1.3 Representing Motion

1. AA straight line that slopes upward and stays above the t axis
   A student who thinks the acceleration points in the direction of motion picks this: the cart moves in +x, so they give it a positive acceleration and a rising graph. The shrinking gaps show the cart slowing, so its acceleration is in −x and the graph slopes down.
2. BA curve that rises quickly at first and then levels off flat
   A student who draws the position graph when asked for the velocity graph picks this: the cart's position does rise quickly and then level off. The velocity is the slope of that position graph, which decreases steadily, so the v_x–t graph is a falling straight line.
3. CA straight line sloping down that stays above the t axis Correct
   Every gap is to the right, so v_x is positive throughout and the graph stays above the time axis. The gaps (8, 6, 4 and 2 m) shrink by the same 2 m in each equal interval, so v_x falls by equal amounts in equal times: a straight line with negative slope.
4. DA horizontal straight line that lies just below the t axis
   A student who draws the acceleration graph in place of the velocity graph picks this: the cart's constant, negative acceleration does give a horizontal line below the axis on an a_x–t graph. The velocity is positive and decreasing, so the v_x–t graph slopes down above the axis.

Working Gaps 8, 6, 4, 2 m in equal intervals: moving in +x (v_x > 0 throughout) and the gap falls by 2 m each interval, so v_x decreases by equal amounts in equal times: constant negative a_x. v_x–t graph: straight line sloping down, staying above the t axis.

CED 1.3.A.1 · Read this in Fix

1. A30 m Correct
   From rest, x = (1/2)a t². At 4.0 s, x = 32 m; at 1.0 s, x = 2.0 m. The displacement during the interval is 32 m − 2.0 m = 30 m. Equivalently, the cart already moves at 4.0 m/s at t = 1.0 s, so Δx = (4.0)(3.0) + (1/2)(4.0)(3.0)² = 30 m.
2. B32 m
   A student who takes the displacement during an interval to be the position at the end of it picks this: x(4.0 s) = 32 m. The cart had already moved 2.0 m by t = 1.0 s, so the displacement between 1.0 s and 4.0 s is 32 − 2.0 = 30 m.
3. C18 m
   A student who treats the interval as if the cart started from rest at t = 1.0 s picks this: (1/2)(4.0)(3.0)² = 18 m. At t = 1.0 s the cart is already moving at 4.0 m/s, which adds (4.0 m/s)(3.0 s) = 12 m.
4. D48 m
   A student who multiplies one velocity by the time picks this: the velocity at 4.0 s, 16 m/s, times 3.0 s gives 48 m. The cart moves slower than 16 m/s for most of the interval; its average velocity is (4.0 + 16)/2 = 10 m/s, giving 30 m.

Working x(t) = (1/2)a t² from rest at x₀ = 0. x(4.0 s) = 0.5 × 4.0 × 16 = 32 m; x(1.0 s) = 0.5 × 4.0 × 1.0 = 2.0 m. Δx = 32 − 2.0 = 30 m. (Check: v(1.0 s) = 4.0 m/s, so Δx = 4.0 × 3.0 + 0.5 × 4.0 × 3.0² = 12 + 18 = 30 m.)

CED 1.3.A.2 · Read this in Fix

1. AIts position changes by 10 m downward in every single second
   A student who reads 10 m/s² as 'moves 10 m each second' picks this. The acceleration fixes the change in velocity per second; the distance covered in each second keeps changing (5 m in the first second after release from rest, then 15 m, 25 m, …).
2. BIts velocity changes by 10 m/s downward in every second Correct
   In free fall the acceleration is g = 10 m/s² downward, whatever the object's mass and whether it is rising or falling. Acceleration is the change in velocity per unit time, so v_y changes by 10 m/s downward in every second.
3. CIts acceleration is larger than g if its mass is large
   A student who thinks heavier objects fall with a greater acceleration picks this. With air resistance negligible, every object near Earth's surface has the same acceleration, g ≈ 10 m/s² downward, whatever its mass.
4. DIts speed drops by 10 m/s in each second of flight
   A student who thinks a negative acceleration always means slowing down picks this, since a_y = −10 m/s². The object slows only while it moves upward; when it moves downward, its velocity and acceleration are both downward and it speeds up.

Working a_y = −10 m/s² throughout free fall, for any mass and either direction of motion: Δv_y = −10 m/s in each second.

CED 1.3.A.3 · Read this in Fix

1. AIts velocity is zero, and so its acceleration is zero as well.
   A student who thinks zero velocity means zero acceleration picks this. The acceleration is the slope of the v_x–t graph, which the stem says is not zero; the velocity is passing through zero, not staying there.
2. BIt stays at rest for a short time and then moves on again.
   A student who thinks an object pauses at a turning point picks this. With a nonzero slope the graph crosses the axis at a single instant, so v_x is zero for that instant only; staying at rest would need the graph to lie along the axis.
3. CIts velocity is zero, but its acceleration is not zero. Correct
   The height of the v_x–t graph is zero, so v_x = 0 at that instant. The slope of the graph is the acceleration, and it is not zero: the velocity is changing sign, so the object is reversing direction at that instant.
4. DIt is passing back through its own starting position now.
   A student who reads the velocity graph as a position graph picks this, taking the crossing to mean x = 0. The height of a v_x–t graph is the velocity, so the crossing means v_x = 0; where the object is depends on the area under the graph.

Working Crossing: v_x = 0. Slope ≠ 0: a_x = dv_x/dt ≠ 0. The object reverses direction instantaneously.

CED 1.3.A.4.ii · Read this in Fix

1. A1
   A student who thinks constant acceleration means equal distances in equal times picks this. Equal distances in equal times is constant velocity; with constant acceleration from rest the distances in successive intervals are d, 3d, 5d, …
2. B5 Correct
   From rest, x = (1/2)a t² grows as t². By 2T the cart has traveled 4d and by 3T it has traveled 9d, so during the third interval it travels 9d − 4d = 5d.
3. C9
   A student who gives the distance from the start, x(3T) = 9d, as the distance traveled during the third interval picks this. The cart had already traveled 4d by 2T, so the third interval contributes 9d − 4d = 5d.
4. D6
   A student who multiplies the velocity at the end of the interval by the time picks this: v(3T) = 3aT, and 3aT × T = 3aT² = 6d. The cart moves slower than 3aT for most of the interval; its average velocity during it is 2.5aT, giving 5d.

Working x = (1/2)a t², so d = (1/2)aT². x(3T) − x(2T) = (1/2)a(9T² − 4T²) = (1/2)aT² × 5 = 5d. k = 5.

CED 1.3.A.2 · Read this in Fix

1. A0.33 v₀²/g
   A student who thinks the speed falls by equal amounts over equal distances picks this: having lost 2/3 of its speed, the ball would be at 2/3 of its maximum height v₀²/(2g), that is v₀²/(3g). The speed falls by equal amounts in equal times; v² falls linearly with height, so most of the speed is lost in the lower part of the rise.
2. B0.06 v₀²/g
   A student who leaves out the initial velocity picks this: (v₀/3)² = 2gΔy gives v₀²/(18g) ≈ 0.06 v₀²/g. That is the height the ball still has to rise above this point, not its height above the launch point; the v₀² term must stay in the equation.
3. C0.67 v₀²/g
   A student who multiplies the launch speed by the time picks this: the time to slow from v₀ to v₀/3 is 2v₀/(3g), and v₀ × 2v₀/(3g) = 0.67 v₀²/g. The ball moves slower than v₀ throughout; with its average velocity, (2v₀/3) × 2v₀/(3g) = 0.44 v₀²/g.
4. D0.44 v₀²/g Correct
   With up positive and a_y = −g, v_y² = v₀² − 2gΔy. Setting v_y = v₀/3 gives Δy = (v₀² − v₀²/9)/(2g) = (4/9)v₀²/g ≈ 0.44 v₀²/g, which is 8/9 of the maximum height v₀²/(2g).

Working Up positive, a_y = −g. v_y² = v_y0² + 2a_y Δy: (v₀/3)² = v₀² − 2gΔy, so Δy = (v₀² − v₀²/9)/(2g) = (8/9)v₀²/(2g) = (4/9)v₀²/g ≈ 0.44 v₀²/g.

CED 1.3.A.3 · Read this in Fix

1. AIt is zero at the instant when the ball's velocity is zero
   A student who thinks zero velocity means zero acceleration picks this. The data show v_y changing from +1.0 to 0.0 to −1.0 m/s at the same rate across that instant, so the velocity is still changing there: the acceleration is not zero.
2. BIt is upward while the ball is rising and downward while it falls
   A student who thinks the acceleration points along the motion picks this. While the ball rises, v_y decreases (from +2.0 to 0.0 m/s), so Δv_y, and hence the acceleration, is downward while it rises too.
3. CIt is constant and downward throughout, even when v_y is zero Correct
   v_y falls by 1.0 m/s in every 0.10 s interval, including the two intervals either side of the instant when v_y = 0. So a_y = (−1.0 m/s)/(0.10 s) = −10 m/s² throughout: constant and downward, at the top as well.
4. DIt is negative while the ball rises and positive as it falls
   A student who thinks a negative acceleration means slowing down and a positive one speeding up picks this: the ball slows on the way up and speeds up on the way down. But v_y keeps decreasing (from 0.0 to −2.0 m/s) on the way down too, so a_y is negative throughout.

Working Δv_y = −1.0 m/s in every 0.10 s interval, including the intervals either side of v_y = 0: a_y = −1.0/0.10 = −10 m/s², constant and downward throughout.

CED 1.3.A.3 · Read this in Fix

1. AA rising straight line, then a rising curve that flattens, then a flat line Correct
   The slope of the x–t graph is the velocity. From 0 to 2 s the velocity is a constant +6 m/s, so x rises along a straight line. From 2 to 5 s the velocity is still positive but decreasing, so x keeps rising with a slope that falls to zero. From 5 to 6 s the velocity is zero, so x stays constant.
2. BA horizontal line, then a line sloping downward, then a line along the t axis
   A student who draws the position graph with the same shape as the velocity graph picks this. The x–t graph shows the velocity as its slope: a constant positive velocity gives a rising straight line, not a horizontal one.
3. CA horizontal line, then a rising curve that flattens, then another flat line
   A student who thinks a horizontal line on any motion graph means the object is at rest picks this, treating the object as at rest from 0 to 2 s. A horizontal v–t line at 6 m/s means constant velocity: the object moves 12 m in those 2 s.
4. DA rising straight line, then a falling curve, then a horizontal flat line
   A student who thinks a downward-sloping velocity graph means the object moves backward picks this. From 2 to 5 s the velocity is still positive (the graph is above the time axis), so the object keeps moving in +x while slowing down; x keeps rising.

Working 0–2 s: v constant +6 m/s → x–t straight line rising. 2–5 s: v positive, decreasing to 0 → x still rising, slope decreasing → curve rising and flattening. 5–6 s: v = 0 → x constant → horizontal line.

CED 1.3.A.4 · Read this in Fix

1. AP > R > Q
   A student who ranks by the height of the graph picks this: P is highest, then R, then Q. The height is the position; the speed is the steepness of the tangent, which is greatest at R.
2. BR > Q > P
   A student who treats a negative velocity as a smaller speed picks this: the slope at P is negative, so they rank P below Q's zero. Speed is the magnitude of the velocity; the object at P is moving (in the −x direction), faster than at Q.
3. CQ > P > R
   A student who calculates slope as the change in t divided by the change in x picks this: the inverted ratio is largest for the horizontal tangent at Q and smallest for the steepest one, at R. Slope is Δx/Δt, which is zero at Q and greatest in magnitude at R.
4. DR > P > Q Correct
   Speed is the magnitude of the slope of the tangent. The tangent at R is the steepest, about twice as steep as at P; the tangent at P slopes down (the object moves in −x) but is still steeper than the horizontal tangent at Q, where the object is momentarily at rest.

Working Speed = |slope of tangent|. Q: horizontal tangent, 0. P: moderate downward slope. R: steeper upward slope (about twice P's). Speeds: R > P > Q. (Heights: P > R > Q; signed velocities: R > Q > P.)

CED 1.3.A.4.i · Read this in Fix

1. A0.33 b/c Correct
   v_x = dx/dt = 2bt − 3ct², the slope of the x–t graph. It is greatest where dv_x/dt = 2b − 6ct = 0, at t = b/(3c) ≈ 0.33 b/c, where v_x = b²/(3c); at the ends of the interval v_x is 0 and −b²/c.
2. B0.67 b/c
   A student who thinks the velocity is greatest where the position is greatest picks this: x is greatest where dx/dt = 0, at t = 2b/(3c). There the slope of the x–t graph, and so v_x, is zero; the velocity is greatest where the graph is steepest upward.
3. C0.50 b/c
   A student who takes the velocity to be x/t picks this: x/t = bt − ct² is greatest at t = b/(2c). x/t is the average velocity from t = 0, not the instantaneous velocity, which is dx/dt.
4. D1.00 b/c
   A student who ignores the sign of the velocity picks this: at t = b/c the object moves fastest, with v_x = −b²/c, so its speed is greatest there. Its velocity component is then negative, less than the positive maximum b²/(3c) at t = b/(3c).

Working v_x = dx/dt = 2bt − 3ct². dv_x/dt = 2b − 6ct = 0 at t = b/(3c); d²v_x/dt² = −6c < 0, so this is a maximum: v_x,max = b²/(3c). Ends: v_x(0) = 0, v_x(b/c) = −b²/c. So v_x is greatest at t = (1/3)(b/c) ≈ 0.33 b/c. (x greatest at 2b/(3c); x/t = bt − ct² greatest at b/(2c); |v_x| greatest at b/c.)

CED 1.3.A.4.i · Read this in Fix

1. A3.0 m/s²
   A student who divides the velocity at P by the time, 9 m/s ÷ 3 s, picks this. That is the slope of the line from the origin to P, the average acceleration over the first 3 s; the curve is getting less steep, so the acceleration at P is less than that average.
2. B2.0 m/s² Correct
   The instantaneous acceleration is the slope of the tangent at P. Using the two marked points on the tangent, (15 − 3) m/s ÷ (6 − 0) s = 2.0 m/s².
3. C0.5 m/s²
   A student who divides the change in time by the change in velocity, 6 s ÷ 12 m/s, picks this. Slope is the change in the vertical-axis quantity divided by the change in the horizontal-axis quantity, Δv/Δt, in m/s².
4. D9.0 m/s²
   A student who reads the height of the graph at P as the acceleration picks this. The height of a velocity–time graph is the velocity, 9 m/s; the acceleration is the slope of the tangent.

Working Tangent through (0 s, 3 m/s) and (6 s, 15 m/s): a = (15 − 3) m/s ÷ (6 − 0) s = 2.0 m/s². (Curve v = 4t − t²/3: dv/dt at 3 s = 4 − 2 = 2.)

CED 1.3.A.4.ii · Read this in Fix

1. A+7.5 m
   A student who adds the areas as positive amounts picks this: 6 m + 1.5 m = 7.5 m. That is the distance traveled. From 2 s to 3 s the velocity is negative, so that 1.5 m is traveled back toward the start and is subtracted for the displacement.
2. B−9.0 m
   A student who multiplies the final velocity by the time picks this: (−3 m/s)(3 s) = −9.0 m. The velocity is not constant; the displacement is the signed area under the graph, which is +4.5 m.
3. C−3.0 m
   A student who takes the slope of the graph for the displacement picks this: (−3 − 6) m/s ÷ 3 s = −3.0 (in m/s², the acceleration). The displacement is the area under a velocity–time graph, not its slope.
4. D+4.5 m Correct
   The displacement is the area between the graph and the time axis, with area below the axis counted as negative: (1/2)(2 s)(6 m/s) = +6 m from 0 to 2 s, and (1/2)(1 s)(−3 m/s) = −1.5 m from 2 to 3 s. Δx = +6 − 1.5 = +4.5 m.

Working Area above axis, 0–2 s: (1/2)(2 s)(6 m/s) = +6 m. Area below axis, 2–3 s: (1/2)(1 s)(−3 m/s) = −1.5 m. Δx = 6 − 1.5 = +4.5 m. (Distance = 7.5 m.)

CED 1.3.A.4.iii · Read this in Fix

1. A0.68 v₀τ
   A student who takes the average velocity to be the mean of the initial and final velocities picks this: (v₀ + v₀/e)/2 × τ ≈ 0.68 v₀τ. That mean is the average velocity only for constant acceleration; this v_x–t graph is curved, so the area must be found by integrating.
2. B0.63 v₀τ Correct
   The displacement is the integral of the velocity: Δx = ∫₀^τ v₀e^−t/τ dt = v₀τ[1 − e⁻¹] ≈ 0.63 v₀τ. This is the area under the v_x–t curve from 0 to τ.
3. C1.00 v₀τ
   A student who multiplies the initial velocity by the time picks this: v₀ × τ. The object slows down throughout the interval, so it covers less than v₀τ; the displacement is the area under the falling curve.
4. D0.50 v₀τ
   A student who uses x = x₀ + v_x0 t + (1/2)a_x t² with the initial acceleration, a_x(0) = −v₀/τ, picks this: v₀τ − (1/2)(v₀/τ)τ² = 0.50 v₀τ. The acceleration here changes with time, so the constant-acceleration equations do not apply.

Working Δx = ∫₀^τ v₀e^−t/τ dt = [−v₀τ e^−t/τ]₀^τ = v₀τ(1 − e⁻¹) = 0.632 v₀τ ≈ 0.63 v₀τ. Distractors: v₀ × τ = 1.00 v₀τ; (v₀ + v₀e⁻¹)/2 × τ = 0.684 v₀τ; constant a = a(0) = −v₀/τ: v₀τ − (1/2)(v₀/τ)τ² = 0.50 v₀τ.

CED 1.3.A.4.iii · Read this in Fix

1. Av₀ + bT²/2 Correct
   The change in velocity is the area under the a_x–t graph: Δv_x = ∫₀^T bt dt = bT²/2, a triangle of base T and height bT. Adding the initial velocity, v_x(T) = v₀ + bT²/2.
2. BbT²/2
   A student who leaves out the initial velocity picks this. The integral gives the change in velocity, bT²/2; the cart was already moving at v₀, so v_x(T) = v₀ + bT²/2.
3. Cv₀ + bT²
   A student who uses v_x = v_x0 + a_x t with the acceleration at the end, bT, picks this: v₀ + (bT)T. The acceleration rises from 0 to bT, so it is not constant; integrating gives half of bT².
4. Dv₀ + bT
   A student who takes the change in velocity to be the acceleration at the end of the interval, bT, picks this. The change in velocity is the area under the a_x–t graph, ∫a_x dt = bT²/2; bT has units of acceleration, not velocity.

Working v_x(T) = v₀ + ∫₀^T bt dt = v₀ + bT²/2. Distractors: from rest, bT²/2; constant a = a(T) = bT: v₀ + bT·T = v₀ + bT²; Δv = a(T): v₀ + bT.

CED 1.3.A.4.iv · Read this in Fix

1. A+3.0 m/s
   A student who treats the area under the a–t graph as the final velocity picks this, leaving out the initial velocity. The area, +3.0 m/s, is the change in velocity; the cart started at −1.0 m/s, so it ends at +2.0 m/s.
2. B+8.0 m/s
   A student who adds both areas as positive amounts picks this: −1.0 + 6.0 + 3.0 = +8.0 m/s. From 2 s to 4 s the acceleration is negative, so that area, −3.0 m/s, reduces the velocity.
3. C+2.0 m/s Correct
   The change in velocity is the signed area under the a–t graph: (+3.0 m/s²)(2 s) + (−1.5 m/s²)(2 s) = +6.0 − 3.0 = +3.0 m/s. Adding the initial velocity: −1.0 + 3.0 = +2.0 m/s.
4. D−1.5 m/s
   A student who reads the height of the graph at t = 4 s picks this. That height is the acceleration at that instant, −1.5 m/s²; the velocity comes from the initial velocity plus the area under the graph.

Working Δv = area: (3.0 m/s²)(2 s) + (−1.5 m/s²)(2 s) = 6.0 − 3.0 = +3.0 m/s. v(4 s) = −1.0 + 3.0 = +2.0 m/s.

CED 1.3.A.4.iv · Read this in Fix

1. AAt t = 4 s, cart Q is 3 m in front of cart P
   A student who reads the velocity graph as a position graph picks this: the heights at t = 4 s, 8 and 5, are taken as positions 8 m and 5 m. Those heights are velocities; the positions come from the areas, 16 m for Q and 20 m for P.
2. BAt t = 4 s, cart Q is 12 m ahead of cart P
   A student who multiplies each cart's velocity at t = 4 s by the time picks this: Q 8 × 4 = 32 m and P 5 × 4 = 20 m. Q has been slower than 8 m/s for the whole interval; its displacement is the triangle area, 16 m.
3. CAt t = 2.5 s, P and Q are side by side
   A student who thinks crossing velocity graphs means the carts are level picks this. Where the lines cross, the carts have equal velocities. By then P has gone (5)(2.5) = 12.5 m and Q only (1/2)(2.5)(5) = 6.25 m, so P is well ahead.
4. DAt t = 4 s, cart P is 4 m ahead of cart Q Correct
   Each cart's displacement is the area under its graph. From 0 to 4 s, P's area is (5 m/s)(4 s) = 20 m and Q's is (1/2)(4 s)(8 m/s) = 16 m, so P is 4 m ahead at t = 4 s even though Q is now moving faster.

Working Areas 0–4 s: P = 5 × 4 = 20 m; Q = (1/2)(4)(8) = 16 m. P is 4 m ahead at t = 4 s (Q draws level at t = 5 s). At t = 2.5 s (where the lines cross): P 12.5 m, Q 6.25 m — not side by side.

CED 1.3.A.4.iii · Read this in Fix

1. A1
   A student who assumes the acceleration is constant, as in the kinematic equations, picks this. Here a_x = 6Ct grows with time; the kinematic equations apply only when a_x is constant, which a cubic x(t) does not have.
2. B4
   A student who thinks the acceleration grows in proportion to the speed picks this: v_x = 3Ct² grows by a factor of 4. Acceleration is the rate of change of velocity, a_x = 6Ct, which grows only by a factor of 2.
3. C2 Correct
   v_x = dx/dt = 3Ct² and a_x = dv_x/dt = 6Ct, the slope of the v_x–t graph. The acceleration is proportional to t, so doubling the time doubles it: k = 2.
4. D8
   A student who reads the acceleration off the size of the position function itself, as one would read a height off a graph of x, picks this: x = Ct³ grows by a factor of 8. The acceleration is the second derivative, a_x = 6Ct, which grows only by a factor of 2.

Working v_x = 3Ct², a_x = dv_x/dt = 6Ct. a(2t₁)/a(t₁) = 2. (Velocity grows ×4, position ×8.)

CED 1.3.A.4.ii · Read this in Fix

1. A1.73 v₀/g
   A student who leaves out the initial velocity, as if the ball were dropped from rest, picks this: 3v₀²/(2g) = (1/2)gt² gives t = √3 v₀/g ≈ 1.73 v₀/g. The ball starts with upward velocity v₀, so it first rises and stays in the air longer.
2. B3.00 v₀/g Correct
   Take upward as positive: a_y = −g and the ground is at y = −3v₀²/(2g). Then −3v₀²/(2g) = v₀t − (1/2)gt², a quadratic whose positive root is t = [v₀ + √(v₀² + 3v₀²)]/g = 3.00 v₀/g. The ball rises for v₀/g, is back at the launch level at 2v₀/g, and falls the rest of the way in a further v₀/g.
3. C1.00 v₀/g
   A student who ignores the direction of the initial velocity picks this: taking downward as positive but entering +v₀ gives 3v₀²/(2g) = v₀t + (1/2)gt², whose positive root is v₀/g. That is the time for a ball thrown downward at v₀; this ball's initial velocity is upward, −v₀ in that coordinate system.
4. D0.75 v₀/g
   A student who takes the displacement to be the final velocity multiplied by the time picks this: the ball lands at speed √(v₀² + 3v₀²) = 2v₀, and 3v₀²/(2g) = 2v₀t gives t = 0.75 v₀/g. The velocity changes throughout the flight, so the displacement is not one velocity multiplied by the time.

Working Up positive: a_y = −g, y₀ = 0, ground at y = −3v₀²/(2g). y = v₀t − (1/2)gt²: −3v₀²/(2g) = v₀t − (1/2)gt², i.e. (g/2)t² − v₀t − 3v₀²/(2g) = 0, so t = [v₀ ± √(v₀² + 3v₀²)]/g = (v₀ ± 2v₀)/g. The positive root is t = 3v₀/g = 3.00 v₀/g (the root −v₀/g is before the throw). Distractors: from rest, √(2·(3/2)) = √3 ≈ 1.73; v₀ entered downward, (−v₀ + 2v₀)/g = 1.00 v₀/g; final speed √(v₀² + 3v₀²) = 2v₀ times t: (3/2)/2 = 0.75 v₀/g.

CED 1.3.A.2 · Read this in Fix

1. A1.50 a₀T²
   A student who uses the constant-acceleration equations with the acceleration at the start, a₀, for the first interval picks this: (1/2)a₀T² plus a₀T × T = 1.50 a₀T². The acceleration falls to zero during that interval, so the cart's velocity at t = T is only a₀T/2.
2. B0.33 a₀T²
   A student who reads the horizontal line along the t axis as the cart being at rest picks this: only the a₀T²/3 covered before t = T is counted. Zero acceleration means constant velocity; the cart keeps moving at a₀T/2 from T to 2T.
3. C0.83 a₀T² Correct
   The velocity is the running area under the a–t graph: v_x(t) = a₀(t − t²/(2T)), reaching a₀T/2 at t = T. Integrating, the cart moves a₀T²/3 by t = T. After that a = 0, so the cart keeps moving at a₀T/2 and goes a further a₀T²/2. Total (5/6)a₀T² ≈ 0.83 a₀T².
4. D1.00 a₀T²
   A student who multiplies the cart's greatest velocity, a₀T/2, by the whole time 2T picks this. The cart moves more slowly than a₀T/2 throughout the first interval, so its displacement is less than a₀T².

Working From the graph, a_x = a₀(1 − t/T) for 0 ≤ t ≤ T and a_x = 0 for T < t ≤ 2T. v_x(t) = ∫₀ᵗ a₀(1 − t′/T) dt′ = a₀(t − t²/(2T)), so v_x(T) = a₀T/2 (the triangle's area). x(T) = ∫₀^T a₀(t − t²/(2T)) dt = a₀(T²/2 − T²/6) = a₀T²/3. From T to 2T the velocity stays a₀T/2, adding a₀T²/2. Total Δx = a₀T²/3 + a₀T²/2 = (5/6)a₀T² ≈ 0.83 a₀T². Distractors: constant a₀ to T, (1/2)a₀T² + a₀T·T = 1.50 a₀T²; at rest after T, 0.33 a₀T²; greatest velocity a₀T/2 times 2T = 1.00 a₀T².

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