Kinematics deals in tracking the state of an object as encoded by its position, velocity, and acceleration. In this quiz, we'll imagine ourselves a delivery person for Tony's pizza, trying to make it around the city on a series of motorcycles. In analyzing our motion from stop to stop, we'll apply common sense to calculate the details of our motion, and try to generalize in anticipation of a formal derivation of the kinematic relations in the next quiz.

To start, Tony's pizza gives us an entry level motorcycle that's a little peculiar.
Before turning it on, you **set a velocity at which it will move** from the moment you turn it on, until the moment you turn it off.

Let's pick up our first pizza and get started.

You are traveling from your house to the pizzeria on the **entry level motorcycle**. Recall that this motorcycle has a constant velocity from the moment you turn it on until the moment you turn it off.

Which of the following velocity vs. time (\(v\) vs. \(t\)) curves represents your motion through the city?

Suppose you set your motorcycle to velocity \(v\) and travel for a time \(T\).

What is the relationship between your velocity, the time, and the distance you travel \(d\)?

Suppose you're on the way to the pizzeria to start your day. You have thirty minutes to get there, so you set your motorcycle to \(v_1 = \SI[per-mode=symbol]{10}{\kilo\meter\per\hour}\) to get there exactly on time. Fifteen minutes into your trip, you realize you're passing your cousins house, and you stop for five minutes to hang out.

At what speed \(v_2\) (in \(\si[per-mode=symbol]{\kilo\meter\per\hour}\)) do you need to travel in the remaining ten minutes to get there in time?

Let's represent our calculation from the last problem in abstract terms. Suppose the total time for your trip is \(T=\SI{30}{\minute}\), the time at which you stopped at your cousin's house is \(t_1=\SI{15}{\minute}\), and your break lasts for \(t_\textrm{break}=\SI{5}{\minute}\).

How can we express the total distance travelled?

Having made your first delivery on time (despite the pit-stop at your cousin's house), Tony's Pizza upgrades your motorcycle.

Now you can

- travel at a constant speed that you set at the start of your trip.
- use a single speed boost that you can trigger at any time. The boost accelerates you from \(v\) to \(v + \Delta v\) over the course of \(\Delta t\) seconds, followed by a deceleration from \(v + \Delta v\) back down to \(v\) over \(\Delta t\) seconds as in the \(v\) vs. \(t\) plot below.

Suppose you're making a delivery traveling at speed \(v\). You notice that the traffic light is about to switch from yellow to red. In order to make it past the traffic light, you trigger the speed boost at time \(t_i\).

What distance do you cover from time \(0\) to time \(T=t_i+\Delta t\) in the diagram above?

Seeing how smoothly you handled the traffic light, Tony decides to express his appreciation by upgrading your motorcycle a second time. Now you can adjust your speed at will, like with a normal motorcycle.

Before, we showed that the distance traveled was given by the area under the \(v\) vs. \(t\) curve. In the boosted acceleration case, we broke the curve into simple shapes whose areas we know how to calculate. We can do the same for any arbitrary \(v\) vs. \(t\) curve by using rectangles as below.

In general, our approximation becomes

\[\begin{align}d = \textrm{Area}(v\ \textrm{vs.}\ t) &= \lim_{N\rightarrow\infty}\sum\limits_{i=1}^N v(t_i)\Delta t \\ &= \lim_{N\rightarrow\infty} \sum\limits_{i=1}^N v(t_i)\frac{T}{N}.\end{align}\]

If we take the limit as \(N\) goes to \(\infty\), the discrepancy between the area of our set of rectangles and the area under the \(v\) vs. \(t\) curve goes to zero, and our estimate becomes perfectly accurate.

In fact, this limit is one of the definitions for the Riemann integral of a function \(v(t)\), and we can say \[\lim_{N\rightarrow \infty}\sum\limits_{i=1}^N \Delta t_i \longleftrightarrow \int dt.\] With this in hand, we can state that \[d = \int v(t) dt.\] We notice that the relationship between \(d\) and \(v\) is the same as that between \(v\) and \(a\), so we can also write \[v = \int a(t) dt.\]

Suppose your motorcycle has a maximum acceleration of \(a=\SI[per-mode=symbol]{5}{\meter\per\second\squared}\) and a maximum deceleration of \(d=\SI[per-mode=symbol]{3}{\meter\per\second\squared}\).

What is the maximum speed \((\)in \(\si[per-mode=symbol]{\meter\per\second})\) it can achieve in time \(T=\SI{200}{\second}\), assuming that it starts **and** stops at rest?

Hint: Draw a \(v\) vs. \(t\) plot.

In this quiz we used common sense and arithmetic to find out how our motorcycle moves in the simple cases of constant velocity or constant acceleration. By carefully examining the implication of our analysis, we were able to generalize our ideas, postulate a general relationship between \(d\) and \(v\), and between \(v\) and \(a\).

In particular, we learned the following:

- The distance travelled is equal to the area under a \(v-t\) curve.
- In the case of constant acceleration from rest, the distance travelled is given by \(\frac12 \Delta v \Delta t\).
- In general, the area under a curve \(v(t)\) is given by \(\int v(t) dt\).

In the following quiz we'll formalize these ideas, and use them to derive the general results of kinematics in one dimension.

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