Analyzing Non-Uniform Motion with Graphs

1. Average vs. Instantaneous Velocity on a Position-Time Graph

For motion described by a curved position-time ($x-t$) graph, the velocity is continuously changing. This requires us to be precise about whether we are discussing the velocity over an interval or at an instant.

• Average Velocity $⟨v⟩$: The average velocity between two times, $t_1$ and $t_2$, is the total displacement divided by the time interval.
  $⟨v⟩=\frac{x(t_2)-x(t_1)}{t_2-t_1}=\frac{\Delta x}{\Delta t}$
  Graphically, this corresponds to the slope of the secant line connecting the points $(t_1,x(t_1))$ and $(t_2,x(t_2))$ on the $x-t$ graph.

• Instantaneous Velocity $v$: The instantaneous velocity at a single time $t$ is the rate of change of position at that exact moment. As we've established, this is the derivative of position with respect to time.
  $v(t)={\lim }_{\Delta t\to 0}\frac{\Delta x}{\Delta t}=\frac{dx}{dt}$
  Graphically, this corresponds to the slope of the tangent line to the $x-t$ curve at the point $(t,x(t))$.

The image below provides a clear visual summary of this distinction.

To truly understand how the instantaneous value emerges from the average, it's essential to see it derived from first principles using limits.

2. A Deeper Connection: The Mean Value Theorem

Given your mathematical background, you will appreciate that there is a formal theorem that connects average and instantaneous velocity: the Mean Value Theorem (MVT) for derivatives.

In mathematical terms, the MVT states:

If a function $f(x)$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there exists at least one number $c$ in $(a,b)$ such that:
$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$

Let's translate this directly into the language of kinematics by replacing $f$ with position $x$, and the independent variable with time $t$.

If an object's position $x(t)$ is a continuous and differentiable function of time over the interval $[t_1,t_2]$, then there exists at least one instant of time $c$ within that interval $(t_1,t_2)$ where the instantaneous velocity $v(c)$ is exactly equal to the average velocity $⟨v⟩$ over the entire interval.
$v(c)=\frac{x(t_2)-x(t_1)}{t_2-t_1}$

Graphically, this means that for any secant line drawn across an interval of the $x-t$ graph, there is at least one point on the curve within that interval where the tangent line is parallel to the secant line.

3. Generalizing to Acceleration from a Velocity-Time Graph

The exact same logic applies when we analyze a velocity-time ($v-t$) graph to find acceleration. The relationships are perfectly analogous:

• Average Acceleration $⟨a⟩$: The slope of the secant line on the $v-t$ graph over an interval $\Delta t$.
  $⟨a⟩=\frac{v(t_2)-v(t_1)}{t_2-t_1}$
• Instantaneous Acceleration $a$: The slope of the tangent line on the $v-t$ graph at an instant $t$.
  $a(t)=\frac{dv}{dt}$

The following resource formally defines these concepts and walks you through activities to solidify this understanding.

4. Determining Average Velocity from a V-T Graph

We have defined average velocity as $\frac{\Delta x}{\Delta t}$. But how do we find $\Delta x$ if we are only given a $v-t$ graph? As you know from previous lessons, displacement is the area under the v-t graph.

This principle holds true even for curved graphs. The total displacement is the definite integral of the velocity function.

$\Delta x={\int }_{t_1}^{t_2}v(t)dt$

Therefore, to find the average velocity from a $v-t$ graph over an interval $[t_1,t_2]$, you must:

1. Calculate the net signed area under the curve between $t_1$ and $t_2$. This gives the displacement, $\Delta x$.
2. Divide this displacement by the time interval, $\Delta t=t_2-t_1$.

$⟨v⟩=\frac{1}{t_2-t_1}{\int }_{t_1}^{t_2}v(t)dt$

The workbook LINK provides a rigorous justification for why the area under a continuously changing velocity curve represents displacement, using a limit argument that you will recognize as the foundation of the Riemann integral. I encourage you to read Activity 4-7 ("Determining displacement of an object with constantly changing velocity") in that resource to see this proof laid out.

5. Synthesis: A Worked Example

Let's apply these concepts to a problem. An object's position is described by the function $x(t)=t^3-6t^2+9t+1$, where $x$ is in meters and $t$ is in seconds, for $t\ge 0$.

(a) Find the average velocity between $t=1$ s and $t=4$ s.

First, find the positions at the endpoints of the interval:
$x(1)=1-6+9+1=5$ m
$x(4)=4^3-6(4^2)+9(4)+1=64-96+36+1=5$ m

Now, calculate the average velocity:
$⟨v⟩=\frac{x(4)-x(1)}{4-1}=\frac{5-5}{3}=0\text{ m/s}$
The average velocity is zero, which means the net displacement over this interval is zero.

(b) Find the instantaneous velocity function $v(t)$ and the instantaneous velocity at $t=2$ s.

We differentiate $x(t)$ to find $v(t)$:
$v(t)=\frac{dx}{dt}=3t^2-12t+9$
Now, evaluate at $t=2$:
$v(2)=3(2^2)-12(2)+9=12-24+9=-3\text{ m/s}$

(c) According to the Mean Value Theorem, there must be an instant $c\in (1,4)$ where $v(c)=⟨v⟩$. Find this time $c$.

We need to find $c$ such that $v(c)=0$.
$3c^2-12c+9=0$
$c^2-4c+3=0$
$(c-1)(c-3)=0$
The solutions are $c=1$ and $c=3$. Since the MVT guarantees a point within the open interval $(1,4)$, our answer is $c=3$ s. This is the instant where the object's velocity is momentarily zero, matching the average velocity over the interval.

Conclusion

In this lesson, you have generalized your graphical analysis skills to non-uniform motion, solidifying the link between calculus and kinematics.

Key Takeaways:

• For any kinematic quantity, the average rate of change over an interval is the slope of the secant line on the corresponding graph.
• The instantaneous rate of change at a point is the slope of the tangent line, which is mathematically equivalent to the derivative.
• The net change in a quantity is the area under the curve of its rate of change, which is mathematically equivalent to the definite integral.
• The average value of a function $f(t)$ over an interval is its definite integral divided by the interval's duration. For velocity, this means $⟨v⟩=\frac{\Delta x}{\Delta t}$.

Preview of the Next Lesson:

We have now completed our deep dive into the graphical and calculus-based analysis of one-dimensional motion. In the next module, we will expand our view into a second dimension. The first lesson will be foundational: you will prove the principle of independence for perpendicular components of motion and apply it to describe general 2D trajectories. This principle is the key that allows us to solve complex 2D problems, like projectile motion, by breaking them down into two simpler 1D problems.