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].
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].
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?
$\displaystyle v(t) = 2t^2 + 3t + 1 = 0$
$\displaystyle (2t + 1)(t + 1) = 0$
$\displaystyle t = -\frac{1}{2}$ , $\displaystyle t = -1$
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.
distance and displacement are equal only if velocity is greater than or equal to 0 over the interval in question.
for your problem ...
$\displaystyle v(t) > 0$ 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.
so ... total distance = [x(-1) - x(-3)] - [x(-1/2) - x(-1)] + [x(8) - x(-1/2)]