O Level Additional Mathematics Calculus Quiz

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O-Level Additional Mathematics Quiz - Calculus

Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50

Duration: 1 hour 15 minutes Total Marks: 50

Instructions:

• Answer ALL questions.
• Show all working clearly. Marks are awarded for method.
• Give non-exact answers to 3 significant figures, or 1 decimal place for angles in degrees.
• You may use an approved calculator.
• The number of marks is given in brackets [ ] at the end of each question or part question.

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Section A: Differentiation Techniques (Questions 1–6)

Answer ALL questions in this section.

1. Differentiate with respect to $x$: (a) $y = 3x^4 - 2x^3 + 5x - 7$ [2] (b) $y = \frac{2}{x^3} + \sqrt{x}$ [2]

2. Find $\frac{dy}{dx}$ for each of the following: (a) $y = (2x + 1)(x^2 - 3)$ [3] (b) $y = \frac{x^2 + 1}{x - 2}$ [3]

3. Differentiate with respect to $x$: (a) $y = \sin 3x$ [2] (b) $y = e^{2x+1}$ [2] (c) $y = \ln(5x - 2)$ [2]

4. Given $y = \cos^2 x$, find $\frac{dy}{dx}$. [3]

5. Find $\frac{d^2y}{dx^2}$ when $y = x^3 - 6x^2 + 9x + 4$. [3]

6. The curve $C$ has equation $y = x^3 - 3x^2 + 2$. (a) Find $\frac{dy}{dx}$. [1] (b) Find the equation of the tangent to $C$ at the point where $x = 1$. [3]

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Section B: Applications of Differentiation (Questions 7–13)

Answer ALL questions in this section.

7. Find the coordinates of the stationary points on the curve $y = 2x^3 - 3x^2 - 12x + 7$ and determine the nature of each stationary point. [6]

8. The curve $y = x^2 + \frac{16}{x}$ is defined for $x > 0$. (a) Find $\frac{dy}{dx}$. [2] (b) Find the coordinates of the stationary point on the curve. [3] (c) Determine whether this stationary point is a maximum or a minimum. [2]

9. A curve has equation $y = \frac{2x + 1}{x - 1}$, where $x \neq 1$. (a) Find $\frac{dy}{dx}$. [2] (b) Explain why the curve has no stationary points. [2]

10. The displacement $s$ metres of a particle from a fixed point $O$ at time $t$ seconds is given by $s = t^3 - 6t^2 + 9t$, for $t \geq 0$. (a) Find the velocity of the particle when $t = 2$. [2] (b) Find the acceleration of the particle when $t = 4$. [2] (c) Find the times when the particle is instantaneously at rest. [2]

11. A rectangular box with a square base and an open top is to have a volume of 500 cm³. The base has side length $x$ cm and the height is $h$ cm. (a) Show that the external surface area $A$ cm² is given by $A = x^2 + \frac{2000}{x}$. [2] (b) Find the value of $x$ that minimises the surface area. [3] (c) Find the minimum surface area. [1]

12. The curve $y = x^3 + ax^2 + bx + c$ has a stationary point at $(2, 5)$ and passes through the point $(1, 3)$. Find the values of $a$, $b$, and $c$. [5]

13. A spherical balloon is being inflated such that its volume $V$ cm³ increases at a constant rate of $100$ cm³/s. The volume of a sphere is $V = \frac{4}{3}\pi r^3$, where $r$ cm is the radius. (a) Find $\frac{dV}{dr}$. [1] (b) Find the rate at which the radius is increasing when $r = 5$ cm. [3]

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Section C: Integration (Questions 14–20)

Answer ALL questions in this section.

14. Find the following indefinite integrals: (a) $\int (4x^3 - 6x^2 + 2x - 1) \, dx$ [2] (b) $\int \left( \frac{3}{x^2} + \sqrt[3]{x} \right) dx$ [2]

15. Integrate with respect to $x$: (a) $\int \sin 2x \, dx$ [2] (b) $\int e^{3x-1} \, dx$ [2] (c) $\int \frac{1}{2x + 5} \, dx$ [2]

16. Evaluate: (a) $\int_1^3 (2x^2 - x + 1) \, dx$ [3] (b) $\int_0^{\frac{\pi}{4}} \cos 2x \, dx$ [3]

17. The curve $y = f(x)$ passes through the point $(2, 10)$ and $f'(x) = 3x^2 - 4x + 1$. Find $f(x)$. [4]

18. Find the area of the region bounded by the curve $y = x^2 - 4x + 3$, the $x$-axis, and the lines $x = 0$ and $x = 2$. [5]

19. The diagram shows part of the curve $y = 4x - x^2$.

*(In the diagram, the curve intersects the $x$-axis at the origin and at $x = 4$.)*

Find the area of the shaded region bounded by the curve and the $x$-axis. [4]

20. A particle moves along a straight line. Its velocity $v$ m/s at time $t$ seconds is given by $v = 6t - t^2$, for $0 \leq t \leq 8$. (a) Find the displacement of the particle from its starting point when $t = 4$. [3] (b) Find the total distance travelled by the particle in the first 6 seconds. [4]

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END OF QUIZ

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O-Level Additional Mathematics Quiz - Calculus: ANSWER KEY

Total Marks: 50

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Section A: Differentiation Techniques (Questions 1–6)

1. (a) $y = 3x^4 - 2x^3 + 5x - 7$ $\frac{dy}{dx} = 12x^3 - 6x^2 + 5$ [M1] for differentiating each term correctly; [A1] for fully correct answer. [2]

(b) $y = 2x^{-3} + x^{1/2}$ $\frac{dy}{dx} = -6x^{-4} + \frac{1}{2}x^{-1/2} = -\frac{6}{x^4} + \frac{1}{2\sqrt{x}}$ [M1] for rewriting in index form and differentiating; [A1] for correct simplified answer. [2]

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2. (a) $y = (2x + 1)(x^2 - 3)$ Method 1 (Product Rule): $u = 2x + 1$, $v = x^2 - 3$ $u' = 2$, $v' = 2x$ $\frac{dy}{dx} = u'v + uv' = 2(x^2 - 3) + (2x + 1)(2x)$ $= 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$

Method 2 (Expand first): $y = 2x^3 + x^2 - 6x - 3$ $\frac{dy}{dx} = 6x^2 + 2x - 6$ [M1] for using product rule or expanding; [M1] for correct differentiation; [A1] for correct simplified answer. [3]

(b) $y = \frac{x^2 + 1}{x - 2}$ Quotient Rule: $u = x^2 + 1$, $v = x - 2$ $u' = 2x$, $v' = 1$ $\frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{2x(x - 2) - (x^2 + 1)(1)}{(x - 2)^2}$ $= \frac{2x^2 - 4x - x^2 - 1}{(x - 2)^2} = \frac{x^2 - 4x - 1}{(x - 2)^2}$ [M1] for correct quotient rule setup; [M1] for correct expansion; [A1] for correct simplified answer. [3]

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3. (a) $y = \sin 3x$ $\frac{dy}{dx} = 3\cos 3x$ [M1] for chain rule; [A1] for correct answer. [2]

(b) $y = e^{2x+1}$ $\frac{dy}{dx} = 2e^{2x+1}$ [M1] for chain rule; [A1] for correct answer. [2]

(c) $y = \ln(5x - 2)$ $\frac{dy}{dx} = \frac{5}{5x - 2}$ [M1] for chain rule; [A1] for correct answer. [2]

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4. $y = \cos^2 x = (\cos x)^2$ Using chain rule: let $u = \cos x$, then $y = u^2$ $\frac{dy}{du} = 2u$, $\frac{du}{dx} = -\sin x$ $\frac{dy}{dx} = 2u \cdot (-\sin x) = -2\cos x \sin x = -\sin 2x$ [M1] for recognising chain rule; [M1] for correct application; [A1] for correct simplified answer (either form acceptable). [3]

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5. $y = x^3 - 6x^2 + 9x + 4$ $\frac{dy}{dx} = 3x^2 - 12x + 9$ $\frac{d^2y}{dx^2} = 6x - 12$ [M1] for first derivative; [M1] for second derivative; [A1] for correct answer. [3]

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6. (a) $y = x^3 - 3x^2 + 2$ $\frac{dy}{dx} = 3x^2 - 6x$ [A1] for correct derivative. [1]

(b) At $x = 1$: $y = 1^3 - 3(1)^2 + 2 = 1 - 3 + 2 = 0$ $\frac{dy}{dx} = 3(1)^2 - 6(1) = 3 - 6 = -3$ Gradient of tangent $m = -3$ Point: $(1, 0)$ Equation: $y - 0 = -3(x - 1)$ $y = -3x + 3$ [M1] for finding $y$-coordinate; [M1] for finding gradient; [A1] for correct equation. [3]

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Section B: Applications of Differentiation (Questions 7–13)

7. $y = 2x^3 - 3x^2 - 12x + 7$ $\frac{dy}{dx} = 6x^2 - 6x - 12$

At stationary points, $\frac{dy}{dx} = 0$: $6x^2 - 6x - 12 = 0$ $x^2 - x - 2 = 0$ $(x - 2)(x + 1) = 0$ $x = 2$ or $x = -1$

When $x = 2$: $y = 2(8) - 3(4) - 12(2) + 7 = 16 - 12 - 24 + 7 = -13$ Stationary point: $(2, -13)$

When $x = -1$: $y = 2(-1) - 3(1) - 12(-1) + 7 = -2 - 3 + 12 + 7 = 14$ Stationary point: $(-1, 14)$

Nature: $\frac{d^2y}{dx^2} = 12x - 6$

At $x = 2$: $\frac{d^2y}{dx^2} = 12(2) - 6 = 18 > 0$ → minimum at $(2, -13)$

At $x = -1$: $\frac{d^2y}{dx^2} = 12(-1) - 6 = -18 < 0$ → maximum at $(-1, 14)$

[M1] for finding $\frac{dy}{dx}$; [M1] for setting to zero and solving; [M1] for finding $y$-coordinates; [M1] for finding $\frac{d^2y}{dx^2}$; [M1] for testing nature; [A1] for both points and correct nature. [6]

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8. (a) $y = x^2 + 16x^{-1}$ $\frac{dy}{dx} = 2x - 16x^{-2} = 2x - \frac{16}{x^2}$ [M1] for rewriting; [A1] for correct derivative. [2]

(b) At stationary point, $\frac{dy}{dx} = 0$: $2x - \frac{16}{x^2} = 0$ $2x = \frac{16}{x^2}$ $2x^3 = 16$ $x^3 = 8$ $x = 2$ (since $x > 0$)

When $x = 2$: $y = 2^2 + \frac{16}{2} = 4 + 8 = 12$ Stationary point: $(2, 12)$ [M1] for setting derivative to zero; [M1] for solving; [A1] for correct coordinates. [3]

(c) $\frac{d^2y}{dx^2} = 2 + \frac{32}{x^3}$ At $x = 2$: $\frac{d^2y}{dx^2} = 2 + \frac{32}{8} = 2 + 4 = 6 > 0$ Since $\frac{d^2y}{dx^2} > 0$, the stationary point is a minimum. [M1] for finding second derivative; [A1] for correct conclusion with justification. [2]

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9. (a) $y = \frac{2x + 1}{x - 1}$ Quotient rule: $u = 2x + 1$, $v = x - 1$ $u' = 2$, $v' = 1$ $\frac{dy}{dx} = \frac{2(x - 1) - (2x + 1)(1)}{(x - 1)^2} = \frac{2x - 2 - 2x - 1}{(x - 1)^2} = \frac{-3}{(x - 1)^2}$ [M1] for quotient rule; [A1] for correct simplified derivative. [2]

(b) $\frac{dy}{dx} = \frac{-3}{(x - 1)^2}$ Since $(x - 1)^2 > 0$ for all $x \neq 1$, $\frac{dy}{dx} = \frac{-3}{(x - 1)^2} < 0$ for all $x \neq 1$. Therefore, $\frac{dy}{dx}$ is never equal to zero, so the curve has no stationary points. [M1] for reasoning about sign of derivative; [A1] for clear conclusion. [2]

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10. (a) $s = t^3 - 6t^2 + 9t$ $v = \frac{ds}{dt} = 3t^2 - 12t + 9$ When $t = 2$: $v = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3$ m/s [M1] for differentiating to find velocity; [A1] for correct answer. [2]

(b) $a = \frac{dv}{dt} = 6t - 12$ When $t = 4$: $a = 6(4) - 12 = 24 - 12 = 12$ m/s² [M1] for differentiating to find acceleration; [A1] for correct answer. [2]

(c) Instantaneously at rest when $v = 0$: $3t^2 - 12t + 9 = 0$ $t^2 - 4t + 3 = 0$ $(t - 1)(t - 3) = 0$ $t = 1$ or $t = 3$ [M1] for setting $v = 0$; [A1] for correct times. [2]

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11. (a) Volume: $V = x^2h = 500$ $h = \frac{500}{x^2}$

Surface area (open top): $A = x^2 + 4xh$ (base + 4 sides) $A = x^2 + 4x\left(\frac{500}{x^2}\right) = x^2 + \frac{2000}{x}$ [M1] for expressing $h$ in terms of $x$; [A1] for correct expression. [2]

(b) $A = x^2 + 2000x^{-1}$ $\frac{dA}{dx} = 2x - 2000x^{-2} = 2x - \frac{2000}{x^2}$

At minimum, $\frac{dA}{dx} = 0$: $2x - \frac{2000}{x^2} = 0$ $2x = \frac{2000}{x^2}$ $2x^3 = 2000$ $x^3 = 1000$ $x = 10$

Check: $\frac{d^2A}{dx^2} = 2 + \frac{4000}{x^3} > 0$ for $x > 0$, so minimum. [M1] for differentiating; [M1] for setting to zero and solving; [A1] for $x = 10$. [3]

(c) $A = 10^2 + \frac{2000}{10} = 100 + 200 = 300$ cm² [A1] for correct answer. [1]

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12. $y = x^3 + ax^2 + bx + c$ $\frac{dy}{dx} = 3x^2 + 2ax + b$

Stationary point at $(2, 5)$: (1) Point lies on curve: $5 = 8 + 4a + 2b + c$ → $4a + 2b + c = -3$ (2) Derivative zero at $x = 2$: $3(4) + 2a(2) + b = 0$ → $12 + 4a + b = 0$ → $4a + b = -12$

Passes through $(1, 3)$: (3) $3 = 1 + a + b + c$ → $a + b + c = 2$

From (2): $b = -12 - 4a$ Substitute into (1): $4a + 2(-12 - 4a) + c = -3$ $4a - 24 - 8a + c = -3$ $-4a + c = 21$ ... (4)

Substitute into (3): $a + (-12 - 4a) + c = 2$ $-3a + c = 14$ ... (5)

(4) - (5): $(-4a + c) - (-3a + c) = 21 - 14$ $-a = 7$ $a = -7$

From (5): $-3(-7) + c = 14$ → $21 + c = 14$ → $c = -7$

From (2): $b = -12 - 4(-7) = -12 + 28 = 16$

Therefore: $a = -7$, $b = 16$, $c = -7$ [M1] for derivative; [M1] for equation (1); [M1] for equation (2); [M1] for equation (3); [A1] for all three correct values. [5]

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13. (a) $V = \frac{4}{3}\pi r^3$ $\frac{dV}{dr} = 4\pi r^2$ [A1] for correct derivative. [1]

(b) Given $\frac{dV}{dt} = 100$ cm³/s. Using chain rule: $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$ $100 = 4\pi r^2 \cdot \frac{dr}{dt}$

When $r = 5$: $100 = 4\pi(25) \cdot \frac{dr}{dt}$ $100 = 100\pi \cdot \frac{dr}{dt}$ $\frac{dr}{dt} = \frac{1}{\pi} \approx 0.318$ cm/s [M1] for using chain rule; [M1] for substituting values; [A1] for correct answer. [3]

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Section C: Integration (Questions 14–20)

14. (a) $\int (4x^3 - 6x^2 + 2x - 1) \, dx$ $= \frac{4x^4}{4} - \frac{6x^3}{3} + \frac{2x^2}{2} - x + C$ $= x^4 - 2x^3 + x^2 - x + C$ [M1] for integrating each term; [A1] for correct answer with constant. [2]

(b) $\int (3x^{-2} + x^{1/3}) \, dx$ $= \frac{3x^{-1}}{-1} + \frac{x^{4/3}}{4/3} + C$ $= -3x^{-1} + \frac{3}{4}x^{4/3} + C$ $= -\frac{3}{x} + \frac{3}{4}\sqrt[3]{x^4} + C$ [M1] for rewriting and integrating; [A1] for correct answer with constant. [2]

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15. (a) $\int \sin 2x \, dx = -\frac{1}{2}\cos 2x + C$ [M1] for recognising reverse chain rule; [A1] for correct answer. [2]

(b) $\int e^{3x-1} \, dx = \frac{1}{3}e^{3x-1} + C$ [M1] for reverse chain rule; [A1] for correct answer. [2]

(c) $\int \frac{1}{2x + 5} \, dx = \frac{1}{2}\ln|2x + 5| + C$ [M1] for reverse chain rule; [A1] for correct answer. [2]

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16. (a) $\int_1^3 (2x^2 - x + 1) \, dx$ $= \left[ \frac{2x^3}{3} - \frac{x^2}{2} + x \right]_1^3$ $= \left( \frac{2(27)}{3} - \frac{9}{2} + 3 \right) - \left( \frac{2(1)}{3} - \frac{1}{2} + 1 \right)$ $= \left( 18 - 4.5 + 3 \right) - \left( \frac{2}{3} - 0.5 + 1 \right)$ $= 16.5 - \left( \frac{2}{3} + 0.5 \right)$ $= 16.5 - \frac{7}{6}$ $= \frac{99}{6} - \frac{7}{6} = \frac{92}{6} = \frac{46}{3} = 15\frac{1}{3}$ [M1] for integration; [M1] for substitution of limits; [A1] for correct answer. [3]

(b) $\int_0^{\pi/4} \cos 2x \, dx$ $= \left[ \frac{1}{2}\sin 2x \right]_0^{\pi/4}$ $= \frac{1}{2}\sin\left(\frac{\pi}{2}\right) - \frac{1}{2}\sin(0)$ $= \frac{1}{2}(1) - 0 = \frac{1}{2}$ [M1] for integration; [M1] for substitution; [A1] for correct answer. [3]

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17. $f'(x) = 3x^2 - 4x + 1$ $f(x) = \int (3x^2 - 4x + 1) \, dx = x^3 - 2x^2 + x + C$

Passes through $(2, 10)$: $10 = 2^3 - 2(4) + 2 + C$ $10 = 8 - 8 + 2 + C$ $10 = 2 + C$ $C = 8$

Therefore: $f(x) = x^3 - 2x^2 + x + 8$ [M1] for integration; [M1] for substituting point; [M1] for finding constant; [A1] for correct function. [4]

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18. $y = x^2 - 4x + 3$ Find where curve crosses $x$-axis: $x^2 - 4x + 3 = 0$ $(x - 1)(x - 3) = 0$ $x = 1$ or $x = 3$

Between $x = 0$ and $x = 2$, the curve crosses the $x$-axis at $x = 1$. For $0 \leq x < 1$: $y > 0$ (above $x$-axis) For $1 < x \leq 2$: $y < 0$ (below $x$-axis)

Area = $\int_0^1 (x^2 - 4x + 3) \, dx + \left| \int_1^2 (x^2 - 4x + 3) \, dx \right|$

$\int (x^2 - 4x + 3) \, dx = \frac{x^3}{3} - 2x^2 + 3x$

$\int_0^1 = \left[ \frac{x^3}{3} - 2x^2 + 3x \right]_0^1 = \left( \frac{1}{3} - 2 + 3 \right) - 0 = \frac{4}{3}$

$\int_1^2 = \left[ \frac{x^3}{3} - 2x^2 + 3x \right]_1^2$ $= \left( \frac{8}{3} - 8 + 6 \right) - \left( \frac{1}{3} - 2 + 3 \right)$ $= \left( \frac{8}{3} - 2 \right) - \left( \frac{1}{3} + 1 \right)$ $= \frac{2}{3} - \frac{4}{3} = -\frac{2}{3}$

Total area = $\frac{4}{3} + \left| -\frac{2}{3} \right| = \frac{4}{3} + \frac{2}{3} = 2$ square units. [M1] for finding $x$-intercepts; [M1] for splitting integral; [M1] for integration; [M1] for evaluating; [A1] for correct total area. [5]

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19. $y = 4x - x^2$ Curve meets $x$-axis when $y = 0$: $4x - x^2 = 0$ $x(4 - x) = 0$ $x = 0$ or $x = 4$

Area = $\int_0^4 (4x - x^2) \, dx$ $= \left[ 2x^2 - \frac{x^3}{3} \right]_0^4$ $= \left( 2(16) - \frac{64}{3} \right) - 0$ $= 32 - \frac{64}{3}$ $= \frac{96}{3} - \frac{64}{3} = \frac{32}{3}$ square units. [M1] for finding limits; [M1] for integration; [M1] for substitution; [A1] for correct answer. [4]

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20. (a) $v = 6t - t^2$ Displacement $s = \int_0^4 (6t - t^2) \, dt$ $= \left[ 3t^2 - \frac{t^3}{3} \right]_0^4$ $= \left( 3(16) - \frac{64}{3} \right) - 0$ $= 48 - \frac{64}{3} = \frac{144}{3} - \frac{64}{3} = \frac{80}{3} \approx 26.7$ m [M1] for integrating; [M1] for evaluating; [A1] for correct answer. [3]

(b) Total distance = $\int_0^6 |v| \, dt$ Find when $v = 0$: $6t - t^2 = 0$ $t(6 - t) = 0$ $t = 0$ or $t = 6$

For $0 \leq t \leq 6$: $v \geq 0$ (since $6t - t^2 = t(6 - t) \geq 0$) So $|v| = v$ for $0 \leq t \leq 6$.

Total distance = $\int_0^6 (6t - t^2) \, dt$ $= \left[ 3t^2 - \frac{t^3}{3} \right]_0^6$ $= \left( 3(36) - \frac{216}{3} \right) - 0$ $= 108 - 72 = 36$ m [M1] for finding when $v = 0$; [M1] for determining sign of $v$; [M1] for integrating; [A1] for correct total distance. [4]

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END OF ANSWER KEY