Position and Displacement — Question 10

Question 10

A position history must join at a switching time

Let $u=t/(1\ \mathrm s)$ be dimensionless. A proposed position history is $\vec r=(u,u^2)$ m for $0\leq u\leq2$, and $\vec r=(a+2(u-2),b-(u-2))$ m for $2<u\leq4$. Assume a physical path with no instantaneous jump.

1. Find a and b from position continuity.

2. Find the displacement from 1 s to 4 s.

3. If a programmer instead uses $a=b=0$, identify the jump at 2 s and explain why a correct final endpoint alone would not validate the whole history.

Question 10 - Solution

A position history must join at a switching time

(a) Match the two limiting positions.

$$\lim_{u\to2^-}\vec r=(2,4)\,\mathrm m=\lim_{u\to2^+}\vec r\ \Rightarrow\ \boxed{a=2,\ b=4}$$ The first branch reaches $(2,4)$ m at $u=2$. The second approaches $(a,b)$ m as $u$ decreases toward 2. Thus $a=2$ and $b=4$. Continuity applies separately to every position component.

(b) Evaluate endpoints on their appropriate branches.

$$\vec r(4)=(2+4,4-2)=(6,2)\,\mathrm m,\quad\Delta\vec r=(6,2)-(1,1)=(5,1)\,\mathrm m$$ At 1 s, $\vec r=(1,1)$ m. At 4 s, $u=4$ and $\vec r=(6,2)$ m. Therefore $\Delta\vec r=(5,1)$ m, with magnitude $\sqrt{26}$ m. The calculation spans a branch change without requiring a straight path.

(c) Audit the transition as well as the destination.

$$\vec r(2^+)-\vec r(2^-)=(0,0)-(2,4)=(-2,-4)\,\mathrm m\ne\vec0$$ The faulty right-hand limit is $(0,0)$ m, so the position jumps by $(-2,-4)$ m. That violates the stated no-jump model. Endpoint displacement can still be calculated from a faulty record, but agreement at one endpoint cannot repair an inconsistent intermediate position history.