3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A train moves forward along a straight, level track at a constant 20 m/s. A passenger walks toward the back of the train. A seated passenger measures her velocity as 1.5 m/s toward the back of the train. A person standing on the platform measures her velocity as 18.5 m/s forward. Which statement about these two measurements is correct?
Answer and reasoning
ABoth are correct, since each is measured relative to a different frame.Correct A velocity is always measured relative to a frame. Relative to the train she moves at 1.5 m/s toward the back; the train carries her forward at 20 m/s, so relative to the platform she moves at −1.5 m/s + 20 m/s = 18.5 m/s forward. Each observer's value is correct for that observer's frame.
BThe platform value is the true one, since the platform is at rest. A student who treats the ground as the one true frame picks this. The platform is at rest only in the ground's own frame; the seated passenger's measurement is just as valid, and it is the velocity that passenger actually observes.
CThe train value is the true one, since it is the velocity of her own walking. A student who thinks the motion an object produces itself is its real motion picks this. Her walking velocity is simply her velocity relative to the train floor; relative to the platform the train's motion must be included, and that value is equally correct.
DThey cannot both be correct, since they give opposite directions. A student who thinks direction of motion belongs to the object itself picks this. Direction is measured relative to a frame: she moves toward the back relative to the train and forward relative to the platform, and both statements are true.
Working Train frame: 1.5 m/s backward (−1.5 m/s, forward positive). Ground frame: −1.5 m/s + 20 m/s = +18.5 m/s. Both values follow from the same motion; each is her velocity relative to a different frame.
A moving walkway in an airport moves at constant velocity vW relative to the floor. A passenger walks on the walkway with velocity vP relative to the walkway. A worker walks on the floor beside the walkway at constant velocity vF relative to the floor. All three velocities are components along the same straight line. Which expression gives the passenger's velocity as measured by the worker?
Answer and reasoning
AvP − vW − vF A student who subtracts the frame's velocity at every change of frame picks this. Converting from the walkway to the floor needs vW ADDED: a passenger standing still on the walkway moves at +vW relative to the floor, not −vW.
BvP + vW + vF A student who always adds velocities picks this. The worker's own velocity must be subtracted: a worker walking beside the passenger at the same velocity should measure zero, which this expression does not give.
CvF − vP − vW A student who subtracts in the wrong order picks this. This is the worker's velocity relative to the passenger, which has the same magnitude as the answer but the opposite sign.
DvP + vW − vFCorrect Relative to the floor, the passenger moves at vP + vW, since the walkway carries everything on it at vW. The worker measures that velocity minus the worker's own velocity relative to the floor: vP + vW − vF. A worker keeping pace with the passenger measures zero, as expected.
Working Step 1, walkway frame to floor frame: add the walkway's velocity, vP,floor = vP + vW. Step 2, floor frame to worker's frame: subtract the worker's velocity, vP,worker = (vP + vW) − vF. Check: if the worker walks beside the passenger at the same floor velocity, vF = vP + vW and the result is 0.
A glass elevator moves upward at a constant 3 m/s. A passenger in the elevator holds a coin at rest relative to the elevator and then releases it. Air resistance is negligible. An observer standing on the ground outside watches the coin until it lands on the elevator floor. Which statement about the coin's acceleration is correct?
Answer and reasoning
AThe ground observer measures it upward at first, while the coin is rising. A student who thinks acceleration points along the motion picks this. The ground observer does see the coin rise at first, but it is slowing down, so its change in velocity, and its acceleration, is downward.
BThe ground observer measures it as zero when the coin stops rising. A student who thinks zero velocity means zero acceleration picks this. At the top of its rise the coin's velocity is zero for only an instant and is still changing, so its acceleration there is still g downward.
CBoth measure the same acceleration, g downward, throughout the flight.Correct The elevator moves at constant velocity, so its frame is inertial. The two observers' velocities for the coin always differ by the same 3 m/s, so the change in velocity each second, and hence the acceleration, is g downward for both, at every instant.
DThe passenger measures a larger one, since the elevator moves upward. A student who thinks acceleration depends on the observer's velocity, like velocity itself, picks this. A constant elevator velocity shifts every measured velocity by the same amount and leaves the acceleration unchanged.
Working Both frames are inertial (ground at rest, elevator at constant velocity). Passenger: coin starts at rest and falls; ground: coin starts at 3 m/s upward, rises 0.45 m, stops momentarily and falls. In both frames the velocity changes by g downward each second, so both measure g downward throughout.
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 attached to an observer, with an origin and a chosen positive direction, relative to which positions, velocities and accelerations are measured. The frame chosen determines the direction and magnitude of the values measured: a passenger seated on a moving train is at rest in the train's frame and moving in the ground's frame.
Observer
The person or instrument making a measurement, at rest in its own reference frame. 'The velocity measured by an observer' means the velocity relative to that observer's frame. Every observer's measurement is correct in that observer's frame; the ground frame is not privileged.
Students often think Whether an object is moving, and its speed and direction, are properties of the object itself: an object at rest is at rest for every observer, and two observers who disagree about its velocity cannot both be right. In fact No. A velocity is always measured relative to a reference frame. Observers in different frames can measure different speeds, and even opposite directions, for the same object, and each measurement is correct in its own frame.
Students often think An object's true velocity is its velocity relative to the ground; any observer, moving or not, measures (or should report) that ground-frame velocity, and a different value is only an appearance. In fact No. The ground is one reference frame among many. A velocity measured from a moving train or car is just as valid, and it is the value that observer actually measures; it generally differs from the ground-frame value.
1.4.B.1 Converting a velocity between frames Fix
Converting a velocity between frames
If an object moves with velocity v1 relative to frame 1, and frame 1 moves with velocity u relative to frame 2 (signed components along the same line), the object's velocity relative to frame 2 is v1 + u. Converting the other way, from frame 2 to frame 1, subtracts u: v1 = v2 − u. SI unit of velocity: m/s.
Students often think Relative velocity always means subtracting, so every change of frame is made by subtracting the frame's velocity, including converting a velocity measured on a walkway or train into a velocity relative to the ground. In fact No. To convert from the moving frame to the ground, ADD the frame's velocity: vground = vmeasured in frame + vframe. Subtraction is the step in the other direction, from the ground frame into the moving frame.
1.4.B.2 Observed (relative) velocity Fix
Observed (relative) velocity
The velocity of an object as measured by a particular observer. It results from combining the object's velocity with the velocity of the observer's frame: with both measured in the same frame (for example the ground), the object's velocity relative to the observer is vobject − vobserver. An observer moving with the object measures zero. SI unit: m/s.
Combining velocities along one line
Velocities along one line are written as signed components, opposite directions having opposite signs. Adding or subtracting the signed components is vector addition or subtraction in one dimension; adding or subtracting speeds without their signs is not, and gives wrong answers whenever the objects move in opposite directions.
Inertial reference frame
A frame that is at rest or moves with constant velocity relative to another inertial frame. In AP Physics 1 the ground is treated as inertial, and so is a train, car, walkway or elevator moving in a straight line at constant velocity; every frame is assumed inertial unless stated otherwise.
Acceleration measured in different inertial frames
Two observers whose frames move at constant velocity relative to each other measure velocities for an object that differ by the same constant amount at every instant. The change in velocity in any interval is therefore the same for both, so they measure the same acceleration, in magnitude and direction. On a velocity–time graph the second observer's line is the first shifted up or down, with the same slope. SI unit: m/s².
Students often think Relative velocities are always found by adding velocities, so the velocity an observer measures is the object's velocity plus the observer's own velocity. In fact No. With both velocities measured in the same frame, the observer measures the object's velocity MINUS the observer's velocity: vobject − vobserver. Adding a frame's velocity is the step used to convert a velocity measured in that moving frame back to the ground.
Students often think The speed of one object relative to another is always the difference of their speeds, whatever their directions. In fact Only when they move in the same direction. With signed components the relative velocity is vobject − vobserver; for objects moving in opposite directions this gives the SUM of their speeds.
4 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 4
A ball rests on level ground. Several observers move along the ground in straight lines at constant velocity. Which observer measures the ball's velocity as 5 m/s to the left?
Answer and reasoning
AAn observer moving left at 5 m/s A student who subtracts in the wrong order, finding the observer's velocity relative to the ball, picks this. An observer moving left at 5 m/s measures 0 − (−5 m/s) = +5 m/s: the ball appears to move to the right.
BNo observer, since the ball is at rest A student who thinks being at rest is a property of the ball itself picks this. The ball is at rest relative to the ground only; an observer moving relative to the ground measures a nonzero velocity for it.
CAn observer moving right at 5 m/sCorrect The velocity an observer measures is the object's velocity minus the observer's velocity. With right positive, 0 − (+5 m/s) = −5 m/s: an observer moving right at 5 m/s sees the ball move left at 5 m/s, just as a passenger sees roadside objects move backward.
DAny observer moving at 5 m/s either way A student who combines speeds and ignores directions picks this. The direction of the observer's motion decides the direction measured: moving right gives 5 m/s to the left, but moving left gives 5 m/s to the right.
Working Ground frame, right positive: ball v = 0. Observer with velocity u measures 0 − u = −u. For −5 m/s (5 m/s to the left), u = +5 m/s: the observer moves right at 5 m/s.
A cyclist rides east at 6 m/s relative to the road. Observer X stands beside the road. Observer Y is in a car that moves east along the same road at a constant 10 m/s. Which statement correctly compares the cyclist's velocity as measured by Y with the velocity measured by X?
Answer and reasoning
AY measures the same velocity as X does, 6 m/s east along the road. A student who takes the ground-frame velocity as the one every observer reports picks this. Y moves, so Y's measurement must include Y's own velocity: 6 m/s − 10 m/s = −4 m/s.
BY measures a smaller speed than X does, and in the opposite direction.Correct Y measures the cyclist's velocity minus Y's own velocity: 6 m/s − 10 m/s = −4 m/s, so 4 m/s west. The car pulls ahead of the cyclist, so from the car the cyclist drops back (moves west) at 4 m/s, a smaller speed than the 6 m/s that X measures.
CY measures a larger speed than X does, and in the same direction. A student who adds the observer's velocity instead of subtracting it picks this, getting 6 m/s + 10 m/s = 16 m/s east. An observer moving the same way as the cyclist must see the cyclist move more slowly, not faster.
DY measures a smaller speed than X does, in the same direction. A student who subtracts in the wrong order, 10 m/s − 6 m/s = +4 m/s, picks this. That is Y's velocity relative to the cyclist. The car is faster, so relative to the car the cyclist moves backward, to the west.
Working East positive, ground frame. X: +6 m/s. Y: vcyclist − vY = 6 − 10 = −4 m/s, i.e. 4 m/s west: a smaller speed than X measures (4 < 6) and the opposite direction.
Car A and truck B move along a straight road. The graph shows their positions x as functions of time t, measured by an observer standing beside the road. What is the velocity of truck B as measured by the driver of car A?
Answer and reasoning
A−40 m/sCorrect The slopes give the ground-frame velocities: vA = 100 m ÷ 4 s = +25 m/s and vB = −60 m ÷ 4 s = −15 m/s. The driver of A measures vB − vA = −15 m/s − 25 m/s = −40 m/s: the vehicles approach each other, so their relative speed is the sum of their speeds.
B+40 m/s A student who subtracts in the wrong order, vA − vB, picks this. That is A's velocity measured by the truck driver. From car A the truck approaches from ahead, moving in the −x direction.
C+10 m/s A student who drops the signs and takes the difference of the speeds, 25 m/s − 15 m/s, picks this. The vehicles move in opposite directions, so with signed velocities the relative velocity is −15 m/s − (+25 m/s) = −40 m/s: the truck approaches car A at 40 m/s.
D−15 m/s A student who reports the truck's velocity relative to the road picks this. That is what the roadside observer measures; the driver of A moves at +25 m/s, and that velocity must be subtracted.
Working Slopes (ground frame): vA = (100 m − 0)/(4 s) = +25 m/s; vB = (140 m − 200 m)/(4 s) = −15 m/s. Velocity of B relative to A: vB − vA = −15 m/s − 25 m/s = −40 m/s.
A cart moves along a straight track, speeding up relative to the track. Observer G stands beside the track. Observer R rides on a platform that moves along the track at constant velocity. The table shows the cart's velocity v measured by each observer at four instants t. Which claim is supported by the data?
Answer and reasoning
AR finds that the cart's acceleration reverses at t = 2.0 s. A student who treats acceleration as the rate of change of speed picks this. R sees the speed fall from 4.0 m/s to 0 and then rise, but R's signed velocity increases by 2.0 m/s every second, so R's measured acceleration is constant and positive throughout.
BR finds the same acceleration for the cart as G, 2.0 m/s².Correct Each observer's velocity rises by 2.0 m/s in each 1.0 s interval: G from 1.0 m/s to 7.0 m/s and R from −4.0 m/s to 2.0 m/s over 3.0 s. Both measure 6.0 m/s ÷ 3.0 s = 2.0 m/s². R's values are all 5.0 m/s lower, a constant shift that leaves every change in velocity unchanged.
CR finds a smaller acceleration, since R moves the same way as the cart. A student who thinks acceleration changes from frame to frame like velocity picks this. R's velocities are smaller, but they change by the same 2.0 m/s each second as G's, so R's acceleration is the same.
DR finds zero acceleration at t = 2.0 s, when it sees the cart at rest. A student who thinks zero velocity means zero acceleration picks this. R's velocity passes through zero at 2.0 s but goes from −2.0 m/s to 0 to 2.0 m/s in equal times, so the acceleration there is still 2.0 m/s².
Working G: Δv = 7.0 − 1.0 = 6.0 m/s in 3.0 s, a = 2.0 m/s²; each second +2.0 m/s. R: Δv = 2.0 − (−4.0) = 6.0 m/s in 3.0 s, a = 2.0 m/s²; each second +2.0 m/s. Every R value is the G value minus 5.0 m/s, so R moves at +5.0 m/s. Same acceleration, constant, positive throughout.
Compiled from the AP Physics 1 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
