{"id":1391,"date":"2021-06-30T22:27:15","date_gmt":"2021-06-30T22:27:15","guid":{"rendered":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/?p=1391"},"modified":"2021-12-30T19:14:49","modified_gmt":"2021-12-30T19:14:49","slug":"section-12-1-v2","status":"publish","type":"post","link":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/2021\/06\/30\/section-12-1-v2\/","title":{"rendered":"Chapter 12: Kinematics"},"content":{"rendered":"\n

Kinematics is the study of objects in motion, without asking anything about why<\/em> the object is moving. We will focus on translational motion with constant acceleration.<\/p>\n\n\n\n

12.1 One-dimensional kinematics<\/h2>\n\n\n\n

Say you are driving in a car along a long, straight road. If you know your current velocity, and you know how your velocity is changing (that is, you know your acceleration), you can determine your velocity at some later time. Assuming the acceleration is constant, we can multiply it by the elapsed time to find out how the velocity has changed in that amount of time. Then add that to whatever the velocity was to begin with. In symbols, we have<\/p>\n\n\n

\\[
\nv_{fx} = v_{ix} + a_xt
\n\\]<\/p>\n\n\n\n

We have essentially just used some unit analysis to figure this out: <\/p>\n\n\n

\\[ \\left(\\textrm{m\/s}^2\\right)\\left(\\textrm{s}\\right) = \\textrm{m\/s} \\]<\/p>\n\n\n\n

We can use similar reasoning to find your position after some time, by multiplying velocity by time (since \\((\\textrm{m\/s})(\\textrm{s}) = \\textrm{m}\\)). Unfortunately, unit analysis leaves out any dimensionless<\/em> (or unitless) factors. It turns out that we need to include a factor of \\(\\frac{1}{2}\\) to accurately find position:<\/p>\n\n\n

\\[
\nx_f = x_i + v_{ix} t + \\frac{1}{2}a_x t^2
\n\\]<\/p>\n\n\n\n

For this text, we don’t need to worry about where the \\(\\frac{1}{2}\\) comes from; if you continue on in physics you’ll find out soon enough.<\/p>\n\n\n\n

There is one more relationship that is often helpful:<\/p>\n\n\n

\\[
\n v_{fx}^2 = v_{ix}^2 + 2a_x(x_f – x_i)
\n\\]<\/p>\n\n\n\n

These relationships are known as the kinematic equations<\/em> for constant acceleration. They tell you relationships between the different physical quantities that are used to describe an object’s motion (position, velocity, and acceleration). You can sit down and memorize these equations, but you will find that this happens on its own when you work through enough practice problems.<\/p>\n\n\n\n

Note the use of subscripts: there is \\(i\\) for initial, \\(f\\) for final, and \\(x\\) to represent motion along the \\(x\\) axis. If you are dealing with motion in the \\(y\\) direction, simply replace all \\(x\\)s with \\(y\\)s. It seems tedious now, but when we are working with two dimensions it will be extremely important to keep the \\(x\\) and \\(y\\) components of motion distinguished from each other.<\/p>\n\n\n\n

12.1.1 Problem solving steps<\/h3>\n\n\n\n

As with any physics problem, you should always start by drawing a picture. For kinematics problems, we tend to have lots of variables to keep track of\u2014 it is very helpful to get organized. After you have drawn your picture, make a table of what you know and what you don\u2019t know. Then, you can look at the relationships between these variables to see what kinematics equations you need to use to solve the problem.<\/p>\n\n\n\n

To summarize:<\/p>\n\n\n\n

  1. Draw a picture.<\/li>
  2. Organize what you know and what you don’t know.<\/li>
  3. Use the kinematic equations to identify connections between knowns and unknowns. These equations are $$\\begin{align*} x_f &= x_i + v_{ix}t + \\frac{1}{2}a_xt^2 \\tag{12.1} \\\\ v_{fx} &= v_{ix} + a_xt \\tag{12.2} \\\\ v_{fx}^2 &= v_{ix}^2 + 2a_x(x_f – x_i) \\tag{12.3} \\end{align*}$$<\/li>
  4. Solve with appropriate algebraic steps.<\/li><\/ol>\n\n\n\n

    Example 12.1<\/h4>\n\n\n\n

    A car driving down the road accelerates uniformly in a straight line from 25 m\/s to 35 m\/s in 3.5 s. What is the car’s acceleration? How far does the car travel while it accelerates?<\/p>\n\n\n\n

    Known<\/th>Unknown<\/th><\/tr><\/thead>
    \\(x_i = 0\\)<\/td>\\(a_x\\)<\/td><\/tr>
    \\(v_{ix} = 25\\ \\textrm{m\/s}\\)<\/td>\\(x_f\\)<\/td><\/tr>
    \\(v_{fx} = 35\\ \\textrm{m\/s}\\)<\/td><\/td><\/tr>
    \\(t = 3.5\\ \\textrm{s}\\)<\/td><\/td><\/tr><\/tbody><\/table>
    Table of knowns and unknowns<\/figcaption><\/figure>\n\n\n\n

    Using equation (12.2), we can find the acceleration. Then use either other equation to find the distance traveled. I’ll use equation (12.1)<\/p>\n\n\n

    \\[
    \n \\begin{align*}
    \n v_{fx} &= v_{ix} + a_x t & x_f &= x_i + v_{ix} t + \\frac{1}{2}a_xt^2 \\\\
    \n \\hookrightarrow a_x &= \\frac{v_{fx} – v_{ix}}{t} & &= 0 + (25\\ \\textrm{m\/s})(3.5\\ \\textrm{s}) + \\frac{1}{2}\\left(2.9\\ \\textrm{m\/s}^2\\right)(3.5\\ \\textrm{s})^2 \\\\
    \n &= \\frac{(35\\ \\textrm{m\/s}) – (25\\ \\textrm{m\/s})}{3.5\\ \\textrm{s}} & &= 105\\ \\textrm{m} \\\\
    \n &= 2.9\\ \\textrm{m\/s}^2
    \n \\end{align*}
    \n\\]<\/p>\n\n\n\n

    Example 12.2<\/h4>\n\n\n\n

    A speed skater moving across frictionless ice at 8 m\/s hits a 5 m long patch of rough ice. She slows steadily, then continues on at 6 m\/s. What is her acceleration on the rough ice? How much time did it take her to travel across the rough spot?<\/p>\n\n\n\n

    Known<\/th>Unknown<\/th><\/tr><\/thead>
    \\(x_i = 0\\)<\/td>\\(a_x\\)<\/td><\/tr>
    \\(x_f = 5\\ \\textrm{m}\\)<\/td>\\(t\\)<\/td><\/tr>
    \\(v_{ix} = 8\\ \\textrm{m\/s}\\)<\/td><\/td><\/tr>
    \\(v_{fx} = 6\\ \\textrm{m\/s}\\)<\/td><\/td><\/tr><\/tbody><\/table>
    Table of knowns and unknowns<\/figcaption><\/figure>\n\n\n\n

    Use equation (12.3) to determine the acceleration. Use either other equation to find the time. I’ll use equation (12.2).<\/p>\n\n\n

    \\[
    \n \\begin{align*}
    \n v_{f,x}^2 &= v_{i,x}^2 + 2a_x(x_f – x_i) & v_{f,x} &= v_{i,x} + a_x t \\\\
    \n \\hookrightarrow a_x &= \\frac{v_{f,x}^2 – v_{i,x}^2}{2(x_f – 0)} & \\hookrightarrow t &= \\frac{v_{f,x} – v_{i,x}}{a_x} \\\\
    \n &= \\frac{(6\\ \\textrm{m\/s})^2 – (8\\ \\textrm{m\/s})^2}{2(5\\ \\textrm{m})} & &= \\frac{(6\\ \\textrm{m\/s}) – (8\\ \\textrm{m\/s})}{-2.8\\ \\textrm{m\/s}^2} \\\\
    \n &=-2.8\\ \\textrm{m\/s}^2 & &= 0.71\\ \\textrm{s}
    \n \\end{align*}
    \n\\]<\/p>\n\n\n\n

    Example 12.3<\/h4>\n\n\n\n

    A driver slams on the brakes when they see a tree lying across the road. The car slows uniformly with an acceleration of 5.6 m\/s2<\/sup> for 4.2 s, making straight skid marks 62.4 m long, all the way to the tree. With what speed does the car strike the tree?<\/p>\n\n\n\n

    Known<\/th>Unknown<\/th><\/tr><\/thead>
    \\(x_i = 0\\)<\/td>\\(v_{ix}\\)<\/td><\/tr>
    \\(x_f = 62.4\\ \\textrm{m}\\)<\/td>\\(v_{fx}\\)<\/td><\/tr>
    \\(a_x = -5.6\\ \\textrm{m\/s}^2\\)<\/td><\/td><\/tr>
    \\(t = 4.2\\ \\textrm{s}\\)<\/td><\/td><\/tr><\/tbody><\/table>
    Table of knowns and unknowns<\/figcaption><\/figure>\n\n\n\n

    By defining \\(x_i = 0\\) and \\(x_f = +62.4\\ \\textrm{m}\\) we have defined the \\(+x\\) direction as the direction the car is moving. Since the car is slowing down its, acceleration must be in the opposite direction as its motion; this is why the acceleration is in the \\(-x\\) direction.<\/p>\n\n\n\n

    Use equation (12.1) to find the initial speed, then either other equation to find the final speed. I’ll use equation (12.2). Note that “final” here means “just before impact.”<\/p>\n\n\n

    \\[
    \n \\begin{align*}
    \n x_f &=x_i + v_{ix}t + \\frac{1}{2}a_xt^2 & v_{fx} &= v_{ix} + a_xt \\\\
    \n \\hookrightarrow v_{ix} &= \\frac{x_f – \\frac{1}{2}a_xt^2}{t} & &= (26.6\\ \\textrm{m\/s}) + \\left(-5.6\\ \\textrm{m\/s}^2\\right)(4.2\\ \\textrm{s}) \\\\
    \n &= \\frac{x_f}{t} – \\frac{a_xt}{2} & &= 3.1\\ \\textrm{m\/s} \\\\
    \n &= \\frac{62.4\\ \\textrm{m}}{4.2\\ \\textrm{s}} – \\frac{\\left(-5.6\\ \\textrm{m\/s}^2\\right)(4.2\\ \\textrm{s})}{2} \\\\
    \n &= 26.6\\ \\textrm{m\/s}
    \n \\end{align*}
    \n\\]<\/p>\n\n\n\n

    Practice 12.1<\/h4>\n\n\n\n\n