Back
# Kinematics

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, \(a(t) = a_0.\)

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 \(r_n - r_{n-1}.\)

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.

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 (\(r(t_i) = 0\)).

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

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

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

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?

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 \(a\).

×

Problem Loading...

Note Loading...

Set Loading...