A body moving with uniform acceleration has a velocity of $\,12.0\,\frac{\mathrm{cm}}{\mathrm{s}}$ in the positive $x$ direction when its $x$ coordinate is $\,3.00\,\mathrm{cm}.$ If its $x$ coordinate $\,2.00\,\mathrm{s}$ later is $\,-5.00\,\mathrm{cm},$ what is the magnitude of its acceleration?

The given problem (Serway and Beichner 52, problem 25) is ambiguous and has infinite many solutions depending on the time when the initial coordinate of $3.00\,\mathrm{cm}$ occurs. In particular, there is a different solution (acceleration) for each value of $t_f$ and $t_i$ that satisfies $t_f-t_i=\Delta t=2.00\,\text{s}.$

For example, two distinct accelerations satisfying the conditions of the problem are obtained on the interval $t_i=0$ to $t_f=2,$ and on $t_i=1$ and $t_f=3,$ respectively, even though in both cases $\Delta t=2.$

In the first case, integrating $a=\frac{d^2x}{dt^2}$ for constant $a$ and then solving for $a$ by substitution of known values yields $a=-16\frac{\mathrm{cm}}{\mathrm{s}^2}$ and a position function of \begin{equation} x=3+12t-8t^2 \end{equation} To verify that $a=-16\,\frac{\mathrm{cm}}{\mathrm{s}^2}$ used in $\left(1\right)$ is a solution, let $t_i=0$ and $t_f=2$ in $\left(1\right).$ The conditions $\Delta t=2,$ $x\left(t_i\right)=3$ and $x\left(t_f\right)=-5$ are then satisfied. Therefore, $a=-16\,\frac{\mathrm{cm}}{\mathrm{s}^2}$ is a solution.

In the second case, integration yields $a=-8\,\frac{\mathrm{cm}}{\mathrm{s}^2}$ and a position of \begin{equation} x=-5+12t-4t^2 \end{equation} To verify that $a=-8\,\frac{\mathrm{cm}}{\mathrm{s}^2}$ used in $\left(2\right)$ is a solution, let $t_i=1$ and $t_f=3$ in $\left(2\right).$ Then the conditions $\Delta t=2, x\left(t_i\right)=3$ and $x\left(t_f\right)=-5$ are again satisfied. Therefore, $a=-8\,\frac{\mathrm{cm}}{\mathrm{s}^2}$ is a solution.

Since $-16\neq-8,$ the conclusion is that $\Delta t$ does not uniquely determine displacement (or acceleration), it depends on both $t_i$ and $t_f.$ In solving the problem in the solutions manual, the authors used the kinematic equation \begin{equation} x = x_i + v_i t + 12at^2 \end{equation} under the assumption that $t=\Delta t=2,$ thus assuming that $t_i=0.$ Since other solutions arise if this assumption is not observed, it must be stated as one of the conditions of the problem. In general, this problem arises from an indiscriminate use of $\left(3\right).$ This equation is derived using calculus under the assumption that $x_i=x\left(t_i\right)=x\left(0\right),$ that is, $t_i=0.$ Without this assumption, the kinematic equation derived from calculus would be: \begin{equation} x=x_i+v_i\Delta t+\frac{1}{2}a\left(t^2-t_i^2\right) \end{equation} This equation makes it evident that one cannot simply substitute $t=\Delta t$ into $\left(3\right)$ to obtain \begin{equation} x=x_i+v_i\Delta t+\frac{1}{2}a(\Delta t)^2 \end{equation} One can do this only if $t_i=0.$ For instance, if one wants to know the displacement that a particle makes between the 3rd and 5th seconds of its journey, it is incorrect to let $t=\Delta t=5-3=2$ and conclude from the kinematic equation that the displacement is $x-x_i=2v_i+2a,$ because here $t_i=3\neq0.$ It is incorrect because $x_i$ was assumed to occur at $t_i=0,$ not at $t_i=3.$ It is tempting to make this mistake, to assume that since $\Delta t=2-0=5-3=2,$ then we can just let $t=\Delta t.$ That the answer is wrong is obvious when the displacement is found using the following sure method using the kinematic equation: \begin{align*} x\left(5\right) &=x_i+5v_i+12.5a\\ x\left(3\right) &=x_i+3v_i+4.5a\\ x-x_i &=x\left(5\right)-x\left(3\right)\\ &=2v_i+8a \end{align*} Obviously, $2v_i+2a\neq2v_i+8a.$ This second method is a correct use of the kinematic equation to solve the problem. The conclusion is that when using this equation, the choice to let $t=\Delta t$ should not be used indiscriminately. A comparison of the kinematic equations $\left(3\right)$ and $\left(4\right)$ should convince one of this. Deriving each one from calculus makes it even more clear. As a final geometric persuasion, consider a particle whose $x\text{-}t$ position graph is a parabola with vertex at the origin. Then its acceleration is constant and nonzero, and its $v\text{-}t$ velocity graph must be a non-horizontal straight line that also passes through the origin. Since the particle’s position function is an antiderivative of its velocity, then the area under the curve of the velocity graph on a given time interval determines the particles displacement on that interval. Yet, since the graph is a non-horizontal straight line passing through the origin, then for any two interval’s such that $t_{f_1}-t_{i_1}=t_{f_2}-t_{i_2},$ $\left(\Delta t_1=\Delta t_2\right),$ and $t_{i_1}\lt t_{f_1}\lt t_{i_2}\lt t_{f_2},$ it is geometrically obvious that the areas satisfy $\Delta x_1\Delta t_1 \lt \Delta x_2\Delta t_2.$ This says that displacements on equal time intervals are unequal if the intervals have different initial conditions. On the other hand, if acceleration is $0,$ then $t=\Delta t$ is always a safe assumption, regardless of the initial conditions because then the $v\mathrm{-}t$ graph is a horizontal line, making areas under it (displacements) on equal time intervals, equal. \[\tag*{$\blacksquare$}\]

Constant Acceleration

Two major cases of constant acceleration are considered in chapter 2 of the text. Problems and examples from the text, illustrating both cases are given below.

  • Case 1. Constant Velocity, Zero Acceleration.
  • Case 2. Linear Velocity, Nonzero Acceleration.
Problem 7 (Case 1)
\[ t\left(s\right) \] - 1.00 6.00
\[ x\left(m\right) \] \[ x\left(t\right)=\left(1.6t-4.6\right)m \] -3.00 5.00
\[ v_x\left(\frac{\mathrm{m}}{\mathrm{s}}\right) \] \[ v_x\left(t\right) ={\bar{v}}_x\left(1.00,6.00\right) =1.60\frac{\mathrm{m} }{\mathrm{s}} \]
\[ a_x\left(\frac{\mathrm{m}}{\mathrm{s}^2}\right) \] \[ a_x\left(t\right)=0\frac{\mathrm{m} }{\mathrm{s}^2} \]
Example 2.2 Page 29 (Case 2)
\[ x\left(m\right) \] \[ x\left(t\right)=-4t+2t^2 \] 0 -2 0 6
\[ t(s) \] - 0 1 2 3
\[ v_x\left(\frac{\mathrm{m}}{\mathrm{s}}\right) \] \[ v_x\left(t\right)=-4+4t \]
\[ a_x\left(\frac{\mathrm{m}}{\mathrm{s}^2}\right) \] \[ a_x\left(t\right)=4 \]
Problem 1, 10
\[ t(s) \] - 0 1.0 2.0 3.0 4.0 5.0
\[ x(m) \] \[ x\left(t\right)=2.3t^2 \]
\[ v_x\left(\frac{\mathrm{m}}{\mathrm{s}}\right) \] \[ v_x\left(t\right)=4.6t \] 0 2.3 9.2 20.7 36.8 57.5
\[ a_x\left(\frac{\mathrm{m}}{\mathrm{s}^2}\right) \] \[ a_x\left(t\right)=4.6 \]


First Proof of Equations 2.8 and 2.9 (Serway 35-36)

According to the discussion of equations 2.8 and 2.9 (Serway 35-36), the following is a theorem: \begin{equation} \left. \begin{split} v_f = v_i+at \\ v = v_f = v\left(t_f\right) \\ v_i = v(t_i) \\ t_i = 0 \\ t = t_f \end{split} \right\} \Rightarrow \bar{v} =\frac{1}{2} \left(v_i+v_f\right) =\frac{v_i+v_f}{2} \label{eq_ser3536} \end{equation} However, since the definition given in calculus (Finney 346) of the average value of a function is \begin{equation} \bar{v}=\frac{1}{t_f-t_i}\int_{t_i}^{t_f} v(t)\,dt \end{equation} then combining this definition with the conditions of the hypothesis in \eqref{eq_ser3536} we obtain: \begin{equation} \left. \begin{split} vf =v_i+at\\ v =v_f=v(t_f)\\ v_i =v(t_i)\\ t_i =0\\ t =t_f \end{split} \right\} \Rightarrow \bar{v} =\frac{1}{2}\left(v_i+v_f\right) =\frac{1}{t_f-t_i}\int_{t_i}^{t_f} v(t)\,dt \label{eq_gue3536} \end{equation} Two equivalent forms of this theorem follow: \begin{equation} \left. \begin{split} v_f =v_i+a\Delta t\\ v_f =v(t_f)\\ v_i =v(t_i)\\ \Delta t =tf-ti \end{split} \right\} \Rightarrow \bar{v} =\frac{1}{2}\left(v_i+v_f\right) =\frac{1}{t_f-t_i}\int_{t_i}^{t_f} v(t)\,dt \label{eq_gue3536_1} \end{equation} \begin{equation} \left. \begin{split} v=v_0+at\\ v_f=v(t_f)\\ v_i=v(t_i) \end{split} \right\} \Rightarrow \bar{v} =\frac{1}{2}\left(v_i+v_f\right) =\frac{1}{t_f-t_i}\int_{t_i}^{t_f}vt\,dt \label{eq_gue3536_2} \end{equation}

The intended meaning of \eqref{eq_gue3536} is that if velocity $v$ is a linear function of time $t$ with constant acceleration $a,$ then the average of $v$ on the interval $\left(0,t\right)$ is equal to the arithmetic mean of the final and initial velocities, where the initial velocity occurs at time $t=0$ and the final velocity occurs at time $t=t_f.$

Notice that \eqref{eq_gue3536} requires the initial velocity to be the $v\text{-intercept},$ i.e. to occur at time $t=0.$ By letting the $v\text{-intercept}$ be the new term $v_0,$ theorem \eqref{eq_gue3536_1} says that if velocity $v$ is a linear function of time $t$ with constant acceleration $a,$ then the average of $v$ on any arbitrary interval $\left(t_i,t_f\right)$ is equal to the arithmetic mean of the final and initial velocities on that interval. Note that forms \eqref{eq_gue3536_1} and \eqref{eq_gue3536_2} are equivalent, but \eqref{eq_gue3536} is not equivalent to either. In particular, theorems \eqref{eq_gue3536_1} and \eqref{eq_gue3536_2} are more general and theorem \eqref{eq_gue3536} can be derived from either. Since they are equivalent, we will refer to forms \eqref{eq_gue3536_1} and \eqref{eq_gue3536_2} as “the theorem”.

The theorem says something important not only about velocity but about averages in general. For that reason, this discussion more properly belongs in the domain of calculus than of physics. Specifically, if we simply interpret $t$ as an independent variable, $v$ as a variable dependent on $t,$ and $a$ as a constant, then the theorem says that the average of a linear function $v$ of $t$ on an interval $\left(t_f,t_i\right)$ is the arithmetic mean of that interval. This is not generally the case, i.e. the mean and the average of a function are not generally the same.

We will prove forms \eqref{eq_gue3536_1} and \eqref{eq_gue3536_2} by proving theorem \eqref{eq_gue3536_2} and the equivalence of theorem \eqref{eq_gue3536_1} with it. The proof of theorem \eqref{eq_gue3536} from either theorem \eqref{eq_gue3536_1} or \eqref{eq_gue3536_2} is left as an exercise. \[\tag*{$\blacksquare$}\]

Second Proof of Equations 2.8 and 2.9

In giving these proofs, I follow the fitch style (Bergmann 146).
Theorem. \begin{equation*} \left. \begin{split} v=v_0+at\\ v_f=v(t_f)\\ v_i=v(t_i) \end{split} \right\} \Rightarrow \bar{v} =\frac{1}{2}\left(v_i+v_f\right) =\frac{1}{t_f-t_i}\int_{t_i}^{t_f}vt\,dt \tag{\eqref{eq_gue3536_2}} \end{equation*} Proof.
1 \( v=v_0+at\\ v_f=v\left(t_f\right)\\ v_i=v\left(t_i\right) \) A
2 \( \Delta t=t_f-t_i \) A
3 \( \bar{v}=\frac{1}{t_f-t_i}\int_{t_i}^{t_f}v\left(t\right)dt \) definition
4 \( =\frac{1}{\Delta t}\int_{t_i}^{t_f}\left(v_0+at\right)dt \) $=\text{E}$ 1, 2, 3
5 \( =\frac{1}{\Delta t}\left[\left(v_0t+\frac{1}{2}at^2\right)\right]_{t_i}^{t_f} \) integrate
6 \( =\frac{1}{\Delta t}\left[\left(v_0t_f+\frac{1}{2}at_f^2\right)-\left(v_0t_i+\frac{1}{2}at_i^2\right)\right] \) integrate
7 \( =\frac{1}{\Delta t}\left[v_0t_f+\frac{1}{2}at_f^2-v_0t_i-\frac{1}{2}at_i^2\right] \) distribute
8 \( =\frac{1}{\Delta t}\left[v_0t_f-v_0t_i+\frac{1}{2}at_f^2-\frac{1}{2}at_i^2\right] \) commute
9 \( =\frac{1}{\Delta t}\left[v_0\left(t_f-t_i\right)+\frac{1}{2}a\left(t_f^2-t_i^2\right)\right] \) factor
10 \( =\frac{1}{\Delta t}\left[v_0\Delta t+\frac{1}{2}a\left(t_f-t_i\right)\left(t_f+t_i\right)\right] \) $=\text{E}$ 2, 9, factor
11 \( =\frac{1}{\Delta t}\left[v_0\Delta t+\frac{1}{2}a\Delta t\left(t_f+t_i\right)\right] \) $=\text{E}$ 2,10
12 \( =\frac{\Delta t}{\Delta t}\left[v_0+\frac{1}{2}a\left(t_f+t_i\right)\right] \) factor
13 \( =v_0+\frac{1}{2}a\left(t_f+t_i\right) \) $x/x=1$
14 \( =v_f-at_f+\frac{1}{2}a\left(t_f+t_i\right) \) $\text{Lemma}~\eqref{eq_lemma1}$
15 \( =v_f-at_f+\frac{1}{2}at_f+\frac{1}{2}at_i \) distribute
16 \( {=v}_f-\frac{1}{2}at_f+\frac{1}{2}at_i \) combine
17 \( =v_f-\frac{1}{2}a\left(t_f-t_i\right) \) factor
18 \( =v_f-\frac{1}{2}\left(v_f-v_i\right) \) $\text{Lemma}~\eqref{eq_lemma2}$
19 \( =v_f-\frac{1}{2}v_f+\frac{1}{2}v_i \) distribute
20 \( =\frac{1}{2}v_f+\frac{1}{2}v_i \) combine
21 \( =\frac{1}{2}\left(v_f+v_i\right) \) factor
22 \( \bar{v}=\frac{1}{t_f-t_i}\int_{t_i}^{t_f}v\left(t\right)dt=\frac{1}{2}\left(v_f+v_i\right) \) reiterate
23 \( \bar{v}=\frac{1}{t_f-t_i}\int_{t_i}^{t_f}v\left(t\right)dt=\frac{1}{2}\left(v_f+v_i\right) \) substitution elimination 2-22
24 \( \left. \begin{split} v =v_0+at \\ v_f =v\left(t_f\right) \\ v_i =v\left(t_i\right) \end{split}\ \right\} \Rightarrow\bar{v} =\frac{1}{2}\left(v_i+v_f\right) =\frac{1}{t_f-t_i}\int_{t_i}^{t_f}v\left(t\right)dt \) $\Rightarrow\text{I}$ 1-23
Lemma. \begin{equation} \left. \begin{split} v =v_0+at \\ v_f =v\left(t_f\right) \\ v_i =v\left(t_i\right) \\ \end{split} \right\} \Rightarrow v_0=v_f-at_f \label{eq_lemma1} \end{equation} Proof.
1 \( v=v_0+at\\ v_f=v\left(t_f\right)\\ v_i=v\left(t_i\right) \) A
2 \( v\left(t_f\right)=v_0+at_f \) substitution 1
3 \( v_f=v_0+at_f \) $=\text{E}$ 1,2
4 \( v_0=v_f-at_f \) rearrange 3
5 \( \left. \begin{split} v =v_0+at\\ v_f =v\left(t_f\right)\\ v_i =v\left(t_i\right) \end{split} \right\} \Rightarrow v_0=v_f-at_f \) $\Rightarrow\text{I}$ 1-4
Lemma. \begin{equation} \left. \begin{split} v =v_0+at\\ v_f =v\left(t_f\right)\\ v_i =v\left(t_i\right)\\ \end{split}\ \right\} \Rightarrow v_f-v_i=a\left(t_f-t_i\right) \label{eq_lemma2} \end{equation} Proof.
1 \( v=v_0+at\\ v_f=v\left(t_f\right)\\ v_i=v\left(t_i\right) \) A
2 \( v\left(t_f\right)=v_0+at_f \) substitution 1
3 \( v\left(t_i\right)=v_0+at_i \) substitution 1
4 \( v_f=v_0+at_f \) $=\text{E}$ 1,2
5 \( v_i=v_0+at_i \) $=\text{E}$ 1,3
6 \( v_f-v_i=v_0-v_0+at_f-at_i \) subtract 5 from 4
7 \( v_f-v_i=a\left(t_f-t_i\right) \) add, factor
8 \( \left. \begin{split} v=v_0+at\\ v_f=v\left(t_f\right)\\ v_i=v\left(t_i\right) \end{split} \right\} \Rightarrow v_f-v_i=a\left(t_f-t_i\right) \) $\Rightarrow\text{I}$ 1-7

Works Cited

Bergmann, Merrie, James Moor and Jack Nelson. The Logic Book. Third Edition. New York: McGraw Hill, 1998. Print.

Finney, Ross L., et al. Thomas' Calculus. 10th. Boston: Addison Wesley Longman, 2001.

Serway, Raymond A. and Robert J. Beichner. Physics for Scientists and Engineers. 5th Edition. Brooks/Cole, 2000. Print.