The Apprentice’s Guide to Physics: An Introductory Quest Through the Mechanics of the Universe

Consider the follwoing equation:

If \(v\text{,}\) \(v_0\text{,}\) \(a\) are known quantities, and \(t\) is the unknown quantitiy, then this is a one equation, one unknown 6.1.2 problem. By rearranging the equation, we can isolate (i.e. solve for) \(t\text{.}\)

First subtract \(v_0\) from both sides of the equation:

which results in:

Note that in the above step we "subtracted \(v_0\) from both sides". Another commom way to say this is: "subtract \(v_0\) to the other side", since the end result is "moving" \(v_0\) from one side to the other.

Then divide by sides by \(a\text{:}\)

which results in:

Now that \(t\) is all by itself, we are done. However, folks seem to like our unknown quantities that have been solved for to be on the left side of the quation instead of the right side. So if you want, you could also write the above solution

Consider the follwoing equation:

If \(x_0\text{,}\) \(v\text{,}\) \(a\) are known quantities, and \(t\) is the unknown quantitiy, then this is a one equation, one unknown 6.1.2 problem. More specifically, this is a quadratic equation 6.1.5 problem. We can use the quadratic formula 6.1.6 to isolate (i.e. solve for) \(t\text{.}\)

Note that the above equation is not in the standard form of a quadratic equation, so we will need to rearrange it before we can apply the quadratic formula.

First add \(x_0\) from both sides of the equation:

which results in:

Note that in the above step we "subtracted \(x_0\) from both sides". Another commom way to say this is: "subtract \(x_0\) to the other side", since the end result is "moving" \(x_0\) from one side to the other.

Consider the standard form of a quadratic equation: \(ax^2 + bx + c = 0 \text{.}\) While we don’t have to rewrite our equation in the standard form to use the quadratic formula, it can be helpful to identify the values of \(a\text{,}\) \(b\text{,}\) and \(c\) by comparing our equation to the standard form. Doing this gives:\(\frac{1}{2}at^2 + vt + x_0 = 0\)

Now it is a bit easier to directly identify the values of \(a\text{,}\) \(b\text{,}\) and \(c\text{:}\)

Now we can use the quadratic formula to solve for \(t\text{.}\) The quadratic formula is: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \text{.}\) Substituting in our values of \(a\text{,}\) \(b\text{,}\) and \(c\) gives:

which simplifies to:

The above solution gives two possible values for \(t\text{,}\) one using the plus sign, and one using the minus sign. So the above equation for \(t\) is the answer. However, to be more explicit in this example, we can write out the two possible values for \(t\) as:

Notice that the solution written as \(t = \frac{-v \pm \sqrt{v^2 - 2ax_0}}{a}\) does not use subscripts. In this case, the \(\pm\) symbol already indicates that two possible values for \(t\) exist: one obtained using the plus sign and one obtained using the minus sign.

When these two solutions are written as separate equations, it becomes useful to distinguish between them using subscripts: \(t_1\) and \(t_2\text{.}\) This is a common convention in mathematics and science. In general, unique quantities should be represented by unique symbols. Using the same symbol for multiple values can create ambiguity and make equations more difficult to interpret. By writing the solutions as \(t_1\) and \(t_2\text{,}\) we clearly indicate that they represent two distinct values of time. Also note that "1" and "2" subscritps are arbitrary labels, you could have used \(t_3\) and \(t_4\text{,}\) or \(t_{\text{☺}}\) and \(t_{\text{☹}}\) as long as you are consistent in your labeling.

Consider the following two equations:

These equations form a System of Equations 6.1.9 because they share the quantity \(v\text{.}\)

Let’s let \(v_0\text{,}\) \(a\text{,}\) \(t\text{,}\) \(x_0\) be known quantities, and \(x\text{,}\) \(v\) be the unknown quantities. Note that with this information, we have two independent equations and two unknown quantities, so we can solve for both \(x\) and \(v\text{.}\)

Suppose we would like to solve for \(x\) such that it does not contain the quantity \(v\text{.}\) We can use Substitution 6.1.11.

The first equation tells us that:

Therefore, wherever we see \(v\) in the second equation, we can replace it with \((v_0 + at)\text{:}\)

Distributing \(t\) gives:

The two original equations have now been combined into a single equation that no longer contains the quantity \(v\text{.}\)

Consider the equation:

Suppose \(m\) remains constant while \(a\) is doubled. We can use Covariational Reasoning 6.1.15 to determine how \(F\) changes.

Since \(m\) remains constant, we can focus only on the relationship between \(F\) and \(a\text{:}\)

Next, write down how the changing quantity is changing:

This equation tells us that the final \(a\) is twice the initial \(a\text{.}\)

Recall that our proportional relationship is:

Since \(a_f = 2a_i\text{,}\) the quantity \(a\) changes by a factor of \(2\text{.}\) In the proportional relationship, we can think of replacing \(a\) with this factor:

which tells us that the force must also change by a factor of \(2\text{.}\) Therefore we can write:

or,

Therefore, doubling \(a\) results in doubling \(F\text{.}\)