CLASSICAL-MECHANICS

From Secant to Tangent: Defining Velocity as a Derivative

IKey Formulas

Formula	Name	Description
$v_{\text{ave}} = \dfrac{\Delta x}{\Delta t} = \dfrac{x_f - x_i}{t_f - t_i}$	Average velocity	Slope of the secant line on an $x$ – $t$ graph
$v(t) = \displaystyle\lim_{\Delta t \to 0} \dfrac{x(t+\Delta t) - x(t)}{\Delta t} = \dfrac{dx}{dt}$	Instantaneous velocity	Slope of the tangent line; the derivative of position
$\Delta x = x_f - x_i$	Displacement	Signed change in position (not distance traveled)
$\dfrac{v(t_i) + v(t_f)}{2} = v_{\text{ave}}$	Endpoint mean	Equals $v_{\text{ave}}$ only when $v(t)$ is linear in $t$

IIInteractive Visualizations

Visualization 1 — Displacement vs. Distance

On a there-and-back trip the object comes home, so does its average velocity depend on how far it went?

Distance out (m) 1.00

Visualization 2 — Secant Becomes Tangent

As the interval $\Delta t$ shrinks to zero, what fixed value does the secant slope close in on?

Interval Δt (s) 1.10

💡 This limiting slope — the secant slope as $\Delta t \to 0$ — is the derivative $dx/dt$ , the instantaneous velocity $v(1)$ .

Visualization 3 — Average Velocity and the Midpoint

For linear $v(t)$ , where on $[t_i, t_f]$ does the instantaneous velocity equal the average velocity?

Start t_i (s) 1.0 End t_f (s) 4.0

💡 The midpoint match is special to linear $v(t)$ ; once $v(t)$ bends, the parallel-tangent point drifts off the midpoint and the endpoint-velocity mean no longer equals $v_{\text{ave}}$ .

IIIQuiz Questions

Question 1

An object moves $5$ m in the $+x$ direction, then $5$ m back to its starting point. The round trip takes $2$ s.

What is the object's average velocity over the round trip?

✅ Correct! Net displacement is zero, so the average velocity is zero.

❌ Not quite. That is the average speed (distance / time), not the average velocity, which uses displacement.

❌ Not quite. Average velocity = displacement / time, and the displacement of a round trip is zero.

Show solution

Solution:

Average velocity uses displacement, not distance. The object returns to its starting point, so the net displacement is

$\Delta x = x_f - x_i = 0.$

Therefore

$v_{\text{ave}} = \frac{\Delta x}{\Delta t} = \frac{0}{2\ \text{s}} = 0\ \text{m/s}.$

The total distance is $10$ m and the average speed is $5$ m/s, but speed and velocity are different quantities. The answer is 0 m/s.

Question 2

On a position-versus-time graph, the average velocity over an interval is represented geometrically by which of the following?

✅ Correct! Average velocity is rise over run — the secant slope.

❌ Not quite. The tangent slope is the instantaneous velocity. The average velocity is the secant slope.

❌ Not quite. Average velocity = Δx/Δt = rise/run, which is the slope of the secant line.

Show solution

Solution:

Average velocity is

$v_{\text{ave}} = \frac{\Delta x}{\Delta t} = \frac{\text{rise}}{\text{run}},$

which is exactly the slope of the straight line — the secant — joining the two endpoints $(t_i, x(t_i))$ and $(t_f, x(t_f))$ on the $x$ – $t$ graph.

The slope of the tangent at a single point is the instantaneous velocity, not the average.
The area under an $x$ – $t$ curve has no direct velocity meaning.

The answer is the slope of the secant line connecting the two endpoints.

Question 3

For the position function $x(t) = x_0 + \tfrac{1}{2} b\, t^2$ , the limit definition of the derivative gives the instantaneous velocity $v(t) = b\,t$ .

If $b = 4\ \text{m/s}^2$ , what is the instantaneous velocity at $t = 3$ s?

✅ Correct! v(t) = b·t = 4 × 3 = 12 m/s.

❌ Not quite. 18 comes from ½ b t² (the position). Velocity is v(t) = b·t.

❌ Not quite. Use v(t) = b·t with b = 4 and t = 3, giving 12 m/s.

Show solution

Solution:

From the limit definition,

$v(t) = \lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t)}{\Delta t} = b\,t.$

Substituting $b = 4\ \text{m/s}^2$ and $t = 3$ s:

$v(3) = b\,t = (4)(3) = 12\ \text{m/s}.$

A common trap is to plug into $\tfrac{1}{2} b\,t^2$ (the position) or to forget the factor of $t$ . The velocity is $v(t) = b\,t$ , so the answer is 12 m/s.

Question 4

True or False: The arithmetic mean of the endpoint velocities, $\dfrac{v(t_i) + v(t_f)}{2}$ , always equals the average velocity $v_{\text{ave}} = \dfrac{\Delta x}{\Delta t}$ for any motion.

✅ Correct! The endpoint-mean identity holds only for linear v(t), not for every motion.

❌ Not quite. It is true only when v(t) is linear in t. In general, v_ave ≡ Δx/Δt.

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

The statement is False. The identity

$\frac{v(t_i) + v(t_f)}{2} = v_{\text{ave}}$

holds only when $v(t)$ is a linear function of time, as in $x(t) = x_0 + v_0 t + \tfrac{1}{2} b\, t^2$ , where $v(t) = v_0 + b\,t$ . In that special case both sides reduce to $v_0 + \tfrac{1}{2} b\,(t_f + t_i)$ .

For any nonlinear velocity function, the arithmetic mean of the endpoint velocities will not match the true average velocity. The reliable definition is always

$v_{\text{ave}} \equiv \frac{\Delta x}{\Delta t}.$

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