1. ## 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].

2. $\displaystyle \int_{a}^b v(t)dt$ gives you total displacement. $\displaystyle \int_{a}^{b}\left|v(t)\right|dt$ gives you the total distance.

3. 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 $\displaystyle v(t) = 2t^2 + 3t + 1$ ?

4. 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?

5. 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.

6. $\displaystyle v(t) = 2t^2 + 3t + 1 = 0$

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

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

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.

7. 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.

8. displacement is just x(8) - x(-3).

do you understand the difference between displacement and distance traveled?

9. 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?

10. 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.