Analyzing motion problems: total distance traveled | AP Calculus AB | Khan Academy

Alexi received the following problem: a particle moves in a straight line with velocity v of t is equal to negative t squared plus 8 meters per second, where t is time in seconds. At t is equal to 2, the particle's distance from the starting point was 5 meters. What is the total distance the particle has traveled between t equals 2 and t equals 6 seconds?

Which expression should Alexi 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. The key question is: what is the total distance the particle has traveled between t equals 2 and t equals 6? So we just care what happens between those points. We don't care that the particle's distance from the starting point was 5 meters at t equals 2. So this right over here is actually unnecessary information.

The first thing that you might want to 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 2 and time equals 6, this would have been the correct answer. So this would be displacement: displacement from t equals 2 to t is equal to 6.

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. One way to think about it is you would integrate not the velocity function. This would—if you integrate velocity, you get displacement. Instead, you would integrate the speed function.

Now, what is speed? It is the magnitude of velocity. In one dimension, it would just be the absolute value of your velocity function. 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 2 to t is equal to 6.

And let's see, we have that choice right over here. The displacement one here, this is an interesting distractor, but that is not going to be the choice. This one right over here, v prime of 6, that gives you the acceleration. If you're taking the derivative of the velocity function, the acceleration at 6 seconds—that's not what we're interested in.

And this gives you the absolute difference in velocity between time six and time two—that's not what we're trying to figure out either.