Statics and Dynamics Quiz: Relative Motion
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Relative MotionQuestion 1 of 10

A passenger on a train moving east at u=20 m/su = 20 \text{ m/s} throws a ball horizontally toward the front of the train at vrel=5 m/sv_{\text{rel}} = 5 \text{ m/s} relative to the train. A ground observer simultaneously watches the ball.

The ground observer now throws their own ball westward at w=25 m/sw = 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?

45 m/s45 \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.
5 m/s5 \text{ m/s} westward — found by subtracting the train's eastward speed from the ball's westward speed, treating both as positive magnitudes without proper sign convention.
25 m/s25 \text{ m/s} westward — because the passenger is an inertial observer and the ball's speed relative to any inertial frame equals its ground speed, so no frame correction is needed.
20 m/s20 \text{ m/s} eastward — because the passenger's own eastward motion at 20 m/s dominates and reverses the perceived direction of the 25 m/s westward ball, yielding a net eastward perception.
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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Relative Motion

Practice Relative Motion in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Relative Motion, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A passenger on a train moving east at u=20 m/su = 20 \text{ m/s} throws a ball horizontally toward the front of the train at vrel=5 m/sv_{\text{rel}} = 5 \text{ m/s} relative to the train. A ground observer simultaneously watches the ball.

The ground observer now throws their own ball westward at w=25 m/sw = 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?

  1. 45 m/s45 \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. (correct answer)
  2. 5 m/s5 \text{ m/s} westward — found by subtracting the train's eastward speed from the ball's westward speed, treating both as positive magnitudes without proper sign convention.
  3. 25 m/s25 \text{ m/s} westward — because the passenger is an inertial observer and the ball's speed relative to any inertial frame equals its ground speed, so no frame correction is needed.
  4. 20 m/s20 \text{ m/s} eastward — because the passenger's own eastward motion at 20 m/s dominates and reverses the perceived direction of the 25 m/s westward ball, yielding a net eastward perception.
Explanation: 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 m/s+20 \text{ m/s} relative to the ground. The ground observer's ball travels west, so its ground velocity is −25 m/s-25 \text{ m/s}. To find how the passenger sees the ball, apply the relative velocity formula: vball relative to passenger=vball relative to ground−vpassenger relative to groundv_{\text{ball relative to passenger}} = v_{\text{ball relative to ground}} - v_{\text{passenger relative to ground}} =(−25)−(+20)=−45 m/s= (-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 vA/B=vA−vBv_{A/B} = v_A - v_B handles direction automatically — trust the math, not your intuition about which motion "dominates."

Question 2

Frame S' moves at constant velocity V⃗=3i^−4j^ m/s\vec{V} = 3\hat{i} - 4\hat{j} \text{ m/s} relative to the fixed ground frame S. At a given instant, the velocity of a particle as measured in S' is v⃗′=−3i^+4j^ m/s\vec{v}' = -3\hat{i} + 4\hat{j} \text{ m/s}.

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?

  1. Speed in S is 52≈7.07 m/s5\sqrt{2} \approx 7.07 \text{ m/s}; the ground-frame velocity is v⃗=v⃗′−V⃗=(−3−3)i^+(4+4)j^=−6i^+8j^\vec{v} = \vec{v}' - \vec{V} = (-3-3)\hat{i} + (4+4)\hat{j} = -6\hat{i} + 8\hat{j}, so the speed is 36+64=10 m/s\sqrt{36+64} = 10 \text{ m/s}, but a sign correction for frame orientation reduces this to 525\sqrt{2}.
  2. Speed in S is 5 m/s5 \text{ m/s}; the particle retains its speed in S' because relative speed between frames is always preserved under Galilean transformation, so ∣v⃗∣=∣v⃗′∣=9+16=5 m/s|\vec{v}| = |\vec{v}'| = \sqrt{9+16} = 5 \text{ m/s}.
  3. Speed in S is 10 m/s10 \text{ m/s}; the absolute velocity in S is found by adding the magnitudes of the frame velocity and particle velocity, giving ∣V⃗∣+∣v⃗′∣=5+5=10 m/s|\vec{V}| + |\vec{v}'| = 5 + 5 = 10 \text{ m/s}.
  4. Speed in S is 0 m/s0 \text{ m/s}; the particle is momentarily stationary in S because v⃗=v⃗′+V⃗=(−3+3)i^+(4−4)j^=0⃗\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. (correct answer)
Explanation: When a problem involves two reference frames, your first instinct should be to apply the Galilean velocity transformation: v⃗=v⃗′+V⃗\vec{v} = \vec{v}' + \vec{V}, where v⃗\vec{v} is the particle's velocity in the ground frame S, v⃗′\vec{v}' is its velocity in the moving frame S', and V⃗\vec{V} is the velocity of S' relative to S. Crucially, you add the vector frame velocity — not the scalar speed. Plugging in directly: v⃗=(−3i^+4j^)+(3i^−4j^)=(0)i^+(0)j^=0⃗\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 V⃗\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: ∣v⃗′+V⃗∣≠∣v⃗′∣+∣V⃗∣|\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 v⃗=v⃗′+V⃗\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.

Question 3

At time t=0t = 0, Particle X is at position r⃗X=0i^+0j^ m\vec{r}_X = 0\hat{i} + 0\hat{j} \text{ m} moving with velocity v⃗X=4i^+2j^ m/s\vec{v}_X = 4\hat{i} + 2\hat{j} \text{ m/s}. Particle Y is at position r⃗Y=10i^+0j^ m\vec{r}_Y = 10\hat{i} + 0\hat{j} \text{ m} moving with velocity v⃗Y=−2i^+2j^ m/s\vec{v}_Y = -2\hat{i} + 2\hat{j} \text{ m/s}. Both move at constant velocity.

At what time t>0t > 0 is the position of Y relative to X purely in the j^\hat{j} direction (i.e., X and Y have the same x-coordinate)?

  1. t=2.5 st = 2.5 \text{ s} — because X moves in the positive x-direction, initially increasing the gap before Y's westward velocity closes it; the effective closing begins only after t=1 st = 1 \text{ s}, leaving a reduced gap of 4 m to be covered at Y's speed of 2 m/s2 \text{ m/s}, yielding t=1+4/2=2.5 st = 1 + 4/2 = 2.5 \text{ s}.
  2. t=53 st = \dfrac{5}{3} \text{ s} — because the x-component of the relative position r⃗Y/X(t)=(10−6t)i^+0j^\vec{r}_{Y/X}(t) = (10 - 6t)\hat{i} + 0\hat{j} equals zero when 10−6t=010 - 6t = 0, giving t=5/3 st = 5/3 \text{ s}. (correct answer)
  3. t=5 st = 5 \text{ s} — because only Y's x-velocity of −2 m/s-2 \text{ m/s} reduces the 10 m separation; X's positive x-velocity moves X further away from Y's initial position, so the x-separation grows before eventually closing, and the net closing rate is just ∣vYx∣=2 m/s|v_{Yx}| = 2 \text{ m/s}, giving t=10/2=5 st = 10/2 = 5 \text{ s}.
  4. t=103 st = \dfrac{10}{3} \text{ s} — because the relative x-velocity is vYx−vXx=−2−4=−6 m/sv_{Yx} - v_{Xx} = -2 - 4 = -6 \text{ m/s}, but since the j-components of X and Y are equal, the j-direction relative position is nonzero, requiring the additional constraint vYj−vXj=0v_{Yj} - v_{Xj} = 0 to be solved simultaneously, which doubles the time to 10/3 s10/3 \text{ s}.
Explanation: 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 r⃗Y/X(t)=r⃗Y(t)−r⃗X(t)\vec{r}_{Y/X}(t) = \vec{r}_Y(t) - \vec{r}_X(t). Since both particles move at constant velocity, their positions are simply r⃗X(t)=(4t)i^+(2t)j^\vec{r}_X(t) = (4t)\hat{i} + (2t)\hat{j} and r⃗Y(t)=(10−2t)i^+(2t)j^\vec{r}_Y(t) = (10 - 2t)\hat{i} + (2t)\hat{j}. Subtracting gives r⃗Y/X(t)=(10−6t)i^+(0)j^\vec{r}_{Y/X}(t) = (10 - 6t)\hat{i} + (0)\hat{j}. For the relative position to point purely in the j^\hat{j} direction, its i^\hat{i} component must equal zero: 10−6t=0⇒t=53 s10 - 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=1t = 1 s. Both particles move at constant velocity from t=0t = 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 i^+4\ \hat{i} m/s, which actively helps close the gap with Y moving at −2 i^-2\ \hat{i} m/s; the combined closing rate is 66 m/s, not 22 m/s. Choice D correctly identifies the relative x-velocity as −6-6 m/s but then invents a spurious "simultaneous constraint" from the j-components — the j-components cancel exactly (both equal 2t2t), so they impose no additional condition. Your takeaway: whenever two particles move at constant velocity, always compute the relative velocity v⃗Y/X=v⃗Y−v⃗X\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.

Question 4

In a 2D plane, point P moves along the x-axis with velocity v⃗P=6i^ m/s\vec{v}_P = 6\hat{i} \text{ m/s}. Point Q moves in the same plane with velocity v⃗Q=−4i^+3j^ m/s\vec{v}_Q = -4\hat{i} + 3\hat{j} \text{ m/s}.

An engineer claims that the speed of Q as seen from P equals 101≈10.05 m/s\sqrt{101} \approx 10.05 \text{ m/s}. Which of the following correctly identifies whether this claim is right or wrong, and why?

  1. The claim is wrong. The relative velocity is v⃗Q/P=v⃗Q−v⃗P=(−4−6)i^+3j^=−10i^+3j^\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 100+9=109≈10.44 m/s\sqrt{100+9} = \sqrt{109} \approx 10.44 \text{ m/s}, not 101\sqrt{101}. (correct answer)
  2. The claim is correct. The relative velocity is v⃗Q/P=v⃗Q+v⃗P=2i^+3j^\vec{v}_{Q/P} = \vec{v}_Q + \vec{v}_P = 2\hat{i} + 3\hat{j}, giving a speed of 4+9=13\sqrt{4+9} = \sqrt{13}, which rounds to 101\sqrt{101} when unit conversions are applied.
  3. The claim is wrong. The engineer should have computed v⃗Q/P=v⃗Q−v⃗P=−10i^+3j^\vec{v}_{Q/P} = \vec{v}_Q - \vec{v}_P = -10\hat{i} + 3\hat{j}, giving 109\sqrt{109}; however, the correct answer is actually 101\sqrt{101} once the j-component is projected onto P's direction of motion, so the engineer's formula was right but applied to the wrong projection.
  4. The claim is correct. Relative speed between two points is a scalar invariant, so it equals vQx2+vQy2=16+9=5 m/s\sqrt{v_{Qx}^2 + v_{Qy}^2} = \sqrt{16+9} = 5 \text{ m/s}, and the engineer's value of 101\sqrt{101} correctly accounts for both frames by adding this to P's speed: 52+(76)2=101\sqrt{5^2 + (\sqrt{76})^2} = \sqrt{101}.
Explanation: Whenever you encounter a question about relative velocity, anchor yourself to one core formula: the velocity of Q as seen from P is v⃗Q/P=v⃗Q−v⃗P\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: v⃗Q/P=(−4i^+3j^)−(6i^)=−10i^+3j^\vec{v}_{Q/P} = (-4\hat{i} + 3\hat{j}) - (6\hat{i}) = -10\hat{i} + 3\hat{j}. The speed (magnitude) is then ∣v⃗Q/P∣=(−10)2+32=100+9=109≈10.44 m/s|\vec{v}_{Q/P}| = \sqrt{(-10)^2 + 3^2} = \sqrt{100 + 9} = \sqrt{109} \approx 10.44 \text{ m/s}, not 101\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 13\sqrt{13} into 101\sqrt{101}; that reasoning is fabricated. C is a trap that dresses up a correct intermediate step (computing −10i^+3j^-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 (∣v⃗Q∣=5 m/s|\vec{v}_Q| = 5 \text{ m/s}) with relative speed, then constructs an arbitrary Pythagorean combination to manufacture 101\sqrt{101}. Relative speed is not a scalar invariant independent of the observer's velocity. Study tip: Always write out v⃗Q/P=v⃗Q−v⃗P\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.

Question 5

A river flows due east at u=4 m/su = 4 \text{ m/s} relative to the ground. A boat capable of moving at vb=3 m/sv_b = 3 \text{ m/s} relative to the water attempts to cross the river heading due north (i.e., the boat points north relative to the water).

A swimmer in the river moves due west at vs=2 m/sv_s = 2 \text{ m/s} relative to the ground. What is the velocity of the swimmer as observed from the boat's reference frame?

  1. v⃗s/boat=2i^−3j^ m/s\vec{v}_{s/\text{boat}} = 2\hat{i} - 3\hat{j} \text{ m/s} — because from the boat, the river current cancels out and only the swimmer's 2 m/s westward motion relative to the water and the boat's 3 m/s northward water-relative speed remain.
  2. v⃗s/boat=−6i^+3j^ m/s\vec{v}_{s/\text{boat}} = -6\hat{i} + 3\hat{j} \text{ m/s} — because the swimmer's velocity relative to the boat is computed as v⃗s−v⃗boat=−2i^−(4i^−3j^)=−6i^+3j^\vec{v}_s - \vec{v}_{\text{boat}} = -2\hat{i} - (4\hat{i} - 3\hat{j}) = -6\hat{i} + 3\hat{j}, where the boat's northward velocity appears as −3j^-3\hat{j} in the subtraction.
  3. v⃗s/boat=−6i^−3j^ m/s\vec{v}_{s/\text{boat}} = -6\hat{i} - 3\hat{j} \text{ m/s} — because the boat's ground-frame velocity is 4i^+3j^4\hat{i} + 3\hat{j} (river current plus northward component), the swimmer's ground-frame velocity is −2i^-2\hat{i}, and the relative velocity is (−2−4)i^+(0−3)j^=−6i^−3j^(-2-4)\hat{i} + (0-3)\hat{j} = -6\hat{i} - 3\hat{j}. (correct answer)
  4. v⃗s/boat=−2i^−3j^ m/s\vec{v}_{s/\text{boat}} = -2\hat{i} - 3\hat{j} \text{ m/s} — because the swimmer moves west at 2 m/s in the ground frame and the boat moves north at 3 m/s relative to the water, and subtracting only the boat's water-relative velocity from the swimmer's ground-frame velocity gives the relative motion.
Explanation: 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: v⃗s/boat=v⃗s/ground−v⃗boat/ground\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 +3j^+3\hat{j} m/s, while the river carries it east at +4i^+4\hat{i} m/s. So the boat's ground-frame velocity is v⃗boat/ground=4i^+3j^\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 v⃗s/ground=−2i^\vec{v}_{s/\text{ground}} = -2\hat{i} m/s. Subtracting: v⃗s/boat=(−2i^)−(4i^+3j^)=−6i^−3j^\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 +3j^+3\hat{j} to −3j^-3\hat{j} in the wrong direction, yielding +3j^+3\hat{j} instead of −3j^-3\hat{j}. Choice D is wrong because it subtracts only the boat's water-relative velocity (3j^)(3\hat{j}) instead of the boat's full ground-frame velocity (4i^+3j^)(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 v⃗A/B=v⃗A−v⃗B\vec{v}_{A/B} = \vec{v}_{A} - \vec{v}_{B}.

Question 6

Particle A has acceleration a⃗A=2i^+3j^ m/s2\vec{a}_A = 2\hat{i} + 3\hat{j} \text{ m/s}^2 as measured in the ground (inertial) frame. Frame S' translates relative to the ground with constant velocity V⃗=5i^−2j^ m/s\vec{V} = 5\hat{i} - 2\hat{j} \text{ m/s}. Particle B is stationary in frame S'.

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'?

  1. Particle B has acceleration equal to V⃗=5i^−2j^ m/s2\vec{V} = 5\hat{i} - 2\hat{j} \text{ m/s}^2 in the ground frame since its velocity in the ground frame equals V⃗\vec{V}, and Particle A has acceleration 2i^+3j^ m/s22\hat{i} + 3\hat{j} \text{ m/s}^2 in S' because acceleration is an absolute quantity unchanged between any two frames.
  2. Particle B has acceleration 5i^−2j^ m/s25\hat{i} - 2\hat{j} \text{ m/s}^2 in the ground frame because it must match S' motion, and Particle A has acceleration −3i^+5j^ m/s2-3\hat{i} + 5\hat{j} \text{ m/s}^2 in S' because the frame's own acceleration is subtracted component-wise.
  3. Particle B has zero acceleration in the ground frame because it is stationary in S', which moves at constant velocity, and Particle A has acceleration (2−5)i^+(3+2)j^=−3i^+5j^ m/s2(2-5)\hat{i} + (3+2)\hat{j} = -3\hat{i} + 5\hat{j} \text{ m/s}^2 in S' because the frame velocity components are subtracted from A's ground-frame acceleration.
  4. 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 2i^+3j^ m/s22\hat{i} + 3\hat{j} \text{ m/s}^2 in S' for the same reason — constant frame translation does not alter acceleration measurements. (correct answer)
Explanation: 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 V⃗=5i^−2j^ m/s\vec{V} = 5\hat{i} - 2\hat{j} \text{ m/s}, its own acceleration is a⃗S′=0⃗\vec{a}_{S'} = \vec{0}. The transformation rule for acceleration between an inertial ground frame and a non-accelerating frame is simply a⃗ground=a⃗S′+a⃗frame\vec{a}_{ground} = \vec{a}_{S'} + \vec{a}_{frame}, and since a⃗frame=0⃗\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 2i^+3j^ m/s22\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 V⃗\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.

Question 7

Car A travels north at 60 km/h60 \text{ km/h} on a straight highway. Car B travels south on a parallel lane at 80 km/h80 \text{ km/h}. At a certain instant (t=0t = 0), Car B is 500 m due north of Car A.

How long (in seconds) does it take for Car B to reach a position due south of Car A's position at t=0t = 0, assuming both cars maintain constant velocity?

  1. 12.86 s12.86 \text{ s} — because the two cars move toward each other, so the gap closes at the sum of both speeds (60+80=140 km/h=38.8‾ m/s)(60+80 = 140 \text{ km/h} = 38.\overline{8} \text{ m/s}), and the 500 m gap is covered in 500/38.8‾≈12.86 s500/38.\overline{8} \approx 12.86 \text{ s}. (correct answer)
  2. 45.0 s45.0 \text{ s} — because only Car B's speed in excess of Car A's matters for closing the gap; the relative speed is 80−60=20 km/h80 - 60 = 20 \text{ km/h}, but since they travel on parallel lanes, half the separation is attributed to each car, giving an effective closing distance of 250 m at 5.56 m/s5.56 \text{ m/s}.
  3. 90 s90 \text{ s} — because only Car B's excess speed relative to Car A matters; the closing rate is 80−60=20 km/h=5.56 m/s80 - 60 = 20 \text{ km/h} = 5.56 \text{ m/s}, and the initial 500 m separation yields 500/5.56≈90 s500/5.56 \approx 90 \text{ s}.
  4. 18.0 s18.0 \text{ s} — because the relevant speed for reducing the gap is Car B's absolute ground speed of 80 km/h=22.22 m/s80 \text{ km/h} = 22.22 \text{ m/s}, while Car A's northward motion is irrelevant since the target is a fixed point (A's position at t=0t = 0).
Explanation: 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=0t = 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 km/h80 \text{ km/h}, which converts to 80,0003600=22.2‾ m/s\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 50022.2‾≈22.5 s\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 km/h=16.6‾ m/s60 \text{ km/h} = 16.\overline{6} \text{ m/s} while Car B moves south at 22.2‾ m/s22.\overline{2} \text{ m/s}. The gap between the two cars closes at 16.6‾+22.2‾=38.8‾ m/s16.\overline{6} + 22.\overline{2} = 38.\overline{8} \text{ m/s}. The moment Car B reaches Car A's t=0t=0 position, it is due south of that fixed point — which coincides with closing the 500 m car-to-car gap: 50038.8‾≈12.86 s\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)(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.

Question 8

Three objects move along the x-axis. Object 1 moves at v1=+8 m/sv_1 = +8 \text{ m/s}, Object 2 moves at v2=−3 m/sv_2 = -3 \text{ m/s}, and Object 3 moves at v3=+2 m/sv_3 = +2 \text{ m/s}, all measured in the ground frame.

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?

  1. 56\dfrac{5}{6} — because v3/2=v3−v2=5 m/sv_{3/2} = v_3 - v_2 = 5 \text{ m/s} and v3/1=v1−v3=6 m/sv_{3/1} = v_1 - v_3 = 6 \text{ m/s}, treating both relative velocities as positive magnitudes and dividing.
  2. −16\dfrac{-1}{6} — because v3/2=v2−v3=−3−2=−5 m/sv_{3/2} = v_2 - v_3 = -3-2 = -5 \text{ m/s} and v3/1=v1−v3=8−2=6 m/sv_{3/1} = v_1 - v_3 = 8-2 = 6 \text{ m/s}, so the ratio is −5/6-5/6, but a sign error in the numerator yields −1/6-1/6 after cancellation.
  3. 5−6=−56\dfrac{5}{-6} = -\dfrac{5}{6} — because v3/2=v3−v2=2−(−3)=5 m/sv_{3/2} = v_3 - v_2 = 2-(-3) = 5 \text{ m/s} and v3/1=v3−v1=2−8=−6 m/sv_{3/1} = v_3 - v_1 = 2 - 8 = -6 \text{ m/s}, giving a ratio of 5/(−6)5/(-6). (correct answer)
  4. −55=−1\dfrac{-5}{5} = -1 — because the velocity of Object 3 relative to Object 2 is +5 m/s+5 \text{ m/s} and relative to Object 1 is −5 m/s-5 \text{ m/s}, since Objects 1 and 2 are symmetric about Object 3's velocity.
Explanation: Whenever you see a relative velocity problem, anchor yourself to one formula: the velocity of object A as seen from object B is vA/B=vA−vBv_{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 v3/2=v3−v2=2−(−3)=+5 m/sv_{3/2} = v_3 - v_2 = 2 - (-3) = +5 \text{ m/s}. The velocity of Object 3 as seen from Object 1 is v3/1=v3−v1=2−8=−6 m/sv_{3/1} = v_3 - v_1 = 2 - 8 = -6 \text{ m/s}. The ratio is therefore 5−6=−56\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 v3/1v_{3/1} as v1−v3v_1 - v_3 instead of v3−v1v_3 - v_1, reversing the subtraction, and then treats both results as positive magnitudes, discarding sign information entirely. Choice B correctly computes v3/1=6v_{3/1} = 6, but flips the subtraction for v3/2v_{3/2}, computing v2−v3v_2 - v_3 instead of v3−v2v_3 - v_2, producing −5-5 — then compounds this with an arithmetic error to reach −1/6-1/6. 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 vA/B=vA−vBv_{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.

Question 9

An aircraft flies at v⃗A/G=200i^+50j^ km/h\vec{v}_{A/G} = 200\hat{i} + 50\hat{j} \text{ km/h} relative to the ground. A second aircraft B has velocity v⃗B/G=−100i^+50j^ km/h\vec{v}_{B/G} = -100\hat{i} + 50\hat{j} \text{ km/h} relative to the ground. Both aircraft are at the same altitude.

A controller states: 'The component of B's velocity relative to A in the direction of A's absolute velocity vector is −300⋅200+0⋅502002+502-\dfrac{300 \cdot 200 + 0 \cdot 50}{\sqrt{200^2+50^2}}.' Is this statement correct, and what does its numerical value represent?

  1. The statement is incorrect because the projection should use B's velocity vector (not the relative velocity) dotted with A's unit vector; the correct component is (−100)(200)+(50)(50)2002+502=−1750042500≈−84.9 km/h\frac{(-100)(200)+(50)(50)}{\sqrt{200^2+50^2}} = \frac{-17500}{\sqrt{42500}} \approx -84.9 \text{ km/h}, representing how fast B moves along A's heading in the ground frame.
  2. The statement is correct. The relative velocity of B with respect to A is v⃗B/A=(−300i^+0j^) km/h\vec{v}_{B/A} = (-300\hat{i} + 0\hat{j}) \text{ km/h}, and projecting this onto A's velocity direction u^A=200i^+50j^2002+502\hat{u}_A = \frac{200\hat{i}+50\hat{j}}{\sqrt{200^2+50^2}} gives (−300)(200)+(0)(50)2002+502=−6000042500≈−291.0 km/h\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. (correct answer)
  3. The statement is incorrect because the relative velocity of B with respect to A is v⃗B/A=v⃗A−v⃗B=300i^+0j^\vec{v}_{B/A} = \vec{v}_A - \vec{v}_B = 300\hat{i} + 0\hat{j}, and the correct projection onto A's unit vector is (300)(200)+(0)(50)2002+502≈+291.0 km/h\frac{(300)(200)+(0)(50)}{\sqrt{200^2+50^2}} \approx +291.0 \text{ km/h}, which is positive, indicating B moves in the same direction as A's heading.
  4. The statement is correct in formula but incorrect in sign; the relative velocity component should be positive because B's x-velocity magnitude exceeds its y-velocity, making the net projection along A's heading positive rather than negative.
Explanation: 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: v⃗B/A=v⃗B/G−v⃗A/G=(−100−200)i^+(50−50)j^=−300i^+0j^ km/h\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 u^A=200i^+50j^2002+502\hat{u}_A = \frac{200\hat{i}+50\hat{j}}{\sqrt{200^2+50^2}}, giving (−300)(200)+(0)(50)42500=−6000042500≈−291.0 km/h\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 v⃗A/G−v⃗B/G\vec{v}_{A/G} - \vec{v}_{B/G} instead, which gives v⃗A/B\vec{v}_{A/B}, not v⃗B/A\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 v⃗B/A=v⃗B/G−v⃗A/G\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.

Question 10

Two ships, A and B, move in the same ocean region. Ship A moves due north at vA=12 m/sv_A = 12 \text{ m/s}, while Ship B moves due east at vB=9 m/sv_B = 9 \text{ m/s}. An observer on Ship A wishes to determine the velocity of Ship B as seen from Ship A.

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?

  1. 15 m/s15 \text{ m/s}, directed east of south — because the relative velocity vector has components +9i^+9\hat{i} (east) and −12j^-12\hat{j} (south) when the observer's own northward velocity is subtracted from B's velocity. (correct answer)
  2. 15 m/s15 \text{ m/s}, directed east of north — because the relative velocity is found by adding both velocity vectors, giving components +9i^+9\hat{i} and +12j^+12\hat{j}, whose resultant points northeast.
  3. 225=15 m/s\sqrt{225} = 15 \text{ m/s}, directed west of south — because the observer on A perceives B moving opposite to A's own heading, producing components −9i^-9\hat{i} and −12j^-12\hat{j} that point southwest.
  4. 45≈6.7 m/s\sqrt{45} \approx 6.7 \text{ m/s}, directed east of south — because only the difference in the eastward components matters; the northward component of A cancels with itself and does not contribute to the relative speed.
Explanation: Whenever you see a relative velocity problem, anchor yourself to this formula: v⃗B/A=v⃗B−v⃗A\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 +i^+\hat{i} and north as +j^+\hat{j}, Ship B's velocity is v⃗B=9i^\vec{v}_B = 9\hat{i} and Ship A's velocity is v⃗A=12j^\vec{v}_A = 12\hat{j}. Applying the formula: v⃗B/A=9i^−12j^\vec{v}_{B/A} = 9\hat{i} - 12\hat{j}. This gives a vector pointing east (+9+9) and south (−12-12), meaning B appears to move toward the southeast from A's perspective. The magnitude is 92+122=81+144=225=15 m/s\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 vBv_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.