + C$

you may ignore the constant of integration since it will cancel when using the fundamental theorem to evaluate the definite integrals.
• Dec 22nd 2008, 03:02 PM
vassago
so for [-1/2,8] would be x(8)-x(-1/2), correct? what about displacement? it seems that taking the abs value of that would be the same answer.
• Dec 22nd 2008, 03:05 PM
skeeter
displacement is just x(8) - x(-3).

do you understand the difference between displacement and distance traveled?
• Dec 22nd 2008, 03:13 PM
vassago
i don't understand them in regards to finding them using integrals.

would dist and displacement be the same if we had an equation that had roots outside of the interval we are looking at?
• Dec 22nd 2008, 03:24 PM
skeeter
distance and displacement are equal only if velocity is greater than or equal to 0 over the interval in question.

$\displaystyle v(t) > 0$ for $\displaystyle -3 \leq t < -1$
$\displaystyle v(t) < 0$ for $\displaystyle -1 < t < -\frac{1}{2}$
$\displaystyle v(t) > 0$ for $\displaystyle -\frac{1}{2} < t \leq 8$
this is why I broke up the integral into three pieces ... notice the middle integral has a negative sign before it in order to change the negative displacement over the interval $\displaystyle -1 < t < -\frac{1}{2}$ to a positive value.