Secondary 1 Mathematics Calculus Quiz

Secondary 1 Mathematics Quiz - Calculus (Answer Key)

Total Marks: 40

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Section A: Gradient and Rate of Change (Questions 1–5) [10 marks]

1. Distance-time graph of a car

(a) Speed = gradient of distance-time graph
Gradient = $\frac{\text{change in distance}}{\text{change in time}} = \frac{100 - 0}{10 - 0} = \frac{100}{10} = 10$ m/s
Answer: 10 m/s [1]

(b) The gradient of a distance-time graph represents the speed (or velocity) of the object.
Answer: The gradient represents the speed of the car. [1]

Teaching note: In a distance-time graph, gradient = speed. A straight line means constant speed. The steeper the line, the greater the speed.

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2. Water tank filling: $V = 5t + 10$

(a) Rate of flow = coefficient of $t$ = 5 litres/min
Answer: 5 litres/min [1]

(b) The value 10 is the constant term. When $t = 0$, $V = 10$. This is the initial volume of water in the tank before filling starts.
Answer: The initial volume of water in the tank (at $t = 0$) is 10 litres. [1]

Teaching note: In a linear model $y = mx + c$, $m$ is the rate of change and $c$ is the initial value (when $x = 0$).

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3. Velocity-time graph of a particle

(a) First 4 seconds: horizontal line at $v = 6$ m/s. Gradient = 0.
Acceleration = 0 m/s²
Answer: 0 m/s² [1]

(b) Between $t = 4$ and $t = 8$: velocity decreases from 6 to 0 m/s uniformly.
Acceleration = gradient = $\frac{0 - 6}{8 - 4} = \frac{-6}{4} = -1.5$ m/s²
Answer: -1.5 m/s² [1]

(c) Distance = area under velocity-time graph
Area = area of rectangle (0 to 4 s) + area of triangle (4 to 8 s)
= $(4 \times 6) + \frac{1}{2} \times 4 \times 6$
= $24 + 12 = 36$ m
Answer: 36 m [2]

Marking: 1 mark for correct method (area under graph), 1 mark for correct answer.
Teaching note: Area under velocity-time graph = displacement (distance if velocity doesn't change sign).

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4. Cost function: $C = 2x + 50$

(a) Fixed cost = constant term = $50 **Answer:**$50 [1]

(b) Variable cost per item = coefficient of $x$ = $2 per item **Answer:**$2 per item [1]

(c) Revenue $R = 5x$. Profit when $R > C$:
$5x > 2x + 50$
$3x > 50$
$x > \frac{50}{3} \approx 16.67$
Minimum integer $x = 17$ items
Answer: 17 items [2]

Marking: 1 mark for setting up inequality $5x > 2x + 50$, 1 mark for correct answer 17.
Teaching note: Break-even occurs when revenue = cost. Profit requires revenue > cost.

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5. Car accelerating uniformly from rest

(a) Acceleration = $\frac{\text{change in velocity}}{\text{time}} = \frac{20 - 0}{5} = 4$ m/s²
Answer: 4 m/s² [1]

(b) Sketch: Straight line from (0,0) to (5,20) on velocity-time axes.
Answer: [Graph: straight line through origin with positive gradient, labelled axes] [1]

Teaching note: Uniform acceleration from rest gives a straight line through origin on a $v$-$t$ graph. Gradient = acceleration.

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Section B: Area Under Graphs and Accumulation (Questions 6–12) [15 marks]

6. Cyclist velocity-time graph (triangle)

(a) Maximum velocity = peak of triangle = 8 m/s
Answer: 8 m/s [1]

(b) Distance = area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 8 = 40$ m
Answer: 40 m [2]

Marking: 1 mark for area formula, 1 mark for correct answer.

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7. Water flow rate: $r = 2t + 3$

(a) At $t = 4$: $r = 2(4) + 3 = 8 + 3 = 11$ litres/min
Answer: 11 litres/min [1]

(b) Total volume = area under rate-time graph from $t=0$ to $t=4$
= $\int_0^4 (2t + 3) \, dt$
= $\left[ t^2 + 3t \right]_0^4$
= $(16 + 12) - 0 = 28$ litres

Alternative (trapezium area): At $t=0$, $r=3$; at $t=4$, $r=11$. Area = $\frac{1}{2}(3+11) \times 4 = 28$ litres
Answer: 28 litres [3]

Marking: 1 mark for correct method (integration or trapezium area), 1 mark for correct substitution/limits, 1 mark for correct answer.
Teaching note: Total quantity = area under rate-time graph. For linear rate, area is a trapezium.

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8. Rate of temperature change graph

(a) Fastest rate = highest value on graph = 4°C/min (first 3 minutes)
Answer: 0 to 3 minutes [1]

(b) Temperature increase = area under rate-time graph
= Area of rectangle 1 + Area of rectangle 2
= $(3 \times 4) + (3 \times 2) = 12 + 6 = 18$°C
Final temperature = $20 + 18 = 38$°C
Answer: 38°C [3]

Marking: 1 mark for area calculation, 1 mark for adding initial temperature, 1 mark for correct answer.

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9. Particle velocity: $v = 3t^2 - 12t + 9$

(a) At $t = 0$: $v = 3(0)^2 - 12(0) + 9 = 9$ m/s
Answer: 9 m/s [1]

(b) At rest when $v = 0$:
$3t^2 - 12t + 9 = 0$
$t^2 - 4t + 3 = 0$
$(t - 1)(t - 3) = 0$
$t = 1$ or $t = 3$ seconds
Answer: 1 s and 3 s [2]

Marking: 1 mark for setting $v=0$ and simplifying, 1 mark for correct solutions.

(c) Acceleration $a = \frac{dv}{dt} = 6t - 12$
At $t = 2$: $a = 6(2) - 12 = 0$ m/s²
Answer: 0 m/s² [2]

Marking: 1 mark for differentiating to get $a = 6t - 12$, 1 mark for correct substitution and answer.
Teaching note: Acceleration is the derivative of velocity (gradient of $v$-$t$ graph).

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10. Force-distance graph (rectangle)

Work done = area under force-distance graph
= Force $\times$ distance = $5 \times 8 = 40$ J
Answer: 40 J [2]

Marking: 1 mark for identifying area = work done, 1 mark for correct calculation.

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11. Car velocity: $v = 10 - 2t$

(a) Initial velocity at $t = 0$: $v = 10 - 0 = 10$ m/s
Answer: 10 m/s [1]

(b) At rest when $v = 0$: $10 - 2t = 0 \Rightarrow t = 5$ s
Answer: 5 s [1]

(c) Distance = area under $v$-$t$ graph from $t=0$ to $t=5$
Graph is a triangle: base = 5, height = 10
Area = $\frac{1}{2} \times 5 \times 10 = 25$ m
Answer: 25 m [2]

Marking: 1 mark for correct area method, 1 mark for correct answer.

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12. Bacteria growth: $\frac{dP}{dt} = 50$

(a) Constant rate of change $\Rightarrow$ linear growth.
$P = 50t + c$. At $t=0$, $P=200 \Rightarrow c = 200$.
$P = 50t + 200$
At $t = 6$: $P = 50(6) + 200 = 300 + 200 = 500$ bacteria
Answer: 500 bacteria [2]

Marking: 1 mark for finding $P = 50t + 200$, 1 mark for correct answer.

(b) Sketch: Straight line from (0,200) to (6,500) on population-time axes.
Answer: [Graph: straight line with positive gradient starting at (0,200)] [1]

Teaching note: Constant rate of change $\Rightarrow$ linear graph. Gradient = rate of change.

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Section C: Applications and Problem Solving (Questions 13–20) [15 marks]

13. Stone thrown upwards: $h = 20t - 5t^2$

(a) Velocity $v = \frac{dh}{dt} = 20 - 10t$
Initial velocity at $t = 0$: $v = 20$ m/s
Answer: 20 m/s [1]

Teaching note: Velocity is the derivative of displacement (gradient of $h$-$t$ graph).

(b) Maximum height when $v = 0$: $20 - 10t = 0 \Rightarrow t = 2$ s
$h_{\text{max}} = 20(2) - 5(2)^2 = 40 - 20 = 20$ m
Answer: 20 m [2]

Marking: 1 mark for finding $t=2$, 1 mark for correct height.

(c) Hits ground when $h = 0$: $20t - 5t^2 = 0 \Rightarrow 5t(4 - t) = 0$
$t = 0$ (start) or $t = 4$ s
Answer: 4 s [1]

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14. Train velocity-time graph (trapezoid)

(a) Acceleration = gradient of first section = $\frac{30 - 0}{20 - 0} = 1.5$ m/s²
Answer: 1.5 m/s² [1]

(b) Deceleration = gradient of last section = $\frac{0 - 30}{100 - 80} = \frac{-30}{20} = -1.5$ m/s²
Magnitude of deceleration = 1.5 m/s²
Answer: 1.5 m/s² [1]

(c) Distance = area under graph = area of trapezoid
= $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
= $\frac{1}{2} \times (100 + 60) \times 30$
Wait: parallel sides are the velocities at start and end of constant velocity phase?
Better: Area = rectangle + two triangles
= $(60 \times 30) + \frac{1}{2}(20 \times 30) + \frac{1}{2}(20 \times 30)$
= $1800 + 300 + 300 = 2400$ m

Or trapezium formula: Average velocity $\times$ time = $\frac{0+30}{2} \times 20 + 30 \times 60 + \frac{30+0}{2} \times 20 = 300 + 1800 + 300 = 2400$ m
Answer: 2400 m [3]

Marking: 1 mark for correct area method, 1 mark for correct calculation of areas, 1 mark for correct answer.

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15. Oil leak: $R = 100 - 5t$

(a) At $t = 4$: $R = 100 - 5(4) = 100 - 20 = 80$ litres/hour
Answer: 80 litres/hour [1]

(b) Leak stops when $R = 0$: $100 - 5t = 0 \Rightarrow t = 20$ hours
Answer: 20 hours [1]

(c) Total oil leaked = area under $R$-$t$ graph from $t=0$ to $t=20$
Triangle: base = 20, height = 100
Area = $\frac{1}{2} \times 20 \times 100 = 1000$ litres
Answer: 1000 litres [3]

Marking: 1 mark for identifying triangle area, 1 mark for correct base and height, 1 mark for correct answer.

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16. Rocket: $v = 50t - 5t^2$

(a) Maximum velocity when acceleration $a = 0$:
$a = \frac{dv}{dt} = 50 - 10t = 0 \Rightarrow t = 5$ s
$v_{\text{max}} = 50(5) - 5(5)^2 = 250 - 125 = 125$ m/s
Answer: 125 m/s [2]

Marking: 1 mark for setting $a=0$ and finding $t=5$, 1 mark for correct $v_{\text{max}}$.

(b) Acceleration $a = 50 - 10t$
At $t = 3$: $a = 50 - 30 = 20$ m/s²
Answer: 20 m/s² [1]

(c) Height = displacement = area under $v$-$t$ graph from 0 to 10
$h = \int_0^{10} (50t - 5t^2) \, dt = \left[ 25t^2 - \frac{5}{3}t^3 \right]_0^{10}$
= $25(100) - \frac{5}{3}(1000) = 2500 - \frac{5000}{3} = \frac{7500 - 5000}{3} = \frac{2500}{3} \approx 833.33$ m
Answer: $\frac{2500}{3}$ m or 833.33 m [2]

Marking: 1 mark for correct integration, 1 mark for correct evaluation and answer.

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17. Water flow rate graph (triangle)

(a) Maximum flow rate = peak = 10 L/min
Answer: 10 L/min [1]

(b) Total volume = area of triangle = $\frac{1}{2} \times 12 \times 10 = 60$ litres
Answer: 60 litres [2]

Marking: 1 mark for area formula, 1 mark for correct answer.

(c) Flow rate = 5 L/min occurs at two times (rising and falling).
By symmetry, at $t = 3$ min and $t = 9$ min.
Or calculation: Rising: gradient = $\frac{10}{6} = \frac{5}{3}$ L/min². $5 = \frac{5}{3}t \Rightarrow t = 3$.
Falling: $5 = 10 - \frac{5}{3}(t-6) \Rightarrow t = 9$.
Answer: 3 min and 9 min [1]

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18. Particle displacement: $s = t^3 - 6t^2 + 9t$

(a) Velocity $v = \frac{ds}{dt} = 3t^2 - 12t + 9$
At $t = 2$: $v = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3$ m/s
Answer: -3 m/s [2]

Marking: 1 mark for correct differentiation, 1 mark for correct substitution and answer.

(b) Acceleration $a = \frac{dv}{dt} = 6t - 12$
At $t = 2$: $a = 6(2) - 12 = 0$ m/s²
Answer: 0 m/s² [1]

(c) Changes direction when $v = 0$:
$3t^2 - 12t + 9 = 0 \Rightarrow t^2 - 4t + 3 = 0 \Rightarrow (t-1)(t-3) = 0$
$t = 1$ s and $t = 3$ s
Answer: 1 s and 3 s [2]

Marking: 1 mark for setting $v=0$ and solving, 1 mark for both correct times.

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19. Cost function: $C = 0.5x^2 + 20x + 500$

(a) Marginal cost = $\frac{dC}{dx} = x + 20$
At $x = 10$: Marginal cost = $10 + 20 = 30$
Answer: $30 [2]

Marking: 1 mark for correct differentiation, 1 mark for correct substitution and answer.
Teaching note: Marginal cost is the derivative of total cost — the cost of producing one additional unit.

(b) Marginal cost represents the instantaneous rate of change of total cost with respect to quantity — the approximate cost of producing the next (11th) unit when 10 units have already been produced.
Answer: The cost of producing one additional unit (the 11th unit) when 10 units have been produced. [1]

(c) Average cost = $\frac{C}{x} = \frac{0.5x^2 + 20x + 500}{x} = 0.5x + 20 + \frac{500}{x}$
At $x = 10$: Average cost = $0.5(10) + 20 + \frac{500}{10} = 5 + 20 + 50 = 75$
Answer: $75 [1]

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20. Car journey: three phases

(a) Sketch:

• 0 to 10 s: straight line from (0,0) to (10,20)
• 10 to 30 s: horizontal line at $v = 20$
• 30 to 35 s: straight line from (30,20) to (35,0)
  Answer: [Graph with three phases, labelled axes and key points] [1]

(b) Maximum velocity = velocity after acceleration phase = $a \times t = 2 \times 10 = 20$ m/s
Answer: 20 m/s [1]

(c) Total distance = area under $v$-$t$ graph
= Area of triangle (0-10) + rectangle (10-30) + triangle (30-35)
= $\frac{1}{2} \times 10 \times 20 + 20 \times 20 + \frac{1}{2} \times 5 \times 20$
= $100 + 400 + 50 = 550$ m
Answer: 550 m [2]

Marking: 1 mark for correct area breakdown, 1 mark for correct answer.

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End of Answer Key