Unit 1: worked solutions

Find the first leg: Δx₁ = +5 − (−2) = +7 m.

Why this step: A displacement uses final minus initial position; subtracting the negative starting position adds 2 m.

Find the return leg: Δx₂ = +1 − (+5) = −4 m.

Why this step: The robot ends left of its turnaround point, so this leg has a negative signed change.

Total displacement is +7 − 4 = +3 m; distance is |7| + |−4| = 11 m.

Why this step: Vector changes combine with signs, whereas every traveled leg contributes a positive length.

Check the endpoint result: +1 − (−2) = +3 m.

Why this step: The net displacement must depend only on the start and end; agreement catches a sign error in either leg.

From 0 to 1 s: v_avg = (2 − 0)/(1 − 0) = 2 m/s.

Why this step: These are the position change and duration of the first interval, not of the entire record.

From 1 to 3 s: v_avg = (8 − 2)/(3 − 1) = 3 m/s.

Why this step: The second displacement is 6 m over 2 s; using 3 s here would incorrectly include the earlier interval.

For the full trip: v_avg = (8 − 0)/(3 − 0) = 8/3 ≈ 2.67 m/s.

Why this step: A whole-trip average must use whole-trip displacement and time, so the longer interval carries more weight.

Equivalently, (2·1 + 3·2)/(1 + 2) = 8/3 m/s.

Why this step: Time weighting adds the two interval displacements; the unweighted mean 2.5 m/s assumes equal durations that were not given.

Draw a 4 m/s horizontal segment from 0 to 2 s, then a straight line reaching 0 m/s at 4 s.

Why this step: The first interval has fixed velocity; the stated linear fall in the next interval sets a constant negative slope.

The first displacement is the rectangle area: (4 m/s)(2 s) = 8 m.

Why this step: Velocity is positive throughout this interval, so its area contributes positive displacement.

The second displacement is the triangle area: ½(2 s)(4 m/s) = 4 m. Total displacement is 12 m.

Why this step: The velocity falls but remains nonnegative, so this area adds to rather than subtracts from the first displacement.

Acceleration is 0 for 0–2 s and (0 − 4)/(4 − 2) = −2 m/s² for 2–4 s.

Why this step: Acceleration comes from each segment’s slope; the first is horizontal and the second loses 4 m/s in 2 s.

Keep the walkway’s +1.5 m/s velocity and add the traveler’s signed walkway-relative velocity.

Why this step: The walkway carries its own origin through the building, while the traveler also moves within that origin’s frame.

Walking forward gives v_building = +2.0 + 1.5 = +3.5 m/s.

Why this step: The two motions share a direction, so they reinforce rather than cancel.

Walking backward gives v_building = −2.0 + 1.5 = −0.5 m/s.

Why this step: Turning reverses only the traveler’s walkway-relative component; the walkway still moves forward at 1.5 m/s.

The final result means 0.5 m/s backward through the building.

Why this step: The backward walking magnitude exceeds the walkway’s forward speed, making the building-relative resultant negative.

Resolve v₀ₓ = 10 cos 37° = 8 m/s and v₀ᵧ = 10 sin 37° = 6 m/s.

Why this step: The angle is measured above horizontal, so the adjacent component uses cosine and the upward component uses sine.

At the peak, vᵧ = 0 while vₓ = 8 m/s; speed is therefore 8 m/s.

Why this step: Gravity removes the upward component but no horizontal acceleration removes the 8 m/s component.

Use 0 = 6 − 10t to find t = 0.60 s.

Why this step: With upward positive, gravity is −10 m/s²; the peak occurs when that change cancels the initial +6 m/s.

Draw a 4 cm arrow one way, then a 4 cm arrow in the opposite direction, head to tail.

Why this step: The stated scale turns each 4 m displacement into 4 cm; equal opposite arrows bring the final head back to the first tail.

The resultant is zero, so final position equals initial position.

Why this step: The resultant joins the first tail to the final head, which coincide; total traveled distance is still 8 m.

Travel time cannot be determined from these arrows.

Why this step: Their lengths represent meters, not seconds; many different speeds could produce the same out-and-back displacement.

With right positive, positive acceleration adds a positive change to the velocity.

Why this step: A single fixed axis lets us compare acceleration with each cart’s initial motion rather than with its position.

For a rightward cart, positive velocity grows more positive, so speed increases.

Why this step: The velocity and acceleration point together, increasing the velocity magnitude.

For a leftward cart, negative velocity initially moves toward zero, so speed decreases. It can later reverse and speed up rightward.

Why this step: Acceleration initially opposes the motion; after velocity passes through zero, it instead points with the new rightward motion.

For a straight v–t line, the signed area over Δt is a rectangle v_iΔt plus a triangle ½(v_f − v_i)Δt.

Why this step: Straight-line interpolation makes the excess velocity above v_i a triangle; a curved graph would not have that triangular area.

Combine terms: Δx = [v_i + ½(v_f − v_i)]Δt = ½(v_i + v_f)Δt.

Why this step: Both geometric pieces span the same interval, so their algebra reduces to the endpoint average velocity times duration.

The condition is constant acceleration during this interval.

Why this step: A fixed slope is exactly a fixed acceleration; endpoint averaging is not generally valid for a curved velocity graph.

Use v′(t) = v_ground(t) − 2 m/s at every plotted time.

Why this step: The moving observer has a constant +2 m/s velocity, so its frame transformation is one fixed vertical graph shift.

The endpoints become 4 − 2 = 2 m/s and 10 − 2 = 8 m/s over the same 3 s.

Why this step: Each endpoint must be transformed with the same observer velocity; changing only one would invent a different acceleration.

Both slopes equal (10 − 4)/3 = (8 − 2)/3 = 2 m/s².

Why this step: Subtracting equal offsets leaves the numerator unchanged, and inertial frames share the interval in this model.

Draw x–t as a rising straight line and vₓ–t as a positive horizontal line.

Why this step: A horizontal launch with negligible drag has one constant positive horizontal velocity, equal to the slope of x–t.

Draw y–t starting with zero slope and bending downward; draw vᵧ–t as a line falling from zero with slope −g.

Why this step: With upward positive and no initial vertical motion, constant downward acceleration makes vertical velocity increasingly negative.

Match each position graph’s local slope to the corresponding velocity graph’s height.

Why this step: That check catches a mistaken straight y–t graph or changing vₓ graph, neither of which fits these component accelerations.

Negative acceleration means that the signed velocity is becoming more negative.

Why this step: Acceleration describes a velocity change along the chosen axis, so its sign alone does not describe the velocity already present.

For v = +5 m/s and a = −1 m/s², one second later v = +4 m/s and speed falls from 5 to 4 m/s.

Why this step: Here acceleration opposes the initial velocity, reducing its magnitude before any reversal.

For v = −5 m/s with the same acceleration, one second later v = −6 m/s and speed rises from 5 to 6 m/s.

Why this step: Here acceleration points with the initial motion, so the magnitude grows even though acceleration remains negative.

Draw the secant joining the common start and end points. Its slope is the same for both curves.

Why this step: The two records have identical full-interval displacement and duration, which are the two inputs to average velocity.

At a chosen time, compare the tangent slopes. The straight line has one fixed slope; the upward-bending curve has changing slope.

Why this step: A local tangent uses nearby motion rather than just the endpoints, so it reveals differences that a full-interval average discards.

Both trips can therefore have the same average velocity and different instantaneous velocities.

Why this step: Agreement of an endpoint ratio does not require agreement at every intermediate time.

With downward positive, the net force is mg − F_drag and a = g − F_drag/m.

Why this step: For a downward-falling ball, drag points upward, so increasing drag subtracts from the gravitational force.

Sketch v positive and increasing with a progressively smaller slope.

Why this step: As long as mg exceeds drag, acceleration remains downward, but its declining magnitude makes velocity grow more slowly.

Sketch a starting near +g and decreasing toward zero as drag approaches mg.

Why this step: At terminal motion the opposing forces balance, giving zero acceleration while downward velocity stays nonzero.

A constant +g line would omit the increasing drag.

Why this step: The net force changes in the stated situation; only a qualitative variable-acceleration description is needed for this AP task.

At t = 0, B is 14 − 2 = 12 m ahead. A gains on B at 4 − 2 = 2 m/s.

Why this step: The constant coefficients of t are velocities; subtracting them gives the rate at which their separation closes.

Set the remaining separation to zero: 12 − 2t = 0, giving t = 6 s.

Why this step: A meeting occurs when the initially positive separation has been fully removed, assuming the stated constant velocities continue.

Substitute into both positions: x_A = 2 + 4·6 = 26 m and x_B = 14 + 2·6 = 26 m.

Why this step: Matching absolute positions verifies that the relative-motion result identifies the same event for both cars.

For horizontal launch, solve h = ½gt² to get t = √(2h/g).

Why this step: The fall begins with zero vertical velocity; only the vertical height and gravity determine when the floor is reached.

Substitute into R = ut: R = u√(2h/g).

Why this step: The same time must describe both components; u remains constant because horizontal acceleration is zero.

Plot R against √h, or square the equation and plot R² against h with slope 2u²/g.

Why this step: Each transformed equation is linear when u and g are held fixed, so the slope can test that dependence rather than assuming R is proportional to h.

For R² versus h, slope units are m²/m = m.

Why this step: The axis-unit ratio must match 2u²/g, whose units are (m²/s²)/(m/s²) = m; a dimensionless slope would indicate mislabeled axes.

Set east positive: the first displacement is +7 m and the return displacement is −3 m.

Why this step: The question fixes the axis; westward travel must have the opposite component sign.

Add signed displacements: Δx = 7 − 3 = +4 m. Add traveled lengths: distance = 7 + 3 = 10 m.

Why this step: The return leg partly cancels the position change, but it still adds 3 m to the path traveled.

Choose distance 10 m and displacement +4 m.

Why this step: The positive resultant means the person finishes east of the start; neither 10 m of displacement nor a westward resultant matches these endpoints.

A return to the starting point gives Δx = 0.

Why this step: The initial and final positions match, whatever route the runner took between them.

Average velocity is Δx/Δt = 0/20 = 0 m/s.

Why this step: The full-interval definition uses net displacement; a positive traveled distance cannot replace that numerator.

Average speed and instantaneous velocity need not be zero.

Why this step: Running out and back includes motion even though the endpoint changes cancel, so only average velocity is guaranteed.

The graph below zero means v is negative; its upward slope means a is positive.

Why this step: Graph height gives the signed velocity, while the rate that height changes gives acceleration; these are separate readings.

While v stays negative, positive acceleration moves it toward zero and reduces |v|.

Why this step: The acceleration is opposite the motion in this interval, so the object initially slows rather than speeding up.

Choose moving negative and initially slowing down.

Why this step: A negative velocity is still motion, and an upward graph slope does not make that velocity positive.

Use v_rider,road = v_rider,bus + v_bus,road with the shared positive road direction.

Why this step: The bus displacement and the rider’s displacement within the bus add during the same elapsed time.

Substitute −1 + 10 = +9 m/s.

Why this step: The rider walks backward relative to the bus, so that component reduces rather than increases the forward road velocity.

The rider still moves forward relative to the road.

Why this step: The bus’s 10 m/s motion is larger than the rider’s 1 m/s backward walk; −11 and +11 use incorrect signs.

After horizontal release, gravity is the only force in the negligible-drag model.

Why this step: The launch contact has ended; no continuing horizontal push is included in projectile motion.

Therefore aₓ = 0 and vₓ stays equal to its launch value.

Why this step: Only a horizontal velocity change would require horizontal acceleration; downward gravity has no x component.

Vertical velocity changes and the speed generally grows during the fall.

Why this step: Gravity changes vᵧ, so the full speed √(vₓ² + vᵧ²) is not held fixed by the constant horizontal component.

The component vₓ = −5 m/s describes motion in the negative x direction.

Why this step: The minus sign carries direction information; it is not a negative amount of speed.

Take the magnitude: speed = |−5| = 5 m/s.

Why this step: Speed answers how fast the object moves, so the direction sign is removed rather than changing the magnitude.

Compute the signed change: Δv = −7 − (−3) = −4 m/s.

Why this step: Use final minus initial; subtracting the initial negative velocity prevents treating −7 and −3 as unsigned speeds.

Divide by elapsed time: a_avg = (−4 m/s)/(2 s) = −2 m/s².

Why this step: The 2 s interval measures how quickly the velocity changed; the unit becomes velocity per second.

The motion is speeding up in the negative direction.

Why this step: The speed magnitude grows from 3 to 7 m/s, so the negative acceleration does not mean slowing down.

For a short constant-acceleration interval, area = aΔt and Δv = aΔt.

Why this step: The acceleration definition links a rectangular graph area directly to the change in velocity over that same interval.

Add signed areas for successive intervals to obtain the total Δv.

Why this step: Below-axis intervals subtract velocity change; summing the pieces keeps their direction information.

Final velocity is v_f = v_i + area, so area alone is velocity change.

Why this step: A nonzero initial velocity must still be added; Position information cannot be obtained from this area directly.

Write the second observer’s velocity as v′ = v − V, where V is the constant relative frame velocity.

Why this step: Two inertial observers in this question differ by a constant-velocity motion, not an accelerating coordinate system.

The change is v_f′ − v_i′ = (v_f − V) − (v_i − V) = v_f − v_i.

Why this step: The same constant offset appears at both endpoints and cancels rather than altering the velocity change.

Divide by the same time interval: a′ = a.

Why this step: In the ordinary nonrelativistic model the observers share elapsed time, so equal velocity changes give equal acceleration; positions and velocities can differ.

At the peak, vᵧ = 0 because vertical motion changes from upward to downward.

Why this step: The highest point is a vertical turning point, not necessarily a turning point in the horizontal direction.

The nonzero horizontal component continues while acceleration remains downward at g.

Why this step: No horizontal force changes vₓ, and gravity persists at the peak just as it does elsewhere in the flight.

Choose vertical velocity zero with downward acceleration.

Why this step: Zero total velocity would ignore the angled launch’s horizontal component; zero gravity would prevent the downward turn.

Let an unchanged displacement have component Δx in the original axis. In the reversed axis its component is Δx′ = −Δx.

Why this step: Reversing the positive direction relabels the same physical arrow, so its projection changes sign.

Its magnitude remains |Δx′| = |Δx|.

Why this step: The coordinate choice does not move either endpoint or stretch the physical displacement arrow.

Compare the car’s velocity arrows at two different points on the bend. They have equal lengths but different directions.

Why this step: Fixed speed fixes arrow length, while the bend changes the direction in which the car moves.

The velocity change is nonzero, so the car accelerates.

Why this step: Acceleration depends on the vector change; keeping its magnitude constant does not keep the vector constant.

At the highest point of a vertical toss, the upward velocity has reached zero before reversing downward.

Why this step: A turning point is where the one-dimensional position slope changes sign, so the instantaneous vertical velocity vanishes.

Gravity is still downward, giving a = −g if up is positive.

Why this step: With air resistance excluded, Earth is still the only force source; being momentarily at rest does not remove its force.

Choose velocity zero and acceleration downward.

Why this step: Confusing a zero value of velocity with a zero rate of velocity change would incorrectly predict that the ball stays at the peak.

The truck is the observer, so v_car,truck = v_car,road − v_truck,road.

Why this step: A truck-fixed origin advances with the truck; subtracting that motion leaves the car’s motion within that frame.

Calculate 6 − 10 = −4 m/s.

Why this step: Both road velocities are positive, but the car is slower, so its relative coordinate decreases.

Interpret −4 m/s as backward relative to the truck, not backward on the road.

Why this step: The car’s road velocity remains +6 m/s; changing observers can reverse a component without reversing the actual road motion.

For horizontal launch from height h, the fall time satisfies h = ½gt², so t = √(2h/g).

Why this step: Both trials have zero initial vertical velocity and the same height and gravity, so changing horizontal speed does not enter the time equation.

Range is R = u t; replacing u by 2u gives R_new = 2R at the same t.

Why this step: Horizontal acceleration is zero, and the common flight time multiplies the constant horizontal launch velocity.

The time is unchanged and the range doubles.

Why this step: Doubling flight time would require changing the vertical motion, which the question explicitly holds fixed.