Position and Displacement — Question 3

Question 3

Read a trajectory from a position function

For $0\leq t\leq3$ s, a particle has $x(t)=(2\ \mathrm{m/s})t$ and $y(t)=(1\ \mathrm{m/s^2})t^2$. The equations specify position in fixed axes.

1. Find the positions at 1 s and 3 s, and the displacement between them.

2. Eliminate t to obtain y as a function of x, retaining units.

3. Does the straight displacement arrow lie along the path between those times? Explain and give its length.

Question 3 - Solution

Read a trajectory from a position function

(a) Evaluate both coordinates at each time.

$$\vec r(1)=(2,1),\quad\vec r(3)=(6,9),\quad\boxed{\Delta\vec r=(4,8)\,\mathrm m}$$ $\vec r(1\ \mathrm s)=(2,1)$ m and $\vec r(3\ \mathrm s)=(6,9)$ m. Their difference is $(4,8)$ m. Subtracting the time labels without evaluating the nonlinear coordinate would miss the vertical change.

(b) Eliminate the parameter dimensionally.

$$t=\frac{x}{2\,\mathrm{m/s}}\ \Rightarrow\ y=(1\,\mathrm{m/s^2})\left(\frac{x}{2\,\mathrm{m/s}}\right)^2=\frac{x^2}{4\,\mathrm m}$$ $t=x/(2\ \mathrm{m/s})$, so $y=x^2/(4\ \mathrm m)$ for $0\leq x\leq6$ m. This is an upward-opening parabola. Writing $y=x^2/4$ without specifying numerical-coordinate conventions would omit a required length unit.

(c) Distinguish a chord from a curve.

Use $u=t/(1\,\mathrm s)$ in the arc-length integral. $$|\Delta\vec r|=\sqrt{4^2+8^2}=4\sqrt5\,\mathrm m,\quad s=\int_1^3\sqrt{4+4u^2}\,du\,\mathrm m>4\sqrt5\,\mathrm m$$ The displacement is the chord joining the two endpoint positions, of length $\sqrt{4^2+8^2}=4\sqrt5\approx8.944$ m. The particle follows the curved segment, not that straight chord. Its traveled distance is strictly greater than the chord length.