Instantaneous Velocity and Speed — Question 9

Question 9

Meeting, matching velocity, and passing are different events

Two point markers can pass through each other on an east-positive straight line. Their exact histories during $0\leq t\leq4$ are $$x_A(t)=t^2,\qquad x_B(t)=4t-3.$$ Here $t$ is numerical seconds and each $x$ numerical metres. Differentiate to find velocities in m/s. No collision or interaction law is imposed.

1. Find all times when the markers occupy the same position. At each meeting, give both velocities and both speeds.

2. Find when the velocities are equal. Do the positions also agree then? Explain why setting $x_A=x_B$ and setting $v_A=v_B$ answer different questions.

3. Factor the separation $d=x_A-x_B$ and determine its sign before, between, and after the meetings. At each meeting, which marker passes the other in the eastward direction?

4. Define the relative velocity as $v_A-v_B$, the derivative of $d$. Evaluate it at both meetings. Compare its magnitude with the difference of the two speeds here; would that shortcut remain valid for oppositely directed velocities?

Question 9 - Solution

Meeting, matching velocity, and passing are different events

(a) A meeting requires equal positions at the same time.

$$t^2=4t-3\quad\Longleftrightarrow\quad(t-1)(t-3)=0.$$ The meetings are at $\boxed{t=1\ \mathrm{s},\ x=1\ \mathrm{m}}$ and $\boxed{t=3\ \mathrm{s},\ x=9\ \mathrm{m}}$. Since $v_A=2t$ and $v_B=4\ \mathrm{m/s}$, at the first meeting their velocities are $+2$ and $+4\ \mathrm{m/s}$; at the second they are $+6$ and $+4\ \mathrm{m/s}$. All are positive, so the speeds are 2 and 4, then 6 and 4 m/s.

(b) Equal derivatives do not require equal coordinates.

Solving $2t=4$ gives $\boxed{t=2\ \mathrm{s}}$. At that time $x_A=4\ \mathrm{m}$ and $x_B=5\ \mathrm{m}$, so they do not meet. Equal positions locate a shared event in space and time; equal velocities compare their local rates of position change. Neither condition generally implies the other.

(c) Use the separation’s sign to establish actual passing.

$$d(t)=(t-2)^2-1,\qquad d(2)=-1\ \mathrm{m},\quad d'(2)=0\ \mathrm{m/s}.$$

The separation is $d=(t-1)(t-3)\ \mathrm{m}$, with $t$ numerical seconds. It is positive before 1 s, negative between 1 and 3 s, and positive after 3 s. Thus A is east of B, then west of B, then east again. At 1 s, B passes A eastward. At 3 s, A passes B eastward. The sign changes establish passing rather than merely touching the same position and separating on the same sides.

(d) Subtract signed velocities before taking a magnitude.

$$v_A-v_B=(2t-4)\ \mathrm{m/s}.$$ The relative velocities are $\boxed{-2\ \mathrm{m/s}}$ at 1 s and $\boxed{+2\ \mathrm{m/s}}$ at 3 s; both relative speed magnitudes are 2 m/s. Here both markers move east, so the absolute difference of their speeds happens to agree. For opposite velocities, such as $+2$ and $-4\ \mathrm{m/s}$, the relative-velocity magnitude is $|2-(-4)|=6\ \mathrm{m/s}$, whereas the difference of speeds has magnitude 2 m/s. Direction information must be retained until after subtraction.