Constant Acceleration - A Special Case — Question 5

Question 5

A delayed start requires its own elapsed time

At $t=0$, cart A passes the origin moving east at a constant $4\ \mathrm{m/s}$. Cart B is at rest at that origin. B remains there until $t=3\ \mathrm{s}$, then accelerates east at a constant $2\ \mathrm{m/s^2}$. The carts are point markers allowed to pass through one another, and A maintains its speed.

1. Write A’s position for $t\geq0$ and B’s piecewise position. Define a local elapsed time $u$ for B’s accelerating stage.

2. Find the first meeting after B starts. Give both the common clock time and B’s elapsed acceleration time, and the meeting position.

3. Find both velocities at that meeting. Does a shared position imply equal velocities? Which cart is gaining position relative to the other then?

4. The quadratic meeting equation has a second algebraic root. Explain why it is not a valid meeting under B’s accelerating-stage formula. Also explain why using $x_B=\tfrac12at^2$ with the common clock would describe a different experiment.

Question 5 - Solution

A delayed start requires its own elapsed time

(a) Keep the start event attached to each formula.

With $t$ numerical seconds, $x_A=4t\ \mathrm{m}$. For B, $$x_B=\begin{cases}0,&0\leq t\leq3,\\(t-3)^2\ \mathrm{m},&t\geq3.\end{cases}$$ Define $u=t-3$ as numerical elapsed seconds after launch. Then $u\geq0$, $x_B=u^2\ \mathrm{m}$, and $x_A=4(u+3)\ \mathrm{m}$.

(b) Solve using a common event time.

At a meeting, $u^2=4u+12$, or $(u-6)(u+2)=0$. The admissible root is $u=6$, giving $$\boxed{t=9\ \mathrm{s},\quad T_B=6\ \mathrm{s},\quad x=36\ \mathrm{m}}.$$ A and B also share the origin at $t=0$, but that is before B starts and is not the requested meeting.

(c) Compare velocities at the meeting event.

A still has $\boxed{v_A=4\ \mathrm{m/s}}$. B has accelerated from rest for 6 s, giving $\boxed{v_B=2(6)=12\ \mathrm{m/s}}$. Thus B gains position relative to A at $\boxed{8\ \mathrm{m/s}}$ and passes it eastward. Equality of positions at one event does not require equality of their rates of change.

(d) Reject a root that violates the branch’s time domain.

$$t^2-10t+9=(t-1)(t-9)=0,\qquad t\geq3\ \mathrm{s}\Rightarrow t=9\ \mathrm{s}.$$

The other root $u=-2$ corresponds to common time $t=1\ \mathrm{s}$, before B’s launch. The accelerating formula is invalid there. The actual B position is zero at 1 s, while A is at 4 m, so they do not meet.

Using $x_B=\tfrac12at^2$ would apply B’s acceleration from $t=0$ rather than from 3 s. That erases the delay and solves a different pursuit problem. The correct time variable is elapsed time since the initial conditions used in the formula.