Acceleration — Question 10

Question 10

Join velocity histories without concealing an acceleration corner

A designer specifies the exact east-positive velocity history $$v(t)=\begin{cases} 1+2t+t^2,&0\leq t\leq2,\\ C+D(t-2)+E(t-2)^2,&2<t\leq4. \end{cases}$$ Here $t$ is numerical seconds, $v$ numerical m/s, and $C,D,E$ numerical coefficients in that convention. The design requires velocity and acceleration to be continuous at 2 s and requires $v(4\ \mathrm{s})=5\ \mathrm{m/s}$. Acceleration is $dv/dt$; compare one-sided derivatives at the join.

1. Determine $C,D,E$ from the three requirements, explaining which condition fixes each coefficient.

2. Find the acceleration at the join and on each open branch. Locate the time of greatest speed during 0–4 s and find that speed. Check the velocity sign before identifying speed with velocity.

3. Find the full-run average acceleration. Does its sign determine the sign of instantaneous acceleration throughout the run?

4. An alternative proposal uses $C=9,D=0,E=-1$. Check its velocity continuity and final velocity. Does acceleration exist at 2 s in that exact alternative model? Is the velocity itself discontinuous there?

5. Explain why matching endpoint velocities and joining the velocity values continuously do not, by themselves, guarantee a smooth acceleration history.

Question 10 - Solution

Join velocity histories without concealing an acceleration corner

(a) Match the value, then the first derivative, then the endpoint.

The left branch gives $v(2)=9\ \mathrm{m/s}$, so continuity requires $C=9$. Its derivative is $2+2t$, approaching $6\ \mathrm{m/s^2}$ at 2 s; the right derivative approaches $D$, so $D=6$. Finally, $5=C+2D+4E$ gives $$\boxed{C=9,\quad D=6,\quad E=-4}.$$

(b) Check velocity and acceleration signs across the full interval.

$$v(t)=\frac{45}{4}-4\left(t-\frac{11}{4}\right)^2\quad(2\leq t\leq4),\qquad v_{\max}=\frac{45}{4}\ \mathrm{m/s}.$$

The acceleration is $2+2t$ for $0<t<2$ and $6-8(t-2)$ for $2<t<4$, in $\mathrm{m/s^2}$. Both sides agree at the join: $\boxed{a(2)=6\ \mathrm{m/s^2}}$.

The right branch reaches zero acceleration at $t-2=3/4$, or $\boxed{t=2.75\ \mathrm{s}}$. Acceleration is positive before this time and negative afterward. Velocity rises from 1 to 9 m/s on the first branch, then rises to its peak and falls to 5 m/s. It remains positive throughout, so speed equals velocity. The greatest speed is $$9+6(0.75)-4(0.75)^2=\boxed{11.25\ \mathrm{m/s}}.$$

(c) Net velocity change does not prescribe every slope.

$$\boxed{\bar a=(5-1)/4=+1\ \mathrm{m/s^2}}.$$ Despite this positive average, instantaneous acceleration is negative after 2.75 s. The final velocity exceeds the initial velocity even though velocity decreases during the last part of the run.

(d) The alternative meets the value constraints but not the derivative constraint.

The proposed right branch is $9-(t-2)^2$. It approaches 9 m/s at 2 s and equals 5 m/s at 4 s. Velocity is continuous at the join. However, the left derivative is 6 while the right derivative is zero, in $\mathrm{m/s^2}$. Acceleration is therefore undefined at 2 s under the two-sided derivative definition. A velocity corner is not a jump in velocity itself.

(e) Continuity and differentiability impose different conditions.

Matching values prevents a break in the velocity graph; matching nearby slopes is an additional requirement for acceleration to exist at the join. Endpoint values constrain only net changes. Neither those values nor an arbitrarily assigned acceleration at one instant can repair unequal limiting slopes of the unchanged velocity history.