Graphical Integration in Motion Analysis — Question 1

Question 1

Signed area can cancel while traveled distance grows

A marker moves on an east-positive line with $x(0)=3\ \mathrm{m}$. Its exact velocity graph consists of straight segments joining $(t,v)=(0,4),(2,4),(6,-4),(8,-4)$, in seconds and metres per second.

For any selected interval, displacement is the signed area under $v(t)$, while distance is the area under $|v(t)|$. A velocity–time area has units $(\mathrm{m/s})\mathrm{s}=\mathrm{m}$.

1. Find the time at which the marker reverses. Divide the graph into simple areas to calculate displacement and distance over 0–8 s.

2. Find the greatest position attained and its time. Is the initial position reached again before 8 s?

3. Find average velocity and average speed for the whole interval. Explain why zero average velocity does not imply rest.

4. A student treats the downward sloping segment as westward motion throughout 2–6 s. Use the graph to correct the claim and identify when the marker is slowing down.

Question 1 - Solution

Signed area can cancel while traveled distance grows

(a) Split at the zero of velocity.

$$\Delta x=\int_0^8v\,dt=8+4-4-8=0,\qquad D=\int_0^8|v|\,dt=8+4+4+8=24\ \mathrm{m}.$$

The segment from $(2,4)$ to $(6,-4)$ crosses $v=0$ at $\boxed{t=4\ \mathrm{s}}$. Velocity changes from positive to negative there. The successive signed areas are $$A_{0,2}=8,\quad A_{2,4}=4,\quad A_{4,6}=-4,\quad A_{6,8}=-8 \quad(\mathrm{m}).$$ Thus $\boxed{\Delta x=0\ \mathrm{m}}$, but $\boxed{D=24\ \mathrm{m}}$. Negative areas subtract from displacement and contribute positively to distance.

(b) Accumulate area from the known initial position.

Position rises until velocity becomes negative. At 4 s, $$\boxed{x_{\max}=3+8+4=15\ \mathrm{m}}.$$ It then decreases continuously to $x(8)=3\ \mathrm{m}$. Before 4 s the accumulated area is positive; between 4 and 8 s the returning negative area has not yet canceled all 12 m. Hence there is no earlier return after departure.

(c) Divide different totals by the same elapsed time.

$$\boxed{\bar v=0\ \mathrm{m/s}},\qquad \boxed{\bar s=24/8=3\ \mathrm{m/s}}.$$ Zero average velocity says the endpoint positions agree. The graph shows nonzero motion over almost the entire interval, so the marker was not at rest.

(d) Height and slope carry different information.

The negative slope during 2–6 s gives negative acceleration, not necessarily negative velocity. The marker still moves east during 2–4 s, when the graph lies above the axis. Its speed decreases on that interval. During 4–6 s it moves west and its speed increases. The two horizontal portions have constant speed; zero velocity at one instant is not a finite stop.