3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A passenger sits in a train moving along a straight, level track at constant velocity. She holds a ball above the floor and releases it from rest relative to herself. Air resistance is negligible. Which describes the path of the falling ball as seen by the passenger and by an observer standing beside the track?
Answer and reasoning
AShe sees it drop vertically; the observer sees a parabolaCorrect Relative to the passenger the ball starts from rest and accelerates straight down, so she sees a vertical line. Relative to the ground the ball starts with the train's horizontal velocity, keeps it (with air resistance negligible its only acceleration is g, downward) and accelerates downward, so the observer sees a parabola. The ball lands directly below the release point in the train.
BShe sees it curve rearward; the observer sees it drop vertically A student who thinks a dropped object falls straight down relative to the ground and is left behind by the train picks this. At release the ball already moves forward with the train and keeps that horizontal velocity, so it stays below the passenger's hand as it falls.
CBoth of them see it fall straight down a vertical line A student who thinks a path is the same for every observer, and that a dropped ball falls straight down, picks this. The observer by the track sees the ball move forward with the train's velocity while it falls, so the path the observer sees is a parabola.
DBoth of them see it follow one and the same parabola A student who treats the ground frame's description as the one every observer shares picks this. In the passenger's frame the ball has no horizontal velocity, so she sees it fall straight down; each frame's description is equally valid.
Working Train frame: ball released from rest, acceleration g down → straight down. Ground frame: initial velocity = train's horizontal velocity, acceleration g down → parabola.
A train moves east along a straight, level track at a constant 6.0 m/s relative to the ground. A passenger walks toward the rear of the train at a constant 2.0 m/s relative to the train for 10 s. Taking east as positive, what is the passenger's displacement relative to the ground during this time?
Answer and reasoning
A+80 m A student who adds the two speeds whatever their directions picks this: (6.0 + 2.0) m/s × 10 s. She walks toward the rear, opposite to the train's motion, so the velocities subtract.
B−20 m A student who gives her displacement relative to the train as her displacement relative to the ground picks this. The train carries her 60 m east in the same 10 s, so relative to the ground she ends up 40 m east of where she started.
C−40 m A student who thinks the change of frame alters only the size of the motion, keeping the direction in which she walks (toward the rear, −x), picks this. Relative to the ground her velocity is +4.0 m/s: she moves east, though more slowly than the train.
D+40 mCorrect The passenger's velocity relative to the ground is her velocity relative to the train plus the train's velocity: −2.0 + 6.0 = +4.0 m/s. In 10 s that gives +40 m. Equivalently, the train moves +60 m while she moves −20 m relative to it: 60 − 20 = +40 m.
Working East +. vground = vrel train + vtrain = −2.0 + 6.0 = +4.0 m/s. Δx = 4.0 × 10 = +40 m. (Displacement relative to train: −20 m; train's displacement: +60 m; sum +40 m.)
The diagram shows three vehicles, P, Q and R, moving along a straight road at constant velocities, shown relative to the road. Which ranks the speeds of the three vehicles as measured by an observer riding in vehicle Q, from greatest to least?
Answer and reasoning
AP > Q > R A student who takes the speeds measured relative to the road as the speeds every observer measures picks this. The observer in Q moves at 15 m/s east, so the speeds she measures are found by subtracting Q's velocity: 5 m/s for P, 25 m/s for R, 0 for Q.
BP = R > Q A student who takes each relative speed as the difference of the speeds, whatever the directions, picks this: 20 − 15 = 5 and 15 − 10 = 5 m/s. R moves west while Q moves east, so their velocities subtract to give 10 + 15 = 25 m/s.
CP > R > Q A student who adds the speeds whatever the directions picks this: 20 + 15 = 35 m/s for P and 10 + 15 = 25 m/s for R. P moves in the same direction as Q, so relative to Q its speed is only 20 − 15 = 5 m/s.
DR > P > QCorrect Subtract Q's velocity from each, with east positive: P: 20 − 15 = +5 m/s; R: −10 − 15 = −25 m/s; Q: 0. The observer in Q measures speeds of 25 m/s for R, 5 m/s for P and zero for Q.
Working Take east +. Relative to Q: vP − vQ = 20 − 15 = +5 m/s; vR − vQ = −10 − 15 = −25 m/s; vQ − vQ = 0. Speeds: R 25 > P 5 > Q 0.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
1.4.A.1 Reference frame Fix
Reference frame
A coordinate system, with an origin, axes and a clock, attached to an observer, relative to which positions, displacements, velocities and accelerations are measured. The values of these quantities depend on the frame chosen.
Observer
A person or instrument that makes measurements in a particular reference frame, usually one at rest relative to that observer.
Students often think An object released from a moving vehicle falls straight down relative to the ground and is left behind by the vehicle, so a passenger sees it move toward the rear. In fact No. At release the object already shares the vehicle's horizontal velocity, and in the absence of air resistance it keeps it. Relative to the ground it follows a parabola; relative to a vehicle moving at constant velocity it falls straight down and lands directly below the release point.
Students often think An object's path is a property of the object alone and is the same for every observer; a dropped object falls straight down for everyone. In fact No. The shape of the path depends on the frame. A ball dropped in a train moving at constant velocity falls straight down relative to the train but follows a parabola relative to the ground.
1.4.B.1 Converting positions between frames Fix
Converting positions between frames
If frame B moves with constant velocity v⃗B relative to frame A and their origins coincide at t = 0, a position measured in A is converted to B by r⃗B = r⃗A − v⃗B t; a displacement in A equals the displacement in B plus the displacement of frame B.
Students often think Velocities measured in different frames can be combined in one calculation (for example, an initial velocity in one frame and a final velocity in another) without converting them to a common frame. In fact No. A change in velocity, or any combination of velocities, must use values measured in the same frame; convert one of them first.
1.4.B.2 Inertial reference frame Fix
Inertial reference frame
A reference frame that is not accelerating. Any frame moving at constant velocity relative to an inertial frame is also inertial. Unless otherwise stated, the frame of any problem may be assumed to be inertial.
Relative velocity
The velocity of an object as measured in a given frame. If an object has velocity v⃗OB relative to frame B, and frame B has velocity v⃗BA relative to frame A, the object's velocity relative to A is v⃗OA = v⃗OB + v⃗BA (a vector sum). Unit: m/s.
Combining velocities as vectors
Velocities measured in different frames combine by vector addition or subtraction, component by component. Their magnitudes add or subtract directly only when the vectors are parallel or antiparallel; perpendicular velocities combine by the Pythagorean theorem.
Velocity relative to a medium
The velocity of a boat relative to the water or of an airplane relative to the air. The velocity relative to the ground is this velocity plus the velocity of the water or air relative to the ground.
Acceleration in inertial frames
Observers in inertial frames measure the same acceleration for an object, because their frames differ by a constant velocity, which adds the same amount to every velocity measured and so cancels from every change in velocity: a⃗A = a⃗B.
Students often think The ground (Earth) frame gives the true motion of an object, so every observer measures the velocity and path that an observer at rest on the ground measures. In fact No. The ground is one inertial frame among many. An observer moving at constant velocity relative to the ground measures different velocities and paths, and those measurements are just as valid.
Students often think Velocities measured in different frames combine by adding their magnitudes, whatever their directions. In fact Only if the two velocities point in the same direction. Velocities combine as vectors: opposite directions subtract, and perpendicular velocities combine by the Pythagorean theorem.
5 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 5
The diagram shows a boat crossing a river with a steady current. The boat is steered upstream at an angle so that it moves straight across the river, along the dashed line. How long does the boat take to cross the river?
Answer and reasoning
A12 s A student who uses the boat's speed relative to the water as its speed across relative to the ground picks this: 60 m ÷ 5.0 m/s. Part of that 5.0 m/s is directed upstream to cancel the current, so only 4.0 m/s carries the boat across.
B15 sCorrect The boat's velocity relative to the ground is its velocity relative to the water plus the current's velocity. For the boat to move straight across, the upstream component of its 5.0 m/s must cancel the 3.0 m/s current, leaving a component across of √(5.0² − 3.0²) = 4.0 m/s. Time = 60 m ÷ 4.0 m/s = 15 s.
C30 s A student who subtracts the current's speed from the boat's speed picks this: 60 m ÷ (5.0 − 3.0) m/s. The current is perpendicular to the crossing direction, so the speeds do not simply subtract; the component across is √(5.0² − 3.0²) = 4.0 m/s.
D10 s A student who always combines two speeds as √(v₁² + v₂²) picks this: √(5.0² + 3.0²) = 5.8 m/s, giving about 10 s. The boat's velocity relative to the water is not perpendicular to the current here; it is the hypotenuse, and the speed across is 4.0 m/s.
Working v⃗ground = v⃗rel water + v⃗current. Moving straight across: upstream component of vrel cancels the current, 3.0 m/s; cross component = √(5.0² − 3.0²) = 4.0 m/s. t = 60 m ÷ 4.0 m/s = 15 s.
An airplane flies with a velocity of magnitude v relative to the air, pointing due north. A steady wind blows toward the east at 0.60v relative to the ground. What is the magnitude of the airplane's velocity relative to the ground?
Answer and reasoning
A1.6v A student who adds the two speeds picks this: v + 0.60v. Magnitudes add directly only for parallel velocities; these are perpendicular, so they combine by the Pythagorean theorem.
B1.0v A student who takes the airplane's velocity relative to the air as its velocity relative to the ground picks this, ignoring the wind. The wind carries the air, and the airplane with it, eastward, so the ground velocity is the sum of the two.
C1.2vCorrect The velocity relative to the ground is the vector sum of the velocity relative to the air (v north) and the wind's velocity (0.60v east). These are perpendicular, so the magnitude is √(v² + (0.60v)²) = 1.17v ≈ 1.2v, directed east of north.
D0.8v A student who puts the velocity relative to the air on the hypotenuse of the triangle picks this: √(v² − (0.60v)²) = 0.8v. The velocity relative to the ground is the sum of the two perpendicular velocities, so it is the hypotenuse and is larger than v.
Working v⃗ground = v⃗air + v⃗wind: north v, east 0.60v (perpendicular). |v⃗ground| = √(v² + 0.36v²) = 1.17v ≈ 1.2v.
A ball is dropped inside a train that moves along a straight, level track at constant velocity. A passenger in the train and an observer beside the track each measure the ball's acceleration as it falls, and they get the same value. Which reasoning correctly explains why their measurements agree?
Answer and reasoning
ABoth observers see the ball fall straight down, so both see exactly the same motion A student who thinks the path is the same for every observer picks this. The observer by the track sees the ball move forward with the train while it falls, along a parabola; the two observers see different paths but the same acceleration.
BAcceleration is a scalar, so its value does not depend on the observer's frame at all A student who treats acceleration as a scalar picks this. Acceleration is a vector, Δv⃗/Δt. It agrees between these observers because their frames differ by a constant velocity, not because it lacks a direction.
CTheir frames differ by a constant velocity, which cancels from every change in velocityCorrect Each velocity the observer measures equals the passenger's value plus the train's constant velocity V⃗. In any change in velocity, V⃗ appears in both the final and initial values and cancels, so Δv⃗, and hence a⃗ = Δv⃗/Δt, is the same in both frames.
DBoth observers measure the same velocity for the ball at every instant of its fall A student who thinks every observer measures the velocity measured from the ground picks this. The observer by the track measures an extra horizontal component equal to the train's velocity; only the changes in velocity, and so the accelerations, agree.
Working v⃗ground = v⃗train frame + V⃗ with V⃗ constant: Δv⃗ is the same in both frames, so a⃗ = Δv⃗/Δt is the same. Velocities and paths differ; acceleration agrees.
Car B moves east along a straight road at a constant 20 m/s relative to the road. A passenger in car B measures the velocity of car A, which moves along the same road: at t = 0 it is 5.0 m/s west, and at t = 5.0 s it is 10 m/s east. A's acceleration is constant. What is car A's acceleration as measured by an observer at rest beside the road?
Answer and reasoning
A3.0 m/s² eastCorrect Car B moves at constant velocity, so its frame is inertial and gives the same acceleration as the roadside frame. With east positive, A's velocity changes from −5.0 to +10 m/s in 5.0 s: a = 15/5.0 = 3.0 m/s² east. (Relative to the road, A goes from 15 to 30 m/s east: the same change.)
B6.0 m/s² east A student who converts A's final velocity to the road frame, 30 m/s east, and divides it by the time picks this. Acceleration is the CHANGE in velocity divided by the time: A's velocity relative to the road goes from 15 to 30 m/s east, a change of 15 m/s.
C1.0 m/s² east A student who uses the change in speed, 10 − 5.0 = 5.0 m/s, picks this. A's velocity reverses direction in B's frame, from 5.0 m/s west to 10 m/s east, which is a change of 15 m/s east.
D7.0 m/s² east A student who subtracts the initial velocity measured in B's frame (5.0 m/s west) from the final velocity converted to the road frame (30 m/s east) picks this: 35/5.0. Both velocities must be in the same frame; in either frame the change is 15 m/s.
Working East +. In B's frame: Δv = 10 − (−5.0) = 15 m/s in 5.0 s → a = 3.0 m/s² east. B's frame is inertial (constant velocity), so the roadside observer measures the same: ground velocities 15 → 30 m/s east, Δv = 15 m/s, a = 3.0 m/s² east.
A passenger in a train moving east along a straight, level track at constant speed u tosses a ball straight up relative to herself. Air resistance is negligible. At the instant the ball reaches its highest point, which describes the ball's velocity as measured by the passenger and by an observer standing beside the track?
Answer and reasoning
APassenger: zero; observer: zero as well A student who thinks a tossed ball is at rest at the top for every observer picks this. Only its vertical velocity is zero there; relative to the ground it still has the train's horizontal velocity u east.
BPassenger: zero; observer: u eastwardCorrect In the passenger's frame the ball moves only vertically, so at the top its velocity is zero. The observer by the track adds the train's velocity: the ball keeps the train's horizontal velocity u east throughout, so at the top it is moving at u toward the east.
CPassenger: u westward; observer: at rest A student who thinks a released ball stops sharing the train's motion and is left behind picks this. The ball keeps the horizontal velocity it had at release, u east relative to the ground, so it stays above the passenger's hand and lands back in it.
DPassenger: u eastward; observer: u eastward A student who takes the velocity measured from the ground as the one every observer measures picks this. In the passenger's frame the ball has no horizontal velocity, so at the top she measures zero.
Working Train frame: ball moves vertically; at top v = 0. Ground frame: add the train's velocity u east; at top vy = 0, vx = u → velocity u east.
Compiled from the AP Physics C: Mechanics Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
