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Learning Introductory Physics with Activities

Section 3.7 Solving Symbolic Equations

Physics uses symbols to represent physical quantities that describe some aspect of the real world. For example, the physical quantity time is typically represented by the letter \(t\text{.}\) Symbols are extremely versatile: the letter \(t\) is used to represent any specific instant in time, such as \(t = 3\) s, or a generic, unspecified instant, time \(t\text{.}\) Different times are further specified by the use of subscripts, such as \(t_i\) for initial time, \(t_f\) for final time, or \(t_4\) for the fourth in a series of related times. Every symbol also carries its own appropriate units! For example, you would not want to write “I turned on the machine at \(t\) seconds”; instead, you would write “I turned on the machine at time \(t\text{.}\)
Physics uses symbolic equations to represent quantitative models of the real world. An example equation is shown below.
\begin{equation*} \Delta x = v_{x,ave}\Delta t \end{equation*}
In this equation, t represents time, x represents the position of an object along the x-axis, and \(v_{x,ave}\) represents the average velocity of the object in the x-direction. The symbol \(\Delta\) transforms both \(x\) and \(t\) to specify change in position and change in time, respectively.
Suppose you know both \(\Delta x\) and \(v_{x,ave}\text{;}\) determine an equation for the amount of time required to travel this distance.
Since \(\Delta t\) is multiplied by \(v_{x,ave}\) on the right hand side of the equation, you can divide both sides by \(v_{x,ave}\) to get \(\Delta t\) alone!
\begin{equation*} \Delta x = v_{x,ave}\Delta t \end{equation*}
\begin{equation*} \frac{\Delta x}{v_{x,ave}} = \frac{v_{x,ave}\Delta t}{v_{x,ave}} \end{equation*}
\begin{equation*} \frac{\Delta x}{v_{x,ave}} = \Delta t \end{equation*}
\begin{equation*} \Delta t = \frac{\Delta x}{v_{x,ave}} \end{equation*}
Physics is full of symbolic equations, and one context might often have several known and unknown quantities and many symbolic equations that relate those quantities. Thus, you will often be asked not only to solve symbolic equations, but to solve systems of symbolic equations.
From elementary chemistry, the pressure \(p\text{,}\) volume \(V\text{,}\) and temperature \(T\) of an ideal gas undergoing adiabatic expansion may be related by the following pair of equations, in which \(n\text{,}\) \(R\text{,}\) \(\gamma\text{,}\) and \(\alpha\) are positive constants:
\begin{equation*} pV = nRT \end{equation*}
\begin{equation*} pV^\gamma = \alpha \end{equation*}
Suppose you want to know the exact temperature of the gas at the end of the expansion. There are only two equations and three unknowns, so it is necessary to specify a temperature: suppose you know the volume at the end of the expansion is \(V_f\text{.}\) Since you want to know temperature, you want to solve one of the equations for the other unknown, pressure, then substitute that expression for pressuring into the other equation. The second equation looks a little more complicated, so start there.
\begin{equation*} pV^\gamma = \alpha \end{equation*}
\begin{equation*} p = \frac{\alpha}{V^\gamma} \end{equation*}
\begin{equation*} pV = nRT \end{equation*}
\begin{equation*} nRT = \frac{\alpha}{V^\gamma}V \end{equation*}
\begin{equation*} nRT = \alpha V^{1 - \gamma} \end{equation*}
\begin{equation*} T = \frac{\alpha V^{1 - \gamma}}{nR} \end{equation*}
At this point, if you also want to know the pressure, take this expression for temperature and plug it back into one of the original equations!
\begin{equation*} pV = nRT \end{equation*}
\begin{equation*} pV = nR\frac{\alpha V^{1 - \gamma}}{nR} \end{equation*}
\begin{equation*} pV = \alpha V^{1 - \gamma} \end{equation*}
\begin{equation*} p = \alpha V^{- \gamma} \end{equation*}
There is one special exception to solving equations without plugging in numbers: the number zero. When a quantity is known to be zero, it is often the result of some special, fundamental property of the situation; thus, you are unlikely to want to go back and change the number away from zero (if you do, you often make the situation more complicated). So if you have a quantity you know is zero, feel free to plug it into any equations early!

Exercises Practice Activities

Figure 3.7.6. A velocity vector and its \(x\)-component.

Activity 3.7.1. Velocity Manipulation.

The diagram above shows a velocity vector and its \(x\)-component, \(v_x\text{.}\)

(a) Vector Components.

Write a symbolic equation that gives \(v_x\) in terms of \(v\) (the magnitude of \(\vec{v}\)) and the given angle \(\theta\text{.}\)
Answer.
\(v_x = v \cos{\theta}\)

(b) Solving for v.

Rewrite your symbolic equation for \(v\) instead.
Answer.
\(v = \frac{v_x}{\cos{\theta}}\)

(c) Solving for theta.

Rewrite your symbolic equation for \(\theta\) instead.
Answer.
\(\theta = \arccos{\frac{v_x}{v}}\)

Activity 3.7.2. The Air Cart.

You have an air cart with mass \(m\) that is subject to two constant forces: thrust and friction. You conduct two experiments with the air cart. In Experiment 1, the thrust force points forward and the friction force points backward, leading to a forward acceleration of \(a_1\text{.}\) In Experiment 2, the thrust and friction forces both point backward, leading to a backward acceleration of \(a_2\text{.}\)

(a) Unknown Quantities.

Identify the unknown quantities of interest and give them an appropriate symbol.
Answer.
The magnitude of the thrust force \(F^t\) and the magnitude of the friction force \(F^f\text{.}\)

(b) Representation.

Sketch a free-body diagram for the air cart during each experiment.
Answer.
Figure 3.7.7. Free-body diagrams of the air cart.

(c) The Law of Motion.

Use your free-body diagrams to write equations relating the forces and the accelerations.
Answer.
\begin{equation*} \text{1:}~ F^t - F^f = ma_1 \end{equation*}
\begin{equation*} \text{2:}~ - F^t - F^f = -ma_2 \end{equation*}

(d) Calculation.

Solve your equations for the unknown quantities.