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

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.

The information given:

- \(s_0=0m\)
- \(v_0=1.2m/s\)
- \(s=?\)
- \(v=3m/s\)
- \(a=\) missing
- \(t=10s\)

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

- \(s_0=0m\)
- \(v_0=1.2m/s\)
- \(v=2.7m/s\)
- \(s=-100m\)
- \(a=?\)

- \(s_0=0m\)
- \(v_0=1.2m/s\)
- \(s=10m\)
- \(v=2.7m/s\)
- \(a=?\)
- \(t=\) missing

- \(s_0=0m\)
- \(v_0=30m/s\)
- \(s=50m\)
- \(a=-8m/s^2\)
- \(v=?\)

- \(s_0=0m\)
- \(v_0=30m/s\)
- \(s=50m\)
- \(a=-8m/s^2\)
- \(v=\) missing
- \(t=?\)

Recall that we found \(v=10m/s\) in the Try It Yourself problem "roller coaster in a tunnel 3", putting all the information together:
**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.

- \(s_0=0m\)
- \(v_0=30m/s\)
- \(s=50m\)
- \(a=-8m/s^2\)
- \(v=10m/s\)
- \(t=?\)

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.

- \(s_0=0m\)
- \(v_0=?\)
- \(s=-100m\)
- \(v=\) missing
- \(a=-g=-9.8m/s^2\)
- \(t=4s\)

- \(s_0=0m\)
- \(v_0=-5.4m/s\)
- \(s=-100m\)
- \(v=?\)
- \(a=-g=-9.8m/s^2\)
- \(t=4s\)

- \(s_0=0m\)
- \(v_0=-1m/s\)
- \(s=\) missing
- \(a=-g=-9.8m/s^2\)
- \(v=-10m/s\)
- \(t=?\)

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
- \(v=-10m/s\)
- \(a=-g=-9.8m/s^2\)
- \(t=?\)

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

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!