Instantaneous Velocity and Speed — Question 10

Question 10

Join two position models without hiding a velocity jump

A point moves on an east-positive line over $0\leq t\leq4\ \mathrm{s}$. In the following piecewise model, $t$ denotes numerical seconds, $x$ numerical metres, and $a,b$ numerical coefficients in that convention: $$x(t)=\begin{cases} 2t,&0\leq t\leq2,\\ 4+a(t-2)+b(t-2)^2,&2<t\leq4. \end{cases}$$ The designer requires position and instantaneous velocity to be continuous at 2 s, and requires $x(4\ \mathrm{s})=0$. To test the derivative at the join, compare the left and right derivative limits. You may use $d[(t-2)^2]/dt=2(t-2)$.

1. Explain why position continuity is already satisfied for every finite $a,b$. Use the velocity condition and the final-position condition to find $a,b$.

2. For your resulting model, find the velocity at the join, every later zero-speed instant, and the position at each such instant. Does reversal occur at the join or later? Justify with signs.

3. Find total distance, net displacement, average speed, and average velocity over 0–4 s without integration.

4. A second designer chooses $a=0,b=-1$, arguing that matching position at 2 s and final position at 4 s is sufficient. Check those two position conditions, then diagnose the velocity at 2 s in this alternative exact model.

5. Why does assigning the point a chosen velocity at the join fail to repair a mismatch between the two position derivatives?

Question 10 - Solution

Join two position models without hiding a velocity jump

(a) Check value and slope as separate conditions.

Both branches approach 4 m at 2 s because every term containing $t-2$ vanishes there. The left derivative is 2 m/s; the right derivative approaches $a\ \mathrm{m/s}$. Matching them requires $a=2$. The final condition is $0=4+2a+4b$, so $$\boxed{a=2,\qquad b=-2}.$$ The matching one-sided derivatives give an existing continuous velocity at the join.

(b) Locate the reversal from the later velocity, not from the join.

At 2 s, $\boxed{v=+2\ \mathrm{m/s}}$. For $t>2$, the velocity is $v=2-4(t-2)\ \mathrm{m/s}$, which vanishes at $\boxed{t=2.5\ \mathrm{s}}$. The position then is $4+2(0.5)-2(0.5)^2=\boxed{4.5\ \mathrm{m}}$. Velocity is positive before this instant and negative afterward. The point continues east through the join, then reverses west at 2.5 s.

(c) Add lengths along the monotone parts.

Positions at 0, 2, 2.5 and 4 s are $0,4,4.5,0\ \mathrm{m}$. Therefore $$\boxed{D=4+0.5+4.5=9\ \mathrm{m}},\qquad\boxed{\Delta x=0\ \mathrm{m}}.$$ Over 4 s, $\boxed{\bar s=9/4=2.25\ \mathrm{m/s}}$ and $\boxed{\bar v=0\ \mathrm{m/s}}$. The zero average velocity describes the return to the start, not continuous rest.

(d) Endpoint positions do not enforce a derivative at the join.

$$\lim_{h\to0^-}\frac{x(2+h)-4}{h}=2,\qquad\lim_{h\to0^+}\frac{x(2+h)-4}{h}=0\quad(\mathrm{m/s}).$$

For $a=0,b=-1$, the later branch is $4-(t-2)^2$. It approaches 4 m at the join and equals zero at 4 s, so both specified position conditions hold. But the left derivative at 2 s is 2 m/s and the right derivative is zero. The two-sided derivative at 2 s does not exist in this alternative model. The right branch alone cannot supply the derivative of the joined history there.

(e) A derivative is determined by nearby position values.

Assigning an arbitrary velocity value at the single instant changes neither one-sided position quotient. It cannot make unequal limits agree. Repairing the mismatch requires changing the local position history or accepting a nondifferentiable idealized join, rather than relabeling its velocity.