PHYS 170

Lecture 02b - Motion in One Dimension (The Equations of Motion)

Introduction

Four equations for motion under constant acceleration:

Linear Equations Missing
\(v = v_0 + at \) \(s\)
\(s - s_0 = v_0t + \frac{1}{2} a t^2 \) \(v\)
\(s - s_0 = \frac{1}{2} (v_0+v)t \) \(a\)
\(v^2 = v_0^2 + 2a(s-s_0) \) \(t\)

There are six variables overall: \(s_0, v_0 \) (intial variables), and \(s, v, a, t\) (final variables).
Each of the equations is labeled on the right by the final variables that is missing. For example, the first equation has the final variables \(v, a, t\) but is missing \(s\), so we will call it the \(s\)-equation.

The meaning of the variables:

Symbol Meaning
\(s_0\) initial displacement
\( v_0 \) initial velocity
\(s\) final displacement
\( v \) final velocity
\(a\) acceleration (constant)
\( t \) time

If the acceleration is not a constant (i.e. it changes over time) then the equations are invalid. However, in this course, you can assume \(a\) is constant unless the question stated otherwise.

Simple Applications

The information given:
  • \(s_0=0m\)
  • \(v_0=1.2m/s\)
  • \(s=?\)
  • \(v=3m/s\)
  • \(a=\) missing
  • \(t=10s\)
Note that "\(s=?\)" above means we are looking for \(s\). Out of the final variables \(s, v, a, t\), only \(a\) does not appear in the list above. This means we should use the \(a\)-equation: $$ \begin{eqnarray} s-s_0 &=& \frac{1}{2}(v_0+v)t \\ \Rightarrow s &=& \frac{1}{2}(1.2+3)(10) = 21m \end{eqnarray} $$
Note that since we want the plane to stop in the end, we know \(v=0m/s\) even though the information is not explicitly given in the question. Since the plane is slowing down, its acceleration is actually negative:
  • \(s_0=0m\)
  • \(v_0=75m/s\)
  • \(s=?\)
  • \(v=0m/s\)
  • \(a=-4.5m/s^2\)
  • \(t=\) missing
There is no \(t\) above so we should use the \(t\)-equation: $$ \begin{eqnarray} v^2 &=& v_0^2 + 2a(s-s_0) \\ \Rightarrow s &=& s_0 + \frac{v^2 - v_0^2}{2a} = 625m \end{eqnarray} $$

Try It Yourself (click to show)

  • \(s_0=0m\)
  • \(v_0=1.2m/s\)
  • \(v=2.7m/s\)
  • \(s=-100m\)
  • \(a=?\)
\(t\) is missing, so you should use the \(t\)-equation.
  • \(s_0=0m\)
  • \(v_0=1.2m/s\)
  • \(s=10m\)
  • \(v=2.7m/s\)
  • \(a=?\)
  • \(t=\) missing
From the \(t\)-equation: $$ \begin{eqnarray} v^2 &=& v_0^2 + 2a(s-s_0) \\ \Rightarrow a &=& \frac{v^2-v_0^2}{2(s-s_0)} = \frac{2.7^2-1.2^2}{2(10)} = 0.29m/s^2 \end{eqnarray} $$
  • \(s_0=0m\)
  • \(v_0=30m/s\)
  • \(s=50m\)
  • \(a=-8m/s^2\)
  • \(v=?\)
\(t\) is missing. From the \(t\)-equation: $$ \begin{eqnarray} v^2 &=& v_0^2 + 2a(s-s_0) \\ \Rightarrow v &=& \sqrt{v_0^2+2a(s-s_0)} = \sqrt{30^2+2(-8)(50)} = 10m/s \end{eqnarray} $$
  • \(s_0=0m\)
  • \(v_0=30m/s\)
  • \(s=50m\)
  • \(a=-8m/s^2\)
  • \(v=\) missing
  • \(t=?\)
\(v\) is missing. From the \(v\)-equation: $$ \begin{eqnarray} s - s_0 &=& v_0 t + \frac{1}{2}at^2 \\ \Rightarrow 50 &=& 30t -4t^2 \\ \Rightarrow 4t^2-30t+50 &=& 0 \\ \Rightarrow 2t^2-15t+25 &=& 0 \\ \Rightarrow t &=& \frac{15\pm \sqrt{15^2-4(2)(25)}}{2(2)} = 2.5s \text{ or } 5s \text{(rejected)} \end{eqnarray} $$ We reject the second solution (\(t=5s\)) to the quadratic equation because the equation assumes constant negative acceleration. When taken literally, means the roller coaster will continue to decelerate even after it comes out of the tunnel at \(2.5s\). In this case, the roller coaster will eventually reverse direction and move backward, and finally arriving at the exit of the tunnel for a second time (in the reversed direction) at \(5s\). In real life, of course the roller coaster will at some point change its acceleration (perhaps to go at constant velocity after it comes out of the tunnel), therefore the mathematical assumption that \(a\) is constant forever is not really valid. As a result the second solution for the return of the roller coaster should be ignored.
Recall that we found \(v=10m/s\) in the Try It Yourself problem "roller coaster in a tunnel 3", putting all the information together:
  • \(s_0=0m\)
  • \(v_0=30m/s\)
  • \(s=50m\)
  • \(a=-8m/s^2\)
  • \(v=10m/s\)
  • \(t=?\)
With this extra information, there is no variable missing from the list above. When no variable is missing, it means you have more information than you need to solve the problem. You can then use any of the four equations of motion that contains \(t\) to solve it. The shortest equation, namely the \(s\)-equation, will do nicely: $$ \begin{eqnarray} v &=& v_0 + at \\ \Rightarrow t &=& \frac{v-v_0}{a} \\ &=& \frac{10-30}{-8} = 2.5s \end{eqnarray} $$ This alternative approach gives the same answer as before. While this method is faster, it assummes (dangerously) that your own calculation of \(v=10m/s\) is correct. In the exam, it is perhaps unwise to make such a bet. I personally prefer the quadratic method despite it being more tedious because that way I could have gotten \(v\) wrong and yet I would still calculate the time correctly.

Acceleration due to Gravity

Gravity pulls everything downward in such a way that the vertical acceleration of an object falling on the surface of the Earth is given by: $$ a = -9.8m/s^2 $$ The negative sign denotes the fact that gravity always ponts downward. The value of \(a\) does not depend on the mass.

The magnitude of the acceleration due to gravity is defined to be \(g\): $$ g = |a| = 9.8m/s^2 $$ Note that \(g\) is by definition positive (because it is the magnitude), so never write something like \(g=-9.8m/s^2\). Be aware of the difference between the notations of \(a\) and \(g\), the former is negative and the latter is negative:
  • \(a = - g= -9.8m/s^2\)
  • \(g=|a|=9.8m/s^2\)
  • \(a = 9.8m/s^2\)
  • \(g=-9.8m/s^2\)
Below is a picture of an apple and a piece of feather under free fall inside a vacuum chamber (to eliminate the air resistance). Even though the feather is much lighter than the apple, they both reach the ground at exactly the same time because their downward acceleration is exactly the same.
two simple vectors

Video of Free Fall in Air and in Vacuum (click to show)

  • \(s_0=0m\)
  • \(v_0=?\)
  • \(s=-100m\)
  • \(v=\) missing
  • \(a=-g=-9.8m/s^2\)
  • \(t=4s\)
Note that the displacement \(s\) is negative because you want the ball to fall down, not up! Since \(v\) is missing, we use the \(v\)-equation: $$ \begin{eqnarray} s - s_0 &=& v_0t + \frac{1}{2}at^2 \\ \Rightarrow v_0 &=& \frac{s-s_0}{t} - \frac{1}{2}at \\ &=& \frac{-100}{4} - \frac{1}{2} (-9.8)(4) = -5.4m/s \end{eqnarray} $$ The negative sign means the initial velocity should be pointing downward.
  • \(s_0=0m\)
  • \(v_0=-5.4m/s\)
  • \(s=-100m\)
  • \(v=?\)
  • \(a=-g=-9.8m/s^2\)
  • \(t=4s\)
None of the variables are missing, so we can use any of the equations that contains \(v\). We choose the shortest one, namely the \(s\)-equation: $$ v = v_0 + at = -5.4 + (-9.8) (4) = -44.6m/s $$ The negative sign means the final velocity points downward, as expected.

Try It Yourself (click to show)

  • \(s_0=0m\)
  • \(v_0=-1m/s\)
  • \(s=\) missing
  • \(a=-g=-9.8m/s^2\)
  • \(v=-10m/s\)
  • \(t=?\)
\(s\) is missing, so we choose the \(s\)-equation: $$ \begin{eqnarray} v &=& v_0 +at \\ \Rightarrow t &=& \frac{v-v_0}{a} \\ &=& \frac{(-10)-(-1)}{-9.8} = 0.92s \end{eqnarray} $$

In the exam, if you miss any of the negative signs in the calculation, you will lost most of the points of that question even if you have done everything else correctly (and even if you have written down the right answer!). The signs here are of paramount importance and must be handled very carefully!

  • \(s_0=0m\)
  • \(v_0=+1m/s\)
  • \(s=\) missing
  • \(v=-10m/s\)
  • \(a=-g=-9.8m/s^2\)
  • \(t=?\)
\(s\) is missing, so we choose the \(s\)-equation: $$ \begin{eqnarray} v &=& v_0 +at \\ \Rightarrow t &=& \frac{v-v_0}{a} \\ &=& \frac{(-10)-1}{-9.8} = 1.12s \end{eqnarray} $$

In the exam, if you miss any of the negative signs in the calculation, you will lost most of the points of that question even if you have done everything else correctly (and even if you have written down the right answer!). The signs here are of paramount importance and must be handled very carefully!

  • \(s_0=0m\)
  • \(v_0=1m/s\)
  • \(s=?\)
  • \(v=-10m/s\)
  • \(a=-g=-9.8m/s^2\)
  • \(t=\) missing
Even though you calculated \(t\) from the last problem, it is a good idea to not use it because you might have made a mistake in your calculation of \(t\). Without using \(t\) from the last problem, we treat \(t\) as missing, so we choose the \(t\)-equation: $$ \begin{eqnarray} v^2 &=& v_0^2 + 2 a (s-s_0) \\ \Rightarrow s &=& \frac{v^2-v_0^2}{2a} \\ &=& \frac{(-10)^2-1^2}{2(-9.8)} = -5.05m \end{eqnarray} $$ The negative sign means the ball moved downward by \(5.05m\). This implies that the height of the cliff is \(5.05m\). While the displacement of the ball is negative, you should not give a negative answer as the height of the cliff (how could a cliff be negatively tall?).

In the exam, if you miss any of the negative signs in the calculation, you will lost most of the points of that question even if you have done everything else correctly (and even if you have written down the right answer!). The signs here are of paramount importance and must be handled very carefully!