Graphical Integration in Motion Analysis — Question 5

Question 5

Recover a missing graph height from a displacement constraint

A velocity graph on an east-positive line joins $(0,0),(2,U),(5,U),(7,-2),(8,-2)$ by straight segments, with time in seconds and velocity in m/s. The unknown height $U$ is positive. Independent endpoint measurements give $x(8)-x(0)=16\ \mathrm{m}$. All data are exact.

1. Express total signed area in terms of $U$ and infer its value. Explain why the trapezoid formula still works when one endpoint velocity is negative.

2. Locate the reversal and calculate the total negative area. Find the traveled distance over 0–8 s.

3. Find the maximum position relative to $x(0)$. Does the endpoint displacement measurement determine the absolute value of that maximum position?

4. A student fits $U$ by setting the sum of unsigned areas equal to 16 m. Explain which measured quantity is being misused and why agreement with the endpoints would be lost.

Question 5 - Solution

Recover a missing graph height from a displacement constraint

(a) Impose the signed displacement constraint.

Writing $U$ numerically in m/s, the four areas in metres are $$\tfrac12(2)U=U,\qquad 3U,\qquad\tfrac12(2)(U-2)=U-2,\qquad -2.$$ Their sum is $5U-4=16$, giving $\boxed{U=4\ \mathrm{m/s}}$. For a straight graph segment, average height is the arithmetic mean of endpoint heights. This computes signed area even across the time axis; it does not compute distance across a reversal.

(b) Split the crossing segment before taking magnitudes.

Let $A_+\geq0$ and $A_-\leq0$ denote the total positive and negative signed velocity areas. $$D=A_++|A_-|=(\Delta x+|A_-|)+|A_-|=\Delta x+2|A_-|.$$

During 5–7 s the slope is $(-2-4)/2=-3\ \mathrm{m/s^2}$. Velocity reaches zero $4/3$ s after 5 s: $$\boxed{t_{\mathrm{rev}}=19/3\ \mathrm{s}}.$$ The negative part from $19/3$ to 7 s is a triangle of magnitude $\tfrac12(2/3)(2)=2/3\ \mathrm{m}$. The final rectangle adds 2 m, so the total negative area is $\boxed{-8/3\ \mathrm{m}}$. Therefore $$\boxed{D=\Delta x+2(8/3)=64/3\ \mathrm{m}}.$$

(c) The maximum precedes the backward part.

The marker moves east until the reversal. Its accumulated positive area is the full net area plus the magnitude later subtracted: $$\boxed{x_{\max}-x(0)=16+8/3=56/3\ \mathrm{m}}.$$ The maximum occurs at $19/3$ s. Absolute position requires $x(0)$, which is not supplied. A common spatial shift changes no graph area or displacement.

(d) An endpoint difference is not an odometer reading.

The measured 16 m is net displacement. Unsigned areas count both eastward and westward travel positively. For the correct graph their sum is $64/3\ \mathrm{m}$, not 16 m. Fitting the unsigned sum to 16 would select a different graph whose signed area fails the supplied displacement constraint. Every fit must preserve the meaning and sign convention of the measurement used.