Uniformly accelerated motion

Acceleration measures how fast the velocity changes. Over an interval it is the average acceleration \(\bar a = \dfrac{\Delta v}{\Delta t}\); making the interval shrink gives the instantaneous acceleration \(a = \dfrac{dv}{dt}\), the slope of the graph \(v(t)\).

Its sign says nothing by itself about speeding up: an elevator going up (\(v > 0\)) that brakes has \(a < 0\). The speed grows when \(a\) and \(v\) have the same sign and shrinks when they have opposite signs; the latter is a deceleration.

If \(a\) is constant, the average and instantaneous accelerations coincide and \(a = \dfrac{v - v_0}{t - 0}\), so

The graph \(v(t)\) is a straight line, so the average velocity is the mean of its ends, \(\bar v = \tfrac12(v_0 + v)\), and \(\Delta x = \bar v\,t\) gives \(\Delta x = \tfrac12(v_0 + v)\,t\). Replacing \(v\):

Isolating \(t = (v - v_0)/a\) from the first equation and substituting it in the second eliminates time:

From \(a = dv/dt\) we have \(dv = a\,dt\); integrating between \(0\) and \(t\) with \(a\) constant, \(\int_{v_0}^{v} dv = \int_0^t a\,dt\), so \(v - v_0 = a\,t\). Doing it again with \(v = dx/dt\):

Differentiating this result gives back \(v\) and \(a\): the three quantities are tied by derivatives one way and by integrals (areas) the other way.

A stretch with constant acceleration involves five quantities, and each equation of the invariant leaves out one of them:

• \(v = v_0 + at\) has no \(\Delta x\)
• \(\Delta x = v_0 t + \tfrac12 a t^2\) has no \(v\)
• \(v^2 = v_0^2 + 2a\Delta x\) has no \(t\)
• \(\Delta x = \tfrac12(v_0 + v)t\) has no \(a\)
• \(\Delta x = vt - \tfrac12 a t^2\) has no \(v_0\)

To find an unknown, use the equation that contains it and lacks the other unknown. When several parts are independent, go back to the original data rather than to a result you computed before, so that an error does not propagate.

In the course notes each case is obtained by applying the conditions of the variant to the invariant. If \(a = 0\) we get uniform motion, \(V = V_o\), \(S = V_o t\). If the speed grows we have MUA, \(V = V_o + at\), \(V^2 = V_o^2 + 2aS\), \(S = V_o t + \tfrac{at^2}{2}\). If it decreases we have MUR, with the same equations and the sign of \(a\) changed: \(V = V_o - at\), \(V^2 = V_o^2 - 2aS\), \(S = V_o t - \tfrac{at^2}{2}\), where \(a\) is the magnitude.

The laboratory works with signed quantities (a single positive direction), which is equivalent and avoids having to decide the sign beforehand.

Before writing any equation choose an origin and a positive direction, and keep them for the whole problem. Displacements, velocities and accelerations that point the positive way are positive and the others negative. Usually \(x_0 = 0\) is taken where the stretch begins, since only the difference \(x - x_0\) appears in the equations.

A negative result is not an error: it says the quantity points the other way. If you assume the motion goes to the right and obtain \(a < 0\), the body is braking.

For constant acceleration \(x(t)\) is a parabola, \(v(t)\) a straight line and \(a(t)\) a horizontal line. The slope of each graph is the next quantity (\(v = dx/dt\), \(a = dv/dt\)) and the area under each one is the change of the previous quantity: the area under \(v(t)\) between two instants is the displacement, counted negative below the axis, and the area under \(a(t)\) is \(\Delta v\).

A motion diagram, the positions at equal intervals of time, shows the same thing: equally spaced dots for uniform motion, growing gaps when it speeds up and shrinking gaps when it slows down.

The invariant is valid only while the acceleration is constant. A pendulum, a rocket that burns fuel or a raindrop falling against the air have variable accelerations; then one must go back to \(a = dv/dt\) and integrate. A motion made of several stretches with different constant accelerations is handled stretch by stretch, the final state of one being the initial state of the next, as in the module of motion by stretches below.