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Current time:0:00Total duration:2:53

AP.CALC:

CHA‑4 (EU)

, CHA‑4.C (LO)

, CHA‑4.C.1 (EK)

- [Instructor] Alexey received
the following problem: A particle moves in a straight line with velocity v of t is
equal to negative t squared plus eight meters per second, where t is time in seconds. At t is equal to two,
the particle's distance from the starting point was five meters. What is the total distance
the particle has traveled between t equals two and
t equals six seconds? Which expression should Alexey
use to solve the problem? So we don't actually have to
figure the actual answer out, we just have to figure out what is the appropriate expression. So like always, pause this video and see if you can work
through it on your own. So now let's tackle this together. So the key question is
what is the total distance the particle has traveled
between t equals two and t equals six? So we just care what happens
between those points, we don't care that the particle's distance from the starting point was
five meters at t equals two. So this right over here is
actually unnecessary information. So the first thing that
you might wanna think about is well maybe distance
is just the integral of the velocity function; We've seen that multiple times. If you want to find the
change in a quantity, you just say the starting
time and the ending time and then you integrate the rate function. So wouldn't it just be that? Now we have to be very very careful. If the question was
what is the displacement for the particle between time equals two and time equals six, this would have been the correct answer. So this would be displacement. Displacement from t equals two to t is equal to six. But they're not saying displacement. They're saying total distance
the particle has traveled. So this is the total path
length for the particle. So one way to think about it, you would integrate not
the velocity function, if you integrate velocity,
you get displacement, instead, you would integrate
the speed function. Now what is speed? It is the magnitude of velocity and in one dimension, it would just be the absolute value of
your velocity function. And so the absolute value
of the velocity function, this would give you,
integrating the speed, this would give you the distance. Distance from t equals two to t is equal to six, and let's see, we have that
choice right over here. The displacement one here, this is an interesting distracter but that is not going to be the choice. This one right over here, v prime of six, that gives you the acceleration. If you're taking the derivative
of the velocity function, the acceleration at six seconds, that's not what we're interested in. And this gives you the absolute
difference in velocity, when in between time six and time two, that's not what we're
trying to figure out either.

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