AP PHYSICS C: MECHANICS › UNIT 1, KINEMATICS › TOPIC 1.2
Displacement, Velocity, and Acceleration
This is where the course parts company with AP Physics 1. The definitions are the same three you already know. The difference is that here the interval is allowed to close, and when it does the slope becomes a derivative and the area becomes an integral.
1.2.A Describe a change in an object’s position.
1.2.B Describe the velocity and the acceleration of an object, both averaged over an interval and taken at an instant.
Wording here is mine, written from the Day 3 and Day 4 Cornell notes. Check it against your own copy of the CED.
1. First, we throw away almost everything
Before any of this works we adopt the object model. The size, the shape and the internal structure of the thing are ignored, and what is left is a single point that still carries the properties we need, such as mass and charge.
That is a deliberate lie, and it is the most useful one in physics. You stop asking which part of the car you are timing, and the question becomes answerable. The model discards exactly the things it discards: ask how a gymnast spins or where a car crumples, and the point model has already thrown the answer away.
2. Average first, because the average is what you can measure
Δr = r − r0vavg = Δr / Δtaavg = Δv / Δt
Every one of those is built from two endpoints and nothing else. Average velocity is not the average of the velocities, and the object could have done anything at all in between. On a graph of position against time it is the slope of the straight line joining the two endpoints, which is called a secant.
3. Then let the interval close
Shrink the interval and the secant pivots. In the limit it touches the curve at one point, and the number it reports is no longer an average over anything:
v(t) = limΔt→0 Δx/Δt = dx/dta(t) = dv/dt = d2x/dt2
That limit is the definition of the derivative, so instantaneous velocity is the derivative of position and instantaneous acceleration is the derivative of velocity. Nothing else in this topic is new. Everything below is a consequence of those two lines.
One motion, three graphs, one cursor
A cart moves along a track with position x(t) = 0.4t3 − 3t2 + 6t + 4, in meters and seconds. Drag the cursor, or press play. The straight line on each panel is the tangent at that instant, and the height of the dot on the panel below it is that tangent’s slope. That is the whole relationship, drawn twice.
Three readings are worth hunting for by hand before you accept them from the simulation.
Where the position graph turns, the velocity is zero. At t = 1.38 s and again at t = 3.62 s the tangent on the top panel is horizontal, and the dot on the middle panel sits exactly on its axis. Those are the two instants when the cart is momentarily at rest, and they are the only two.
Where the velocity graph turns, the acceleration is zero. That happens once, at t = 2.5 s. It is the moment the cart is moving backward fastest. The acceleration is zero there and the velocity is at its most negative, which is a pairing worth staring at until it stops feeling contradictory.
The cart is at rest twice and its acceleration is never zero at either of those instants. At t = 1.38 s the acceleration is about −2.7 m/s2, which is exactly why the cart does not stay at rest.
4. Speeding up, slowing down, and the one line that settles it
An object accelerates when the magnitude of its velocity changes, or its direction changes, or both. Along a single axis there is no direction left to turn, so the question becomes about signs, and in this course you do not have to memorize the answer. Differentiate the square of the speed:
d/dt (½v2) = v (dv/dt) = v a
The left side is the rate at which the speed is growing, up to a positive factor. So the speed increases exactly when the product v a is positive, and decreases exactly when that product is negative. The speed is momentarily neither when the product is zero.
The simulation above shades the interval whenever v a > 0. Notice that the shading does not line up with the sign of a, and does not line up with the sign of v either. It lines up with their product, which is the only thing that was ever in question.
5. Running the derivative backwards is accumulation
If velocity is the derivative of position, then position has to be the antiderivative of velocity. Written with limits, that is the Fundamental Theorem of Calculus wearing physics clothes:
Δx = ∫t1t2 v(t) dtΔv = ∫t1t2 a(t) dt
The integral is a signed area. Area below the axis counts as negative, which is what makes the result a displacement rather than a distance. If you want the distance traveled you have to integrate the speed instead, breaking the interval at every instant the velocity changes sign:
distance = ∫ |v(t)| dt
Those two integrals differ for our cart, and they differ by a lot. Over the full five seconds the displacement is +5.0 m and the distance traveled is 9.47 m, because the cart went forward, came back, and went forward again.
The initial condition is the constant of integration
The top panel is the same velocity graph. Drag the right hand edge of the shaded region to choose an interval starting at t equal to zero. The shaded signed area is the displacement. The lower panel adds that area to a starting position you control, and the slider is the only thing that moves the curve.
Move the slider and watch what changes. The shape of the position curve does not. Its slope at every instant does not. Only its height does, and it moves rigidly, all of it together.
Written out, with the initial condition doing its job:
v(t) = v0 + ∫0t a(t’) dt’x(t) = x0 + ∫0t v(t’) dt’
The constant acceleration equations you memorized in AP Physics 1 are just these two integrals done once each with a held constant. They are not laws. They are the answer to one particular integral, and they are wrong the moment the acceleration varies, which in this course it usually does.
6. Average velocity is an average value
The two definitions of average velocity are the same statement, and the Fundamental Theorem is what makes them the same:
vavg = Δx / Δt = (1 / Δt) ∫t1t2 v(t) dt
For our cart over the whole five seconds the displacement is 5.0 m, so the average velocity is 1.0 m/s exactly. The cart never once had a velocity of 1.0 m/s for any stretch of time, and it spent more than two seconds moving backward. An average value does not have to be a value the function ever settles at.
The same construction for acceleration gives something sharper. Over the whole five seconds:
aavg = Δv / Δt = (6.0 − 6.0) / 5.0 = 0
The average acceleration is exactly zero, and the acceleration itself is zero at only one instant in the entire five seconds. The cart was being accelerated the whole time. The negative first half cancelled the positive second half, and averaging destroyed the information.
Check yourself
1. For x(t) = 0.4t3 − 3t2 + 6t + 4, find v(t) and a(t), then find every instant in 0 ≤ t ≤ 5 at which the cart is at rest.
v = 1.2t2 − 6t + 6 and a = 2.4t − 6. Setting v = 0 and dividing through by 1.2 gives t2 − 5t + 5 = 0, so t = (5 ± √5)/2, which is 1.38 s and 3.62 s. Both lie in the interval, so both count.
2. At t = 2.0 s the cart has v = −1.2 m/s and a = −1.2 m/s2. Is it speeding up or slowing down, and how would you answer without computing anything?
Speeding up. Both are negative, so v a > 0, so d(½v2)/dt > 0. The cart is moving backward and gaining backward speed. Answering “slowing down, because the acceleration is negative” is the standard mistake, and it is wrong for half of all negative accelerations.
3. Find the displacement of the cart over 0 ≤ t ≤ 5 two ways: from the endpoints, and by integrating the velocity.
Endpoints: x(5) − x(0) = 9.0 − 4.0 = +5.0 m. Integral: ∫05(1.2t2 − 6t + 6) dt = [0.4t3 − 3t2 + 6t]05 = 50 − 75 + 30 = +5.0 m. They agree, and they have to: the antiderivative of the velocity is the position, minus whatever constant you dropped.
4. The distance the cart travels over the same five seconds is 9.47 m, not 5.0 m. Set up the integral that gives it, and say why it needs to be split.
Distance is ∫05|v| dt. The absolute value is not something you can antidifferentiate directly, so split the interval where v changes sign, at t = 1.38 and t = 3.62, integrate on each piece, and add the magnitudes: 3.62 + 2.24 + 3.62 = 9.47 m. Average speed is then 9.47 / 5.0 = 1.89 m/s, against an average velocity of 1.0 m/s.
5. A different cart has a(t) = 6t, starts at x0 = 2 m with v0 = −4 m/s. Find v(t) and x(t), and say what goes wrong if you use x = x0 + v0t + ½at2.
v = −4 + 3t2 and x = 2 − 4t + t3. The constant acceleration equation is the result of integrating a constant twice, and this acceleration is not constant, so the equation does not apply at all. There is no value of a that makes it right. When the acceleration depends on time, integrate.
Topic 1.3, Representing Motion. The same three graphs read against each other on sight, without a function to differentiate: slopes, areas, and the shapes that go with them.