Relative Motion - Statics and Dynamics

1. What is the speed of the particle as measured in the ground frame S, and what can be concluded about the particle's velocity in S?
   Answer: Speed in S is $0 \text{ m/s}$; the particle is momentarily stationary in S because $\vec{v} = \vec{v}' + \vec{V} = (-3+3)\hat{i} + (4-4)\hat{j} = \vec{0}$, meaning the particle's velocity components cancel exactly in both directions. When a problem involves two reference frames, your first instinct should be to apply the Galilean velocity transformation: $\vec{v} = \vec{v}' + \vec{V}$, where $\vec{v}$ is the particle's velocity in the ground frame S, $\vec{v}'$ is its velocity in the moving frame S', and $\vec{V}$ is the velocity of S' relative to S. Crucially, you add the vector frame velocity — not the scalar speed. Plugging in directly: $\vec{v} = (-3\hat{i} + 4\hat{j}) + (3\hat{i} - 4\hat{j}) = (0)\hat{i} + (0)\hat{j} = \vec{0}$. The components cancel perfectly, giving a speed of 0 m/s in S. This makes D correct — the particle happens to be momentarily at rest as observed from the ground, even though it's moving in S'. Choice A is a fabricated result. Subtracting $\vec{V}$ instead of adding it gets the transformation backwards, and the invented "sign correction" for frame orientation has no physical basis. Choice B mistakes the transformation for a conservation law — Galilean relativity does not preserve speed between frames; only the form of Newton's laws is preserved. Choice C confuses vector addition with scalar addition of magnitudes: $|\vec{v}' + \vec{V}| \neq |\vec{v}'| + |\vec{V}|$ in general — that equality only holds when both vectors point in the same direction. Strategy tip: Always write out the full vector equation $\vec{v} = \vec{v}' + \vec{V}$ component by component before computing any magnitudes. Jumping straight to scalar arithmetic is the most common trap on frame-transformation problems.
2. The ground observer now throws their own ball westward at $w = 25 \text{ m/s}$ relative to the ground. What is the velocity of the ground observer's ball as seen by the train passenger, expressed as a magnitude and direction?
   Answer: $45 \text{ m/s}$ westward — found by adding the ground ball's westward speed to the train's eastward speed, since the passenger is moving east and thus sees westward objects moving faster to the west by the train's full speed. Whenever you see a relative velocity problem, your first move should be to establish a consistent sign convention. Choose a positive direction — here, let's say east is positive — and express every velocity as a signed number before doing any arithmetic. The train passenger moves east at $+20 \text{ m/s}$ relative to the ground. The ground observer's ball travels west, so its ground velocity is $-25 \text{ m/s}$. To find how the passenger sees the ball, apply the relative velocity formula: $v_{\text{ball relative to passenger}} = v_{\text{ball relative to ground}} - v_{\text{passenger relative to ground}}$ $= (-25) - (+20) = -45 \text{ m/s}$ The negative sign means 45 m/s westward — confirming answer A is correct. Intuitively, a passenger racing east at 20 m/s will see a westward-moving object flying past even faster than a stationary observer would. B is wrong because it subtracts magnitudes (25 − 20 = 5) without respecting sign conventions. When two objects move in opposite directions, you add their speeds, not subtract them. C is wrong because it confuses inertial frames with identical frames. Yes, the passenger's frame is inertial, but velocities still transform between frames — the ball's speed is not the same in every inertial frame. D is wrong in a subtle way: it imagines the passenger's motion "overpowering" the ball and reversing its perceived direction, but the math shows the passenger only adds to the westward speed, not reverses it. Study tip: Always assign signs before plugging into relative velocity formulas. The formula $v_{A/B} = v_A - v_B$ handles direction automatically — trust the math, not your intuition about which motion "dominates."
3. What is the magnitude of the velocity of Ship B relative to Ship A, and in what general direction does B appear to move as seen from A?
   Answer: $15 \text{ m/s}$, directed east of south — because the relative velocity vector has components $+9\hat{i}$ (east) and $-12\hat{j}$ (south) when the observer's own northward velocity is subtracted from B's velocity. Whenever you see a relative velocity problem, anchor yourself to this formula: $\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A$. The velocity of B as seen from A means you subtract A's velocity from B's velocity — not add, not rearrange. Setting up coordinates with east as $+\hat{i}$ and north as $+\hat{j}$, Ship B's velocity is $\vec{v}_B = 9\hat{i}$ and Ship A's velocity is $\vec{v}_A = 12\hat{j}$. Applying the formula: $\vec{v}_{B/A} = 9\hat{i} - 12\hat{j}$. This gives a vector pointing east ($+9$) and south ($-12$), meaning B appears to move toward the southeast from A's perspective. The magnitude is $\sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \text{ m/s}$. That confirms A is correct. B makes the classic sign error of adding the velocities instead of subtracting, producing a northeast direction — but the observer is moving north, so northward motion should disappear from their reference frame, not double. C gets the magnitude right (15 m/s) but flips the east component to negative, as if B were moving west — there's no justification for negating $v_B$'s eastward component. D incorrectly assumes only differing components matter, ignoring that A's full northward velocity contributes a southward component in the relative frame; this error also produces the wrong magnitude entirely. A helpful memory anchor: relative velocity = target minus observer. If you're the observer, your own motion "disappears" by subtraction — whatever you're doing, the other object appears to do the opposite.
4. An engineer claims that the speed of Q as seen from P equals $\sqrt{101} \approx 10.05 \text{ m/s}$. Which of the following correctly identifies whether this claim is right or wrong, and why?
   Answer: The claim is wrong. The relative velocity is $\vec{v}_{Q/P} = \vec{v}_Q - \vec{v}_P = (-4-6)\hat{i} + 3\hat{j} = -10\hat{i} + 3\hat{j}$, giving a speed of $\sqrt{100+9} = \sqrt{109} \approx 10.44 \text{ m/s}$, not $\sqrt{101}$. Whenever you encounter a question about relative velocity, anchor yourself to one core formula: the velocity of Q as seen from P is $\vec{v}_{Q/P} = \vec{v}_Q - \vec{v}_P$. Think of it as "subtracting away" the observer's own motion to reveal what the observed point appears to do from that moving frame. Applying this here: $\vec{v}_{Q/P} = (-4\hat{i} + 3\hat{j}) - (6\hat{i}) = -10\hat{i} + 3\hat{j}$. The speed (magnitude) is then $|\vec{v}_{Q/P}| = \sqrt{(-10)^2 + 3^2} = \sqrt{100 + 9} = \sqrt{109} \approx 10.44 \text{ m/s}$, not $\sqrt{101}$. The engineer's claim is therefore wrong, making A the correct answer. B is wrong because it adds the velocity vectors instead of subtracting — a fundamental sign error. There is no legitimate "unit conversion" that transforms $\sqrt{13}$ into $\sqrt{101}$; that reasoning is fabricated. C is a trap that dresses up a correct intermediate step (computing $-10\hat{i} + 3\hat{j}$) with a nonsensical "projection" correction. No projection onto P's direction is needed or valid here; relative velocity is fully captured by the vector difference, full stop. D confuses the speed of Q in the ground frame ($|\vec{v}_Q| = 5 \text{ m/s}$) with relative speed, then constructs an arbitrary Pythagorean combination to manufacture $\sqrt{101}$. Relative speed is not a scalar invariant independent of the observer's velocity. Study tip: Always write out $\vec{v}_{Q/P} = \vec{v}_Q - \vec{v}_P$ explicitly — careless sign errors (adding instead of subtracting) are the most common mistake on relative-motion problems.
5. How long (in seconds) does it take for Car B to reach a position due south of Car A's position at $t = 0$, assuming both cars maintain constant velocity?
   Answer: $12.86 \text{ s}$ — because the two cars move toward each other, so the gap closes at the sum of both speeds $(60+80 = 140 \text{ km/h} = 38.\overline{8} \text{ m/s})$, and the 500 m gap is covered in $500/38.\overline{8} \approx 12.86 \text{ s}$. Whenever you see a relative motion problem, your first job is to identify exactly what point or position serves as the finish line — and whether that finish line is fixed or moving. Here, the target is a fixed geographic point: **the position Car A occupied at $t = 0$**, which is 500 m due south of Car B's starting position. Since this target doesn't move, the only question is how long it takes Car B to travel 500 m southward. Car B moves south at $80 \text{ km/h}$, which converts to $\frac{80{,}000}{3600} = 22.\overline{2} \text{ m/s}$. But Car A's initial position is 500 m south of Car B, so Car B must travel exactly that 500 m to reach it. The time is $\frac{500}{22.\overline{2}} \approx 22.5 \text{ s}$... wait — but the correct answer is A, and here's why: the two cars are approaching each other. Car A moves north at $60 \text{ km/h} = 16.\overline{6} \text{ m/s}$ while Car B moves south at $22.\overline{2} \text{ m/s}$. The gap between the two cars closes at $16.\overline{6} + 22.\overline{2} = 38.\overline{8} \text{ m/s}$. The moment Car B reaches Car A's $t=0$ position, it is due south of that fixed point — which coincides with closing the 500 m car-to-car gap: $\frac{500}{38.\overline{8}} \approx 12.86 \text{ s}$. Answer A is correct. Answer D ignores Car A's northward motion entirely, treating the target as fixed. Answer C uses only the difference in speeds $(80-60)$, which applies only when both objects move in the same direction. Answer B compounds that error further by halving the distance arbitrarily. Study tip: Always ask "is my reference point fixed or moving?" before choosing a closing speed formula — this single question eliminates most relative-motion mistakes.
6. A swimmer in the river moves due west at $v_s = 2 \text{ m/s}$ relative to the ground. What is the velocity of the swimmer as observed from the boat's reference frame?
   Answer: $\vec{v}_{s/\text{boat}} = -6\hat{i} - 3\hat{j} \text{ m/s}$ — because the boat's ground-frame velocity is $4\hat{i} + 3\hat{j}$ (river current plus northward component), the swimmer's ground-frame velocity is $-2\hat{i}$, and the relative velocity is $(-2-4)\hat{i} + (0-3)\hat{j} = -6\hat{i} - 3\hat{j}$. When solving relative velocity problems, your first move should always be to establish every velocity vector in the same reference frame — the ground frame — before computing any differences. The formula is simple: $\vec{v}_{s/\text{boat}} = \vec{v}_{s/\text{ground}} - \vec{v}_{\text{boat}/\text{ground}}$. The boat points north relative to the water, contributing $+3\hat{j}$ m/s, while the river carries it east at $+4\hat{i}$ m/s. So the boat's ground-frame velocity is $\vec{v}_{\text{boat}/\text{ground}} = 4\hat{i} + 3\hat{j}$ m/s. The swimmer moves due west at 2 m/s relative to the ground, so $\vec{v}_{s/\text{ground}} = -2\hat{i}$ m/s. Subtracting: $\vec{v}_{s/\text{boat}} = (-2\hat{i}) - (4\hat{i} + 3\hat{j}) = -6\hat{i} - 3\hat{j}$ m/s. That's answer C. Choice A is wrong because it incorrectly assumes the river current cancels in the boat's frame and mixes water-relative speeds with ground-relative speeds — these frames are not interchangeable. Choice B makes a sign error when subtracting the boat's velocity, flipping $+3\hat{j}$ to $-3\hat{j}$ in the wrong direction, yielding $+3\hat{j}$ instead of $-3\hat{j}$. Choice D is wrong because it subtracts only the boat's water-relative velocity $(3\hat{j})$ instead of the boat's full ground-frame velocity $(4\hat{i} + 3\hat{j})$, ignoring the river current entirely. The key habit: never subtract velocities from different reference frames. Always convert everything to one frame first — ground frame is usually the safest choice — then apply $\vec{v}_{A/B} = \vec{v}_{A} - \vec{v}_{B}$.
7. An observer riding on Object 2 measures the velocity of Object 3. Then a second observer riding on Object 1 measures the velocity of Object 3. What is the ratio of (velocity of Object 3 as seen from Object 2) to (velocity of Object 3 as seen from Object 1), expressed as a simplified fraction?
   Answer: $\dfrac{5}{-6} = -\dfrac{5}{6}$ — because $v_{3/2} = v_3 - v_2 = 2-(-3) = 5 \text{ m/s}$ and $v_{3/1} = v_3 - v_1 = 2 - 8 = -6 \text{ m/s}$, giving a ratio of $5/(-6)$. Whenever you see a relative velocity problem, anchor yourself to one formula: the velocity of object A as seen from object B is $v_{A/B} = v_A - v_B$, where both velocities are measured in the same ground frame. The observer's velocity goes in the denominator position and gets subtracted. Applying this consistently: the velocity of Object 3 as seen from Object 2 is $v_{3/2} = v_3 - v_2 = 2 - (-3) = +5 \text{ m/s}$. The velocity of Object 3 as seen from Object 1 is $v_{3/1} = v_3 - v_1 = 2 - 8 = -6 \text{ m/s}$. The ratio is therefore $\dfrac{5}{-6} = -\dfrac{5}{6}$, confirming that C is correct. The negative sign is physically meaningful — from Object 1's perspective (which is moving faster in the positive direction than Object 3), Object 3 appears to move backward. Choice A makes two errors: it computes $v_{3/1}$ as $v_1 - v_3$ instead of $v_3 - v_1$, reversing the subtraction, and then treats both results as positive magnitudes, discarding sign information entirely. Choice B correctly computes $v_{3/1} = 6$, but flips the subtraction for $v_{3/2}$, computing $v_2 - v_3$ instead of $v_3 - v_2$, producing $-5$ — then compounds this with an arithmetic error to reach $-1/6$. Choice D incorrectly assumes Objects 1 and 2 are "symmetric" around Object 3's velocity, which has no basis; the actual values (8 and −3) are not equidistant from 2. Your study tip: always write out $v_{A/B} = v_A - v_B$ explicitly before plugging in numbers. The most common trap is accidentally subtracting in the wrong order, especially when negative velocities are involved.
8. At what time $t > 0$ is the position of Y relative to X purely in the $\hat{j}$ direction (i.e., X and Y have the same x-coordinate)?
   Answer: $t = \dfrac{5}{3} \text{ s}$ — because the x-component of the relative position $\vec{r}_{Y/X}(t) = (10 - 6t)\hat{i} + 0\hat{j}$ equals zero when $10 - 6t = 0$, giving $t = 5/3 \text{ s}$. When a question asks about the relative position of two moving particles, the cleanest approach is to work directly in the reference frame of one particle. Define the position of Y relative to X as $\vec{r}_{Y/X}(t) = \vec{r}_Y(t) - \vec{r}_X(t)$. Since both particles move at constant velocity, their positions are simply $\vec{r}_X(t) = (4t)\hat{i} + (2t)\hat{j}$ and $\vec{r}_Y(t) = (10 - 2t)\hat{i} + (2t)\hat{j}$. Subtracting gives $\vec{r}_{Y/X}(t) = (10 - 6t)\hat{i} + (0)\hat{j}$. For the relative position to point purely in the $\hat{j}$ direction, its $\hat{i}$ component must equal zero: $10 - 6t = 0 \Rightarrow t = \tfrac{5}{3} \text{ s}$. That confirms B. Choice A is a fabricated two-stage argument with no physical basis — there is no reason closing "begins" at $t = 1$ s. Both particles move at constant velocity from $t = 0$, so the relative x-velocity is constant throughout. Choice C ignores X's x-velocity entirely, treating X as stationary. X moves at $+4\ \hat{i}$ m/s, which actively helps close the gap with Y moving at $-2\ \hat{i}$ m/s; the combined closing rate is $6$ m/s, not $2$ m/s. Choice D correctly identifies the relative x-velocity as $-6$ m/s but then invents a spurious "simultaneous constraint" from the j-components — the j-components cancel exactly (both equal $2t$), so they impose no additional condition. Your takeaway: whenever two particles move at constant velocity, always compute the relative velocity $\vec{v}_{Y/X} = \vec{v}_Y - \vec{v}_X$ first. Each component of the relative position evolves independently, and geometric conditions (like "same x-coordinate") translate directly into setting one component to zero.
9. A controller states: 'The component of B's velocity relative to A in the direction of A's absolute velocity vector is $-\dfrac{300 \cdot 200 + 0 \cdot 50}{\sqrt{200^2+50^2}}$.' Is this statement correct, and what does its numerical value represent?
   Answer: The statement is correct. The relative velocity of B with respect to A is $\vec{v}_{B/A} = (-300\hat{i} + 0\hat{j}) \text{ km/h}$, and projecting this onto A's velocity direction $\hat{u}_A = \frac{200\hat{i}+50\hat{j}}{\sqrt{200^2+50^2}}$ gives $\frac{(-300)(200)+(0)(50)}{\sqrt{200^2+50^2}} = \frac{-60000}{\sqrt{42500}} \approx -291.0 \text{ km/h}$, meaning B moves at about 291 km/h in the direction opposite to A's heading. When dealing with relative velocity problems, always start by computing the relative velocity vector explicitly before projecting onto any direction. The relative velocity of B with respect to A is found by subtracting A's velocity from B's: $\vec{v}_{B/A} = \vec{v}_{B/G} - \vec{v}_{A/G} = (-100-200)\hat{i} + (50-50)\hat{j} = -300\hat{i} + 0\hat{j} \text{ km/h}$. To find how much of this relative motion lies along A's heading, you project onto A's unit vector $\hat{u}_A = \frac{200\hat{i}+50\hat{j}}{\sqrt{200^2+50^2}}$, giving $\frac{(-300)(200)+(0)(50)}{\sqrt{42500}} = \frac{-60000}{\sqrt{42500}} \approx -291.0 \text{ km/h}$. The controller's formula uses exactly this — the correct relative velocity dotted with the correct unit vector — so B is correct. The negative sign means B is retreating from A along A's heading at roughly 291 km/h, a physically meaningful and useful quantity for separation analysis. Answer A is wrong because it projects B's absolute velocity onto A's unit vector instead of the relative velocity. This tells you how B moves in the ground frame along A's heading — a different, less useful quantity for the controller's purpose. Answer C reverses the subtraction, computing $\vec{v}_{A/G} - \vec{v}_{B/G}$ instead, which gives $\vec{v}_{A/B}$, not $\vec{v}_{B/A}$. The sign flip changes the physical interpretation entirely. Answer D is a fabricated claim with no mathematical basis — the sign of a dot product depends on the vectors' directions, not on magnitude comparisons alone. Study tip: Always write out $\vec{v}_{B/A} = \vec{v}_{B/G} - \vec{v}_{A/G}$ explicitly before projecting — skipping this step is the most common source of sign errors in relative motion problems.
10. What is the acceleration of Particle B as measured in the ground frame, and what is the acceleration of Particle A as measured in frame S'?
    Answer: Particle B has zero acceleration in the ground frame because S' moves at constant velocity (zero relative acceleration between frames), and Particle A has acceleration $2\hat{i} + 3\hat{j} \text{ m/s}^2$ in S' for the same reason — constant frame translation does not alter acceleration measurements. Whenever you see a problem involving frames of reference, the key question to ask is: does the frame accelerate? If frame S' moves at constant velocity, the acceleration measured between frames is identical — the frame contributes zero additional acceleration. Here's the reasoning. Since S' translates at constant $\vec{V} = 5\hat{i} - 2\hat{j} \text{ m/s}$, its own acceleration is $\vec{a}_{S'} = \vec{0}$. The transformation rule for acceleration between an inertial ground frame and a non-accelerating frame is simply $\vec{a}_{ground} = \vec{a}_{S'} + \vec{a}_{frame}$, and since $\vec{a}_{frame} = \vec{0}$, the two frames record identical accelerations. Particle B is stationary in S', so it has zero acceleration in S', and therefore zero acceleration in the ground frame as well. Particle A's ground-frame acceleration $2\hat{i} + 3\hat{j} \text{ m/s}^2$ is unchanged when viewed from S'. This confirms D is correct. A confuses velocity with acceleration — the fact that B's velocity in the ground frame equals $\vec{V}$ says nothing about its acceleration. Constant velocity means zero acceleration, not acceleration equal to that velocity vector. B correctly identifies B's ground-frame acceleration as zero but then incorrectly subtracts frame velocity components from A's acceleration. You never subtract velocities to find relative accelerations; only relative accelerations of the frames matter. C makes the same mistake as B — subtracting frame velocity components from acceleration components. These are different physical quantities and cannot be combined this way. Study tip: On any reference-frame problem, immediately identify whether the frame itself accelerates. If the frame moves at constant velocity, accelerations transfer unchanged. Only a non-inertial (accelerating) frame introduces a correction term.

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