UniGeniusAP Calculus

From the graph to the points

One complete AB question and a BC extension, nine points each.

Write your answer on paper first, then type the working to check it.

AB and BC · no calculator

Free response · 9 pointsSection II Part BNo calculatorAB and BC

CED CHA-4.C

A particle moves along the xx-axis with velocity v(t)v(t), in meters per second, shown in the graph. Its position at t=0t=0 is x(0)=5x(0)=5 meters.
(a)
Find the position of the particle at t=6t=6.
(2 points)

Write the setup as on paper, with units where asked.

(b)
Find the total distance traveled from t=0t=0 to t=6t=6. Show your setup.
(3 points)

Write the setup as on paper, with units where asked.

(c)
On what open intervals is the speed of the particle increasing? Justify your answer.
(2 points)

Write the setup as on paper, with units where asked.

(d)
Find the minimum position on 0≤t≤60\le t\le6. Justify your answer.
(2 points)

Write the setup as on paper, with units where asked.

See the complete nine-point guide

Part (a) · 2 points

  1. Add the signed area under the velocity graph to the initial position.
    x(6)=5+∫06v(t) dtx(6)=5+\int_0^6v(t)\,dt
    (a) P1 SetupUses the signed velocity integral on [0,6][0,6] to find displacement.Lost if: Treating the unsigned area as displacement.
  2. The two outer intervals have zero net area; the middle rectangle contributes −4-4.
    x(6)=5−4=1x(6)=5-4=1
    (a) P2 AnswerPosition 11 meter, supported by signed areas.Lost if: Reporting displacement −4-4 as position.A bare answer with no supporting work does not earn this point.Needs a-P1 first.

Part (b) · 3 points

  1. Velocity changes sign at t=1t=1 and t=5t=5.
    (b) P1 JustificationIdentifies both sign changes, t=1t=1 and t=5t=5.
    • Identifies both velocity zeros: t=1 and t=5
    Lost if: Splitting only at the graph corners.
  2. Add all the geometric areas, reversing the sign of the negative area.
    ∫06∣v(t)∣ dt=∫01v(t) dt−∫15v(t) dt+∫56v(t) dt\int_0^6|v(t)|\,dt=\int_0^1v(t)\,dt-\int_1^5v(t)\,dt+\int_5^6v(t)\,dt
    (b) P2 SetupIntegrates ∣v∣|v|, or adds the positive areas and magnitude of the negative area.Lost if: Adding signed areas without changing the negative sign.
  3. The two positive triangles each have area 11; the negative region has area 66.
    1+6+1=81+6+1=8
    (b) P3 AnswerTotal distance 88 meters with supporting work.Lost if: Using the magnitude of displacement, 44.A bare answer with no supporting work does not earn this point.Needs b-P2 first.

Part (c) · 2 points

  1. On (1,2)(1,2), both vv and v′v\prime are negative. On (5,6)(5,6), both are positive.
    (c) P1 AnswerBoth intervals (1,2)(1,2) and (5,6)(5,6), with no other intervals.Lost if: Using increasing velocity alone.
  2. Speed increases where velocity and acceleration have the same sign.
    (c) P2 JustificationVelocity and acceleration have the same sign on each stated interval.
    • v < 0 and acceleration < 0 on (1,2)
    • v > 0 and acceleration > 0 on (5,6)
    Lost if: Ignoring the sign of velocity.

Part (d) · 2 points

  1. Check the endpoints and the times when velocity is zero.
    x(0)=5,x(1)=6,x(5)=0,x(6)=1x(0)=5,\quad x(1)=6,\quad x(5)=0,\quad x(6)=1
    (d) P2 JustificationCompares positions at t=0,1,5,6t=0,1,5,6, or uses a global sign argument with endpoints.
    • Identifies minimum candidate t=5
    • Compares with both endpoints and other stationary point
    Lost if: A local minimum argument without checking the whole interval.
  2. The smallest position is 00 meters at t=5t=5.
    (d) P1 AnswerMinimum position 00 meters.Lost if: Giving the time without the position.A bare answer with no supporting work does not earn this point.

BC only · graphing calculator required

Free response · 9 pointsSection II Part AGraphing calculator requiredBC only

CED FUN-8.B

A particle moves in the plane with x′(t)=v(t)x\prime(t)=v(t), shown in the graph, and y′(t)=2y\prime(t)=2. Its position at t=0t=0 is (5,0)(5,0), with coordinates measured in meters.
(a)
Find the position of the particle at t=6t=6.
(2 points)

Write the setup as on paper, with units where asked.

(b)
Find the speed at t=3t=3.
(2 points)

Write the setup as on paper, with units where asked.

(c)
Find the total distance traveled from t=0t=0 to t=6t=6. Show your integral setup.
(3 points)

Write the setup as on paper, with units where asked.

(d)
Write an equation of the tangent line to the path at t=0t=0.
(2 points)

Write the setup as on paper, with units where asked.

See the complete nine-point guide

Part (a) · 2 points

  1. Integrate each velocity component and add the initial coordinate.
    x(6)=5+∫06v(t) dt=1,y(6)=∫062 dt=12x(6)=5+\int_0^6v(t)\,dt=1,\qquad y(6)=\int_0^62\,dt=12
    (a) P1 Answerx(6)=1x(6)=1 with initial position and signed area.Lost if: Reporting displacement as the coordinate.A bare answer with no supporting work does not earn this point.
    (a) P2 Answery(6)=12y(6)=12 with integration or constant-rate reasoning.Lost if: Using y′(6)=2y\prime(6)=2 as a position.A bare answer with no supporting work does not earn this point.

Part (b) · 2 points

  1. The velocity vector at t=3t=3 is (−2,2)(-2,2).
    speed=(−2)2+22=22\text{speed}=\sqrt{(-2)^2+2^2}=2\sqrt2
    (b) P1 SetupUses the magnitude of (−2,2)(-2,2).Lost if: Adding velocity components.
    (b) P2 AnswerSpeed 222\sqrt2, or 2.8282.828 to three decimals.Lost if: Giving a negative speed.A bare answer with no supporting work does not earn this point.Needs b-P1 first.

Part (c) · 3 points

  1. Integrate the magnitude of the velocity vector.
    ∫06v(t)2+4 dt\int_0^6\sqrt{v(t)^2+4}\,dt
    (c) P1 SetupIntegrates v(t)2+4\sqrt{v(t)^2+4} on [0,6][0,6].Lost if: Using only ∣v(t)∣|v(t)|.
  2. Use the three line segments to write the calculator integrals.
    ∫02(2−2t)2+4 dt+∫248 dt+∫46(2t−10)2+4 dt\int_0^2\sqrt{(2-2t)^2+4}\,dt+\int_2^4\sqrt8\,dt+\int_4^6\sqrt{(2t-10)^2+4}\,dt
    (c) P2 JustificationUses all three graph pieces correctly to evaluate the speed integral.
    • Uses v=2-2t on [0,2], v=−2 on [2,4], and v=2t-10 on [4,6]
    • Evaluates the speed integral over the whole interval
    Lost if: Using only one segment or omitting the middle interval.
  3. A graphing calculator gives 14.83914.839 meters.
    (c) P3 AnswerDistance 14.83914.839 meters, rounded or truncated correctly to three decimals.Lost if: Rounding intermediate values too early.A bare answer with no supporting work does not earn this point.A decimal must be correct to three places after the decimal point.Needs c-P1 and c-P2 first.

Part (d) · 2 points

  1. Divide the velocity components to find the tangent slope.
    dydx=y′(0)x′(0)=22=1\frac{dy}{dx}=\frac{y\prime(0)}{x\prime(0)}=\frac22=1
    (d) P1 SetupUses dy/dx=y′/x′=1dy/dx=y\prime/x\prime=1.Lost if: Using x′/y′x\prime/y\prime without the correct interpretation.
  2. Use the initial point (5,0)(5,0).
    y=x−5y=x-5
    (d) P2 AnswerAn equivalent equation of y=x−5y=x-5.Lost if: Using the velocity vector as a point on the path.A bare answer with no supporting work does not earn this point.Needs d-P1 first.

Distance is not displacement

The AB graph gives a displacement of −4 meters. Taking its magnitude gives 4, which misses the forward travel. Splitting at both sign changes gives a total distance of 8 meters.

Compare your setup with part (b), then try a new question.

Practice displacement and distance