Instantaneous Velocity and Speed — Question 2

Question 2

Read the tangent, not the height or the origin chord

The solid position graph describes a point moving on an east-positive line. At event P, $t=4\ \mathrm{s}$ and $x=8\ \mathrm{m}$. Its tangent is the dashed straight line through $(2\ \mathrm{s},14\ \mathrm{m})$ and $(6\ \mathrm{s},2\ \mathrm{m})$. The actual curve passes through $(2\ \mathrm{s},16\ \mathrm{m})$.

Instantaneous velocity equals the slope of the position graph’s tangent; instantaneous speed is its magnitude. Use these facts without differentiating an equation.

1. Find the velocity and speed at P. What is the direction of motion there?

2. Diagnose the answer $v=8/4=2\ \mathrm{m/s}$. What geometric line does this ratio describe, and why is it not the specified tangent?

3. Find the average velocity of the actual motion from 2 to 4 s. Explain why the open tangent point at $(2,14)$ must not replace the actual position at 2 s.

4. Is being east of the origin compatible with moving west at P? Would a different horizontal or vertical drawing scale change the physical slope in m/s?

Question 2 - Solution

Read the tangent, not the height or the origin chord

(a) Use coordinate differences along the tangent.

$$\boxed{v(4\ \mathrm{s})=\frac{2-14}{6-2}=-3\ \mathrm{m/s}},\qquad \boxed{s(4\ \mathrm{s})=3\ \mathrm{m/s}}.$$ The negative velocity means motion toward decreasing $x$, or westward.

(b) A position-to-time ratio is not generally a tangent slope.

$$x_{\mathrm{tan}}(t)=8\ \mathrm{m}-(3\ \mathrm{m/s})(t-4\ \mathrm{s}),\qquad x_{\mathrm{tan}}(0)=20\ \mathrm{m}\ne0.$$

The ratio $8/4$ is the slope of the line from the graph’s coordinate origin $(0,0)$ to P. That line is not the given tangent, and $(0,0)$ is not supplied as a motion event. A tangent need not pass through either coordinate origin. Its slope requires two points on that tangent, with coordinate differences and units.

(c) A tangent point need not lie on the actual curve.

The actual endpoints give $$\boxed{\bar v_{2\to4}=\frac{8-16}{4-2}=-4\ \mathrm{m/s}}.$$ Using 14 m instead of 16 m would substitute an auxiliary tangent point for a measured event. The tangent summarizes the curve locally at P; it does not assert a straight-line position history over the whole 2–4 s interval.

(d) Position, direction and drawing angle are distinct.

$$v=\frac{\Delta x_{\mathrm{physical}}}{\Delta t_{\mathrm{physical}}}\quad\text{regardless of printed axis aspect ratio}.$$

At P the point is 8 m east of the origin while moving west at 3 m/s. Its positive coordinate and negative velocity answer different questions. Rescaling either printed axis changes the visual angle but not the coordinate ratio $\Delta x/\Delta t$ calculated with the labeled scales. The physical velocity remains $-3\ \mathrm{m/s}$.