PHYS 170

Three main variables to describe motion:

This chapter is about one dimensional motion only, so we do not need the full mathematics of vectors yet. Instead of writing $\vec{s} = (-5\hat i)m$ for an object $5m$ West of you, we will simply denote the direction using the sign as $s=-5m$. In other words, in one dimension, there is no need to put the vector symbol (the little arrow) on the variables, and no need to use $\hat i, \hat j$. Vectors will become useful in two or higher dimensions.

Sometimes $s$ is denoted as $x$ for horizontal motion, or $y$ for vertical motion.

In calculus, "rate of change" literally means taking the time derivative ($\frac{d}{dt}$). Therefore: $$\begin{eqnarray} v &=& \frac{ds}{dt} \\ a &=& \frac{dv}{dt} = \frac{d}{dt} (\frac{ds}{dt} ) = \frac{d^2s}{dt^2} \end{eqnarray}$$

One can have a rough idea about the meanings using the following approximation: $$\begin{eqnarray} v &=& \frac{ds}{dt} &\approx& \frac{\Delta s}{\Delta t} = \frac{\text{change in displacement}}{\text{change in time}} \\ a &=& \frac{dv}{dt} &\approx& \frac{\Delta v}{\Delta t} = \frac{\text{change in velocity}}{\text{change in time}} \end{eqnarray}$$ where the symbol $\Delta$ means "change". It is always computed by final value minus the initial value, like so: $$\Delta s = s_{final} - s_{initial} \\ \Delta v = v_{final} - v_{initial}$$

Velocity vs Speed

In physics the word "speed" is not the same as "velocity". Speed means the magnitude of velocity, i.e. speed = $|\vec{v}|$. For example, say you calculate $v=-3m/s$ in a 1D problem (negative because the object is moving left), if the question asks you for the "speed", you must answer $3m/s$ instead of $-3m/s$. If your online homework said you had the wrong sign in an answer even though you followed the sign convention correctly, read the question carefully, did it ask you for the velocity or the speed? Many students are not observant enough and it becomes a major source of confusion.

I will assume that you already know how to calculate the slope of a straight line, and that you know the following meanings:

• Positive slope: line going up
• Zero slope: line staying the same height
• Negative slope: line going down

There are multiple ways to calculate the slope, but in our application here the fastest way is the method "Rise over Run": $$slope = \frac{rise}{run} = \frac{\Delta y}{\Delta x}$$ where "rise" stands for the vertical change, and "run" stands for the horizontal change. Below are two videos that refeshes your memory, in particular, you will need to be able to handle negative slope, so pay some attention to the second video.

Slope in physics (unlike in your Math courses) almost always carries its own unit, and is given by: $$\text{unit of the slope} = \frac{\text{unit of the y-axis}}{\text{unit of the x-axis}}$$ For example, if the y-axis is displacement measured in $m$ while the x-axis is time measured in $s$, then the unit of the slope will be $m/s$. If the y-axis is in $m/s$, the x-axis is in $s$, then the slope will be in $m/s^2$.

Try It Yourself (click to show)

motion_graph

Exercise - slope

The graph above is made up of four straight segments. Find their slopes. If you cannot calculate all four slopes correctly, please watch the videos above because you are not ready to continue to the rest of this lecture.

Solution

0

0.02

-2

0.02

1.5

0.02

0

0.02

SubmitShow Solution

Feedback will appear here.

We will learn how to analyze three types of graphs that describe motion, they are:

• $st$ graph (displacement vs time graph)
• $vt$ graph (velocity vs time graph)
• $at$ graph (acceleration vs time graph)

Here is an $st$ graph (displacement $s$ is denoted as $x$) and an explanation of the motion it represents:

Since the y-axis is the horizontal displacement $x$, one could obviously read off the position $x$ from the graph. Less obvious, is that you could also read off the velocity by calculating the slope of the graph as you could see in the figure above. Here is the rule concerning slopes of such motion graphs:

motion_graph

Example - velocity from $st$ graph

Find the velocity at $1s$, and also the velocity at $2s$.

Solution

First note that it is a $st$ graph, so to find the velocity we will need to evaluate the slope. At $1s$: $$ v(1s) = \frac{rise}{run}= \frac{-3m}{1s} = -3m/s $$

At $2s$: $$ v(2s) = \frac{rise}{run}= \frac{2m}{1.5s} = 1.33m/s $$

Try It Yourself (click to show)

motion_graph

Exercise - acceleration from $vt$ graph

Find the acceleration at $2s$ and also at $11s$.

The accerlation is the slope of the $vt$ graph. You need to calculate the slope of the straight lines at $2s$ and at $11s$.

Solution

First note that it is a \(vt\) graph, so to find the acceleration we will need to evaluate the slope. At \(2s\): $$ a(2s) = \frac{rise}{run}= \frac{(12-4)m/s}{10s} = 0.8m/s^2 $$

At \(11s\): $$ a(11s) = \frac{rise}{run}= \frac{-12m/s}{2s} = -6m/s^2 $$

$a(2s)=$

0.8

0.02

$ m/s^2 $, \t\t $a(11s)=$

-6

0.02

$m/s^2 $

$a(2s)=$ $ m/s^2 $, \t\t $a(11s)=$ $m/s^2 $

SubmitHintShow Solution

Feedback will appear here.

As you saw before, slope allows you to start from $s$ to $v$ to $t$. It turns out the area of the graph allows you to go from $a$ to $v$ to $s$:

motion_graph

Example - velocity from a $at$ graph

Suppose a car was at rest at $t=0s$. What is the velocity of the car at $10s$, $20s$, $30s$, and $40s$?

Solution

At $10s$, the only area is the rectangle above the x-axis, giving a positive velocity: $$ v(10s) = (2m/s^2)(5s) = 10m/s $$

At $20s$, there are two rectangles, one above the x-axis, and one below. Note that the area below the x-axis is regarded as a negative area, giving the velocity: $$ \begin{eqnarray} v(20s) &=& (2m/s^2)(5s) - (2m/s^2)(20s-15s) \\ &=& 0m/s \end{eqnarray} $$ Hence the object is at rest at $20s$.

At $30s$, there are two rectangles, one above the x-axis (positive area), and one below (negative area), giving the velocity: $$ \begin{eqnarray} v(30s) &=& (2m/s^2)(5s) - (2m/s^2)(25s-15s) \\ &=& 10m/s-20m/s \\ &=& -10m/s \end{eqnarray} $$ The negative velocity means the car is moving in the negative x direction.

At $40s$, there are three rectangles, two above the x-axis (positive area), and one below (negative area), giving the velocity: $$ \begin{eqnarray} v(40s) &=& (2m/s^2)(5s) - (2m/s^2)(25s-15s) + (2m/s^2)(40s-35s) \\ &=& 10m/s-20m/s +10m/s \\ &=& 0m/s \end{eqnarray} $$ The car is again at rest at $40s$

Try It Yourself (click to show)

motion_graph

Exercise - displacement from a $vt$ graph

Suppose a car was at the origin at $t=0s$. What is its displacement at $5s$, and at $20s$?

Displacement is the area of the \(vt\) graph. In this case the areas are triangles. Watch out for the negative area when the curve drops below the x-axis.

Solution

There are 2 triangles between the curve and the x-axis. At \(5s\), only the triangle above is needed: $$ s(5s) = \frac{1}{2}(5s)(30m/s) = 75m $$ At \(20s\), both triangles are needed, but watch out for the negative area represented by the second triangle: $$ \begin{eqnarray} s(20s) &=& \frac{1}{2}(5s)(30m/s) - \frac{1}{2}(20s-5s)(20m/s) \\ &=& 75m-150m = -75m \end{eqnarray} $$ Negative displcement means the car has moved in the negative x direction (or to the left of the origin).

$s(5s) = $

75

0.2

$m$, $s(20s) = $

-75

0.2

$s(5s) = $ $m$, $s(20s) = $

SubmitHintShow Solution

Feedback will appear here.

motion_graph

Exercise - velocity from a $at$ graph

Suppose a car was at rest at $t=0s$. What is its final velocity at $3s$?

The change of velocity is the area of the $at$ graph. In this case the areas are triangles. Watch out for the negative area when the curve drops below the x-axis.

Solution

There are 2 triangles between the curve and the x-axis. Take the area of the triangle above the x-axis minus the area of the one below $$ \begin{eqnarray} v(3s) &=& \frac{1}{2}(2s)(2m/s^2) - \frac{1}{2}(1s)(1m/s^2) \\ &=& 2m/s- 0.5m/s = 1.5m/s \end{eqnarray} $$

1.5

0.02

$m/s$

$m/s$

SubmitHintShow Solution

Feedback will appear here.

Try It Yourself (click to show)