Classical Mechanics

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 relationships between acceleration and velocity, and between velocity and position.

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

Deriving the Kinematic Relations

When we explored Kinematics in the City we saw that when the delivery person travels at constant velocity, we could describe their motion by \(d=vt\). In general, when \(v\) depends on time, this relation becomes a differential equation for the rate of change of displacement \(d\). Let's see how to do that.

Suppose that we approximate the \(v\) vs. \(t\) graph as a series of rectangles. Then can write \[r_n = v_1\Delta t + v_2\Delta t + \cdots + v_n \Delta t.\] What is the difference in position from moment \(n\) to moment \(n-1\)? In other words, find \(r_n - r_{n-1}.\)

                         

Deriving the Kinematic Relations

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

Deriving the Kinematic Relations

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 \(r\).

Deriving the Kinematic Relations

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 final velocity \(v(t_f)\) given that the initial velocity \(v(t_i)\) is zero.

                         

Deriving the Kinematic Relations

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 \(r(t_f)\).

Deriving the Kinematic Relations

Suppose we're given an arbitrary form for the acceleration \(a(t)\) of a particle.

Find its position \(r(t_f)\) given that it starts from rest (\(r(t_i) = 0\)).

                         

Deriving the Kinematic Relations

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 \(r(t_i)\), we can set \(v(t_i)\) equal to zero, and write the acceleration \(a(t)\) as \(a_0.\)

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

Deriving the Kinematic Relations

Putting it all together, if our motorcycle has an initial velocity \(v(t_i) = v_i\) and starts at \(r(t_i) = r_i,\) we find that the position is given by

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

Deriving the Kinematic Relations

Suppose our motorcyclist starts at their apartment, with initial speed \(v_i\) 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?

                         

Deriving the Kinematic Relations

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 \(v_i\) and then skids over a distance of \(d\), ending the skid at \(v_f\) .

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

                         

Deriving the Kinematic Relations

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 \[r(t) = r_i + v_it + \frac12 a_0 t^2.\] Finally, we showed that we can relate the initial and final velocity without reference to time via \[{v_f}^2 - {v_i}^2 = 2a(r_f - r_i).\]

Deriving the Kinematic Relations

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 \[\begin{align} \mathbf{r}(t) &= \mathbf{r_i} + \mathbf{v_i}t + \frac12 \mathbf{a_0}t^2 \\ \mathbf{v_f}^2 - \mathbf{v_i}^2 &= 2\mathbf{a}\cdot\left(\mathbf{r_f} - \mathbf{r_i}\right). \end{align}\] 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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