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Deriving the kinematic relations


In the last quiz, we were able to relate position, velocity, and acceleration in special cases. We were also able to write down the relationship between the pairs \((a,v)\) and \((v,d)\).

In this quiz we're going to formalize our findings, derive the general kinematic relations, and obtain the relations for the special case of constant acceleration, \(a(t) = a_0\).

When we explored Kinematics in the City we saw that for constant velocity, \(d=vT\). In general, when \(v\) depends on time, this relation is a differential equation for the rate of change of displacement \(d\). How can we write this as a differential equation?

Suppose that we break up the \(v\) vs. \(t\) plot into rectangles. We can write \[r_n = v_1\Delta t + v_2\Delta t + \cdots + v_n \Delta t.\] What is \(r_n - r_{n-1}\)?

In the continuous limit where the differences in time (and position) are infinitesimal, we can allow the differences to become differentials, i.e. \[\begin{align} \Delta r_n = r_n - r_{n-1} &\longleftrightarrow dr \\ \Delta t_n = t_n - t_{n-1} &\longleftrightarrow dt. \end{align}\] At each moment, the change in \(r(t)\) is given by \(v(t)dt\). Therefore, we can write \[\frac{dr}{dt} = v(t),\] or equivalently \[r(t_f) = r(t_i) + \int\limits_{t_i}^{t_f} v(t) dt.\]

Just as \[r_n = v_1\Delta t + v_2 \Delta t + \cdots +v_n \Delta t,\] we have \[v_n = a_1\Delta t + a_2 \Delta t + \cdots + a_n \Delta t.\] Similarly, we have \[\frac{dv}{dt} = a(t) \longleftrightarrow v(t_f) = v(t_i) + \int\limits_{t_i}^{t_f}a(t)dt.\] As we claimed, the relationship between \(a\) and \(v\) is the same as that between \(v\) and \(r\).

Suppose you're given the following form for the acceleration of a particle that starts from rest: \[a(t) = a_0 e^t.\]

Find the velocity at time \(t_f,\) \(v(t_f),\) given that the initial velocity, \(v(t_i),\) is zero.

Thus, given an expression for \(a(t)\), we can find the velocity \(v(t_f)\) at all times. In fact, we can integrate once more to find the position \(r(t_f)\).

Suppose we're given an arbitrary form for the acceleration \(a(t)\) of a particle that starts from rest at \(r(t_i) = 0\).

Find its position \(r(t_f)\).

This is all a bit abstract. Let's apply our formula to a case that we've solved already. Recall our second motorcycle (the one that could briefly boost from \(v\) to \(v+\Delta v\)).

If it starts from rest at \(r(0)\), we can set \(v(0)\) equal to zero, and write the acceleration \(a(t)\) as \(a_0\).

Then, \[\begin{align} r(t_f) &= r(0) + \int\limits_0^{t_f} dt \int\limits_0^{t} dt^\prime a_0 \\&= \int\limits_0^{t_f} dt\ a_0 t \\ &= \frac12 a_0 {t_f}^2. \end{align}\]

Putting it all together, if our motorcycle has an initial velocity \(v(0) = v_0\), then we find that the position is given by

\[\begin{align} r(T) &= r_0 + \int\limits_0^T v_0 dt + \int\limits_0^T dt \left[\int\limits_0^t dt^\prime\ a_0\right] \\ &= r_0 + v_0T + \int\limits_0^T dt\ a_0 t \\ &= r_0 + v_0T + \frac12 a_0T^2. \end{align}\]

Suppose our motorcyclist starts at their apartment, with initial speed \(v_0\) and accelerates at a constant rate so that they're moving at \(v_f\) a time \(T\) later.

How far do they travel during this motion?

Sometimes it is necessary to relate the initial and final velocities of an object in terms of its distance traveled, without reference to the time. As one example, an insurance specialist may come to the scene of a motorcycle crash and know the braking deceleration of the motorcycle along with the distance of its skid, but want to know how fast it was going before the accident.

Suppose a motorcycle starts at \(v=v_0\) and then skids over a distance of \(d\), ending the skid at \(v=v_f\) .

Find \({v_f}^2 - {v_0}^2\) in terms of the distance \(d\) and the deceleration \(a\).

We've shown how to derive the kinematic relations from the fundamental relationships between \(r\), \(v\), and \(a\): \[\frac{dr}{dt}=v, \quad \frac{dv}{dt} = a.\] For the case of constant acceleration, we showed \[r_T = r_0 + v_0T + \frac12 a_0 T^2.\] Finally, we showed that we can relate the initial and final velocity without reference to time via \[{v_T}^2 - {v_0}^2 = 2a(r_T - r_0).\]

In this quiz, we've derived the relations in the context of 1D motion. However, these derivations work just the same in 2 or more dimensions. In general, for \(N\)-dimensional \(\mathbf{r}\), \(\mathbf{v}\), and \(\mathbf{a}\), we have \[\begin{align} \mathbf{r_T} &= \mathbf{r_0} + \mathbf{v_0}T + \frac12 \mathbf{a_0}T^2 \\ \mathbf{v_T}^2 - \mathbf{v_0}^2 &= 2\mathbf{a}\cdot\left(\mathbf{r_T} - \mathbf{r_0}\right). \end{align}\] One common example we'll revisit is that of projectile motion, where the vertical motion of an object in a gravitational field \(g\) is uncoupled from its horizontal motion.


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