01Classical Mechanics
Everything starts with a law \(x(t)\) that tells where a particle is at each instant. From it come the displacement, the average velocity over an interval and, shrinking the interval, the instantaneous velocity: the derivative. Differentiating once more gives the acceleration. Write any law, such as a polynomial or an oscillation, choose two instants, and the laboratory computes every quantity, shows the limit process and separates displacement from distance.
The slope of each graph is the next quantity and the area under each one is the change of the previous one. Drag over a graph to move through the motion.
Four definitions are enough for any straight-line motion; only when the acceleration turns out to be constant does the law become the invariant of the next pages.
How the average velocity becomes the instantaneous velocity, and the instants at which the body turns back.
Kinematics describes motion without asking for its causes. We treat the body as a particle, a point that moves as a whole, and restrict it to a straight line: a bead sliding along a wire. Its motion is completely known once we know the function \(x(t)\), the coordinate measured from an origin along an axis with a positive direction, at every instant.
A motion can be described with an equation or with a graph of \(x\) against \(t\) (which is not the path: the bead always stays on the line). At rest, \(x = A\) is a horizontal line; with constant velocity, \(x = A + Bt\) is a straight line of slope \(B\); accelerated motions give curves, such as \(x = A + Bt + Ct^2\) or the oscillation \(x = A\cos\omega t\). The graph \(x(t)\) is continuous: a body cannot disappear from one place and appear in another.
Between the instants \(t_1\) and \(t_2\) the particle goes from \(x_1\) to \(x_2\); the average velocity is
\[\bar v = \frac{x_2 - x_1}{t_2 - t_1} = \frac{\Delta x}{\Delta t},\]the slope of the straight line that joins both points of the graph. All the details of the motion in between are lost: a racing car that completes a lap has an average velocity of zero.
Making the interval shorter and shorter, the secant tends to the tangent of the curve at \(t_1\), and the average velocity tends to its slope:
\[v = \lim_{\Delta t\to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}.\]For \(x = 3 + t + 2t^2\) at \(t = 1\ \text{s}\) the averages are 7.0, 6.0, 5.5, 5.2… m/s for intervals of 1, 0.5, 0.25, 0.1 s and tend to \(v(1) = 1 + 4t = 5\ \text{m/s}\) (Resnick, table 2-1). From now on "velocity" means instantaneous velocity; its magnitude is the speed.
The velocity can change in turn. The average acceleration over an interval is \(\bar a = \Delta v/\Delta t\), and the instantaneous acceleration
\[a = \frac{dv}{dt} = \frac{d^2x}{dt^2},\]the slope of the graph \(v(t)\). It can be positive while the velocity is negative: the sign of \(a\) and \(v\) together tell whether the body speeds up (same sign) or slows down (opposite signs).
The relations can be read backwards with integrals: the change of velocity is the area under \(a(t)\) and the displacement is the area under \(v(t)\),
\[\Delta v = \int_{t_1}^{t_2} a\,dt,\qquad \Delta x = \int_{t_1}^{t_2} v\,dt.\]When \(a\) is constant these integrals give the invariant \(v = v_0 + at\), \(x = x_0 + v_0 t + \tfrac12 a t^2\); in any other case one must differentiate or integrate the actual law.
The displacement \(\Delta x\) is signed and depends only on the ends; the distance counts every stretch travelled. They differ when the velocity changes sign, that is, when the body turns back at an instant where \(v = 0\). Average speed (distance over time) and average velocity (displacement over time) differ for the same reason.
Problems adapted from Serway (chapters 2, 4, 5 and 6), Resnick and the course notes. Load any of them into the laboratory: the data fill the panel and the solver does the rest, so you can then change any value and see how the answer reacts.
A particle moves along the x axis according to \(x = 2.00 + 3.00t - 1.00t^2\), with x in metres and t in seconds. At \(t = 3.00\ \text{s}\), find (a) its position, (b) its velocity and (c) its acceleration. (Look at the second instant.)
For \(x(t) = 3.000 + 1.000t + 2.000t^2\), compute the average velocity between \(t_1 = 1\ \text{s}\) and later and later instants, and compare it with the instantaneous velocity at \(t = 1\ \text{s}\).
A particle moves according to \(x = 10t^2\). (a) Find the average velocity between 2.00 s and 3.00 s. (b) Between 2.00 s and 2.10 s (change t₂).
An object moves along the x axis according to \(x(t) = (3.00t^2 - 2.00t + 3.00)\ \text{m}\). Determine (a) the average speed between \(t = 2.00\) and \(3.00\ \text{s}\), (b) the instantaneous speed at both instants, (c) the average acceleration and (d) the instantaneous acceleration at both instants.
A particle moves along the x axis with \(x = 2 + 3t - 4t^2\). Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at t = 0.
A particle moves according to \(x = t^3 - 6t^2 + 9t\). When is it at rest? What distance does it travel between 0 and 4 s, and what is its displacement?
A bead oscillates along a wire with \(x = 5\cos 2t\). Study its velocity and acceleration: the acceleration is always opposite to the position.
Theory after R. Resnick, D. Halliday and K. Krane, Física, vol. 1, chapter 2 (sections 2-1 to 2-5); problems from R. Serway and J. Jewett, Física para ciencias e ingeniería, vol. 1, chapter 2, and from the course notes on motion in one dimension.