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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 8

Lesson 2: Connecting position, velocity, and acceleration functions using integrals

# Motion problems with integrals: displacement vs. distance

AP.CALC:
CHA‑4 (EU)
,
CHA‑4.C (LO)
,
CHA‑4.C.1 (EK)
The definite integral of a velocity function gives us the displacement. To find the actual distance traveled, we need to use the speed function, which is the absolute value of the velocity.

## Want to join the conversation?

• Is, is this what people learn at 10th grade in America?
• In America, 10th graders usually learn Algebra 2, but talented 10th graders might learn pre-calculus or even calculus. Usually, calculus is a college level course in America.

Calculus is like algebra, but with the concept of a limit. This concept then leads to the concept of a derivative (think of the slope of a curve at a single point) and the concept of an integral (think of the area under a curve but above the x-axis). Furthermore, taking an integral is essentially the inverse of taking a derivative!

Have a blessed, wonderful day!
• At he starts working out the distance portion using the mentioned |v(t)|, but I am struggling to apply that to this problem: integral |-16t+48|dt from t=2 to 5. How can I use the absolute value formula without using integration by parts? Is integration by parts the only way to go on these un-graphed problems that don't have a range that starts at the origin?

Once worked out using integration by parts, it is supposed to come out to 40, but thats only after you split it into two different integral functions with bounds that I don't exactly understand how to solve for to establish the two integrals. What am I missing?
• The only way to integrate absolute value functions like this is by splitting the integral as you describe. If there is a formula or other such thing, it would be derived by splitting the integral.

Now, the key to splitting the integral is to do it in a way that lets you ignore the absolute value function. You do this by picking the bounds of integration so that the function doesn't change sign in the middle of the integration region.

So for this example, -16t+48 is positive for t<3 and negative for t>3. So we can have one integral from 2 to 3, and on this interval, |-16t+48|=-16t+48. On the other interval, from 3 to 5, |-16t+48|=16t-48. Now these are just simple linear functions, easy to integrate.

(Also, 'integration by parts' is a specific technique that you'll encounter later on, and is very different from this splitting of the interval here. I would advise not mixing the terminology.)
• I was trying to find the distance traveled without a graph; so I integrated the absolute value of 5-t and plugged in 10 then subtracted the value of plugging in 0 and I got 0 which isn't what is said in this video. What'd I do wrong?
• Without seeing your work, I can't know what you did wrong, but it is possible that you integrated the absolute value wrong. You cannot just integrate it like a linear function, you must split the absolute value into two parts.

Since it is absolute value, our answers must always be positive. for t > 5, 5-t is negative, so for the interval [5, 10] the absolute value function will be equal to -(5-t)

for t < 5, 5 - t will be positive, so for the interval [0, 5], the absolute value function will be equal to 5 - t

this leaves you with the definite integral from 0 to 5 of (5 - t), and the definite integral from 5 to 10 of - (5 - t) = (t - 5)

adding the results of these two integrals gives you the correct answer of 25
(1 vote)
• Sal defined displacement as "change in position". But wouldn't that mean that the derivative of displacement is the rate of change of the change in position with respect to time? I can't even understand what that would mean neither geometrically nor algebraically. Wouldn't the rate of change of the change of position with respect to time be acceleration? I'm confused.
• We don't actually use displacement as a function, because displacement requires a time interval, whereas a function gives instants in time. The derivative of the vector-valued position function x(t) is the "rate of change of position", also known as velocity v(t).

I would also be careful using the words "change" and "rate of change" interchangeably; "change" refers to a difference in one quantity, whereas "rate of change" refers to a ratio of changes in different quantities. Acceleration is the "rate of change of the rate of change of position", not the "rate of change of the change of position".
• In regard to the difference between displacement and distance traveled, could it be said that displacement is a state function while distance traveled is a path function?
• Howdy eskry,

That is absolutely correct, and once you study thermodynamics in physics the difference between state and path functions will become very important.

Happy learning! :-)
• Velocity is change in position/change in time, or in other words displacement/change in time. The derivative of position (with respects to time) is displacement/change in time, and so it is velocity.

So why do we say that the derivative of displacement is velocity? Isn't velocity the derivative of position, not change in position? And so how can velocity be the slope of the tangent line to a displacement vs. time graph?

EDIT:
So i'm reading that since displacement is measured as position minus some initial position, and that initial position is constant, the two derivatives are the same.

What really is position? Whenever we indicate position with a number, are we really indicating displacement but with initial position at zero?
(1 vote)
• Position is a vector. It tells us in which direction and at what distance an object is located relative a given reference point.

(Since we're dealing with vectors I decided to let them be two-dimensional.
In one dimension there are only two directions in which a vector can point, which we can represent with ordinary variables being either positive or negative. The calculations are the same, just omit all the 𝑦-coordinates.)

If (ℎ, 𝑘) are the coordinates of the reference point and (𝑥, 𝑦) are the coordinates of the object, then the object's position is
𝒑 = (𝑥, 𝑦) − (ℎ, 𝑘) = (𝑥 − ℎ, 𝑦 −𝑘)

As the object moves over time we get
𝒑(𝑡) = (𝑥(𝑡) − ℎ, 𝑦(𝑡) − 𝑘)
⇒ 𝒑'(𝑡) = (𝑥'(𝑡), 𝑦'(𝑡))

𝒑'(𝑡) is the velocity of the object, which is not the same as its displacement.

Instead, displacement is also a position vector, but tells us the object's position relative to the object's starting point, (𝑥(𝑡₀), 𝑦(𝑡₀)) at some given time 𝑡 = 𝑡₀ :

𝒅(𝑡) = (𝑥(𝑡) − 𝑥(𝑡₀), 𝑦(𝑡) − 𝑦(𝑡₀))

You can verify for yourself that with 𝒑(𝑡₀) = (𝑥(𝑡₀) − ℎ, 𝑦(𝑡₀) − 𝑘) we have
𝒅(𝑡) = 𝒑(𝑡) − 𝒑(𝑡₀)

Either way, we get
𝒅 '(𝑡) = 𝒑'(𝑡)
• At , he says the integral of the rate function (v(t)) is equal to the change in quantity (aka distance traveled). I understand how to use this, but I'm struggling to understand the concept. I know the integral of v(t) is the area under the curve of v(t), but how does that equal the distance traveled?
(1 vote)
• Well, you know that velocity is the derivative of position/distance, since it defines a rate (think meters travelled, distance, changing to m/s, a rate at which an object travels). Velocity also gives the slope of a distance vs. time graph, since you take how many units are travelled over a specific time parameter. Since an integral is the opposite of a derivative, velocity is the antiderivative of position.

To answer your question, looking at the graph of velocity, it is "m/s" vs. seconds. By finding the area, you would essentially multiply the two units of measure, leaving you with just meters or whatever unit is used, leaving you with the "total distance" (finding the total distance would actually require taking the integral of the absolute value of velocity).

Hope this helps!!
• When doing problems that involve taking the derivative or anti-derivative of something, should I have all the derivatives of the all the trigonometric functions memorized?
I have sine, cosine, and tangent memorized, but not the inverses of those nor the reciprocals (secant, cotangent, and cosecant).
I do have the derivatives of them written down in a math scratchpad of mine, but should I definitely have those derivatives of memorized to?
(1 vote)
• If you can derive the derivative/antiderivative fairly quickly, then there is no need to memorize it. It might be useful to memorize the inverse trig derivatives, because I’ve seen a lot of integral problems that simplify to some form of arctan.
Hope this helps!
(1 vote)
• In the next exercise I ran into a problem that was rather confusing:

"A particle moves in a straight line with velocity v(t)=-t^4+t^3 where t is time in seconds. At t=0, the particle's distance from the starting point was 4 meters in the positive direction. What is the particle's position at t=9 seconds?"

The answer was the definite integral from 0 to 9 of v(t)dt + 4. (I wish I could type mathematical symbols in questions). Why is it that the starting distance is added to the integral instead of the starting displacement (which wasn't given?). Also, if it's distance, shouldn't the direction not matter?
(1 vote)
• x = -t^5/5 + t^4/4 + C from integration.

Given t=0 we are told displacement is 4. In other words x=4 when t=0.

Substituting x=4, t=0

yields 4=C.

To find the total distance travelled you would need to use integrational calculus.

You said the starting displacement was not given however it was given: "4 meters in the positive direction". X-Displacement is the how left or right an object is from it origin. In this case the particle was 4 metres to the right of the origin.
(1 vote)
• How does finding the area under curves relate to distance and displacement?
(1 vote)
• Remember the area of a rectangle formula, width times height? You can think about the area of a curve the same way, width times average height. When we take the area of a curve with the y value in m/s and the x value in s, the resulting unit is (m/s)*s, which is just m.

Another way to think about it is by taking advantage of the fact that an integral is an antiderivative. When we differentiate displacement, we get velocity, so when we integrate velocity, we get displacement.
(1 vote)