IB Mathematics: Applications and Interpretation Quiz: Calculus Applications
20 questions · exam conditions
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Calculus ApplicationsQuestion 1 of 20

A particle's velocity is described by the function v(t)=(t−3)e−tv(t) = (t-3)e^{-t} for t≥0t \ge 0.

At t=4t=4, which of the following statements correctly describes the particle's speed?

The particle is changing direction.
The particle's speed is at a local maximum.
The particle's speed is at a local minimum.
The particle is stationary.
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Calculus Applications

Practice Calculus Applications in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Calculus Applications, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A particle's velocity is described by the function v(t)=(t−3)e−tv(t) = (t-3)e^{-t} for t≥0t \ge 0.

At t=4t=4, which of the following statements correctly describes the particle's speed?

  1. The particle is changing direction.
  2. The particle's speed is at a local maximum. (correct answer)
  3. The particle's speed is at a local minimum.
  4. The particle is stationary.
Explanation: First, evaluate velocity at t=4t=4: v(4)=(4−3)e−4=e−4>0v(4) = (4-3)e^{-4} = e^{-4} > 0. Since v≠0v \neq 0, the particle is not stationary and not changing direction. To analyze the speed, we need the acceleration, a(t)=v′(t)a(t) = v'(t). Using the product rule, a(t)=(1)e−t+(t−3)(−e−t)=e−t(1−(t−3))=e−t(4−t)a(t) = (1)e^{-t} + (t-3)(-e^{-t}) = e^{-t}(1 - (t-3)) = e^{-t}(4-t). At t=4t=4, a(4)=e−4(4−4)=0a(4) = e^{-4}(4-4) = 0. Since v(4)>0v(4) > 0 and a(4)=0a(4) = 0, the rate of change of speed is zero. To determine if it's a max or min, we check the sign of a(t)a(t) around t=4t=4. For t<4t < 4, a(t)>0a(t) > 0. For t>4t > 4, a(t)<0a(t) < 0. Since acceleration changes from positive to negative, the velocity (and speed, since v is positive here) is at a local maximum.

Question 2

The rate of consumption of oil in a country is modelled by R(t)=1.2e0.05tR(t) = 1.2e^{0.05t} million barrels per year, where tt is the number of years after the start of 2020.

What is the total amount of oil, in millions of barrels, predicted to be consumed from the start of 2025 to the start of 2030?

  1. 0.437
  2. 8.75 (correct answer)
  3. 9.89
  4. 15.6
Explanation: The total amount consumed is the definite integral of the rate function R(t)R(t). The interval is from the start of 2025 (t=5t=5) to the start of 2030 (t=10t=10). We need to calculate ∫5101.2e0.05t dt\int_5^{10} 1.2e^{0.05t} \,dt. Using a calculator, this integral is approximately 8.75. Analytically, the integral is [1.20.05e0.05t]510=24[e0.5−e0.25]≈8.75[\frac{1.2}{0.05}e^{0.05t}]_5^{10} = 24[e^{0.5} - e^{0.25}] \approx 8.75 million barrels.

Question 3

The rate of change of a company's profit is modelled by P′(t)=5cos⁡(0.2t)+10P'(t) = 5\cos(0.2t) + 10, in thousands of dollars per month, where tt is the number of months from the start of the year. The company's total profit at the start of the year (t=0t=0) was $20,000.

What is the company's total profit, to the nearest thousand dollars, after 6 months?

  1. $32,000
  2. $83,000
  3. $103,000 (correct answer)
  4. $120,000
Explanation: The total profit after 6 months is the initial profit plus the accumulated profit over the 6 months. Accumulated profit = ∫06P′(t) dt=∫06(5cos⁡(0.2t)+10) dt\int_0^6 P'(t) \,dt = \int_0^6 (5\cos(0.2t) + 10) \,dt. We can solve this analytically or with a calculator. Analytically: [5sin⁡(0.2t)0.2+10t]06=[25sin⁡(0.2t)+10t]06=(25sin⁡(1.2)+60)−0≈23.30+60=83.30[\frac{5\sin(0.2t)}{0.2} + 10t]_0^6 = [25\sin(0.2t) + 10t]_0^6 = (25\sin(1.2) + 60) - 0 \approx 23.30 + 60 = 83.30. This is in thousands of dollars. The total profit is P(6)=P(0)+Accumulated Profit=20+83.30=103.30P(6) = P(0) + \text{Accumulated Profit} = 20 + 83.30 = 103.30 thousand dollars. To the nearest thousand, this is $103,000.

Question 4

Water flows into a reservoir at a rate of I(t)=300+20tI(t) = 300+20t litres/hour. Water is released from the reservoir at a constant rate of 400 litres/hour. At t=0t=0, the reservoir contains 50,000 litres.

At what time is the volume of water in the reservoir at a minimum?

  1. 5 hours (correct answer)
  2. 0 hours
  3. 10 hours
  4. 20 hours
Explanation: This problem tests your understanding of optimization using calculus, specifically finding when a rate of change equals zero to locate minimum or maximum values. To find when the volume is minimized, you need to set up the volume function and find where its derivative equals zero. The net rate of water change is: inflow minus outflow = (300+20t)−400=−100+20t(300 + 20t) - 400 = -100 + 20t litres/hour. The volume function is: V(t)=50000+∫0t(−100+20s)ds=50000−100t+10t2V(t) = 50000 + \int_0^t (-100 + 20s) ds = 50000 - 100t + 10t^2 To find the minimum, take the derivative and set it equal to zero: V′(t)=−100+20t=0V'(t) = -100 + 20t = 0 Solving: 20t=10020t = 100, so t=5t = 5 hours. Since V′′(t)=20>0V''(t) = 20 > 0, this confirms a minimum at t=5t = 5. Looking at the wrong answers: B) 0 hours represents the starting time, but the reservoir continues losing water initially since outflow exceeds inflow. C) 10 hours might tempt you if you mistakenly doubled the correct answer or made an algebraic error. D) 20 hours could result from confusing the coefficients in the rate equation or solving t=100/20t = 100/20 incorrectly. A) 5 hours is correct because this is precisely when the net flow rate changes from negative (losing water) to positive (gaining water). Study tip: In reservoir problems, always identify the net rate first, then integrate to get volume. The minimum/maximum occurs when the net rate equals zero, not when individual rates are equal.

Question 5

The displacement, ss metres, of a buoy from a fixed point is given by s(t)=1.5sin⁡(π6t)−0.5t+2s(t) = 1.5\sin(\frac{\pi}{6}t) - 0.5t + 2, for 0≤t≤120 \le t \le 12 hours.

What is the maximum displacement of the buoy from the fixed point in the given interval?

  1. -4.31 m
  2. 1.68 m
  3. 2.00 m
  4. 2.31 m (correct answer)
Explanation: To find the maximum displacement, we need to find the critical points by setting the derivative (velocity) to zero, and then test the displacement at the critical points and the endpoints. v(t)=s′(t)=1.5⋅π6cos⁡(π6t)−0.5=π4cos⁡(π6t)−0.5v(t) = s'(t) = 1.5 \cdot \frac{\pi}{6}\cos(\frac{\pi}{6}t) - 0.5 = \frac{\pi}{4}\cos(\frac{\pi}{6}t) - 0.5. Set v(t)=0v(t)=0: cos⁡(π6t)=0.5π/4=2π\cos(\frac{\pi}{6}t) = \frac{0.5}{\pi/4} = \frac{2}{\pi}. Using a calculator, π6t=arccos⁡(2π)≈0.8806\frac{\pi}{6}t = \arccos(\frac{2}{\pi}) \approx 0.8806, so t≈1.682t \approx 1.682. We must also check the endpoints t=0t=0 and t=12t=12. s(0)=2s(0)=2. s(12)=1.5sin⁡(2π)−0.5(12)+2=−4s(12) = 1.5\sin(2\pi) - 0.5(12) + 2 = -4. s(1.682)=1.5sin⁡(π6(1.682))−0.5(1.682)+2≈2.31s(1.682) = 1.5\sin(\frac{\pi}{6}(1.682)) - 0.5(1.682) + 2 \approx 2.31. Comparing the values s(0)=2s(0)=2, s(12)=−4s(12)=-4, and s(1.682)≈2.31s(1.682) \approx 2.31, the maximum displacement is 2.31 m.

Question 6

The height hh in metres of a roller coaster car is modelled by h(t)=10sin⁡(0.5t)−0.2t2+20h(t) = 10\sin(0.5t) - 0.2t^2 + 20, where tt is the time in seconds.

What is the acceleration of the roller coaster car at t=3t=3 seconds, to three significant figures?

  1. -2.89 m/s2^2 (correct answer)
  2. -10.4 m/s2^2
  3. -0.847 m/s2^2
  4. 28.2 m/s2^2
Explanation: When you encounter a physics problem asking for acceleration given a position function, remember that acceleration is the second derivative of position with respect to time. This connects calculus directly to kinematics. Given the height function h(t)=10sin⁡(0.5t)−0.2t2+20h(t) = 10\sin(0.5t) - 0.2t^2 + 20, you need to find the second derivative and evaluate it at t=3t = 3. First, find the velocity by taking the first derivative: h′(t)=10⋅0.5cos⁡(0.5t)−0.4t=5cos⁡(0.5t)−0.4th'(t) = 10 \cdot 0.5 \cos(0.5t) - 0.4t = 5\cos(0.5t) - 0.4t Then find acceleration by taking the second derivative: h′′(t)=5⋅(−0.5)sin⁡(0.5t)−0.4=−2.5sin⁡(0.5t)−0.4h''(t) = 5 \cdot (-0.5)\sin(0.5t) - 0.4 = -2.5\sin(0.5t) - 0.4 At t=3t = 3: h′′(3)=−2.5sin⁡(0.5⋅3)−0.4=−2.5sin⁡(1.5)−0.4h''(3) = -2.5\sin(0.5 \cdot 3) - 0.4 = -2.5\sin(1.5) - 0.4 Since sin⁡(1.5)≈0.9975\sin(1.5) \approx 0.9975: h′′(3)=−2.5(0.9975)−0.4=−2.494−0.4=−2.894h''(3) = -2.5(0.9975) - 0.4 = -2.494 - 0.4 = -2.894 To three significant figures, this is −2.89-2.89 m/s², which is answer A. Answer B (-10.4) likely comes from incorrectly using the coefficient 10 instead of properly applying the chain rule. Answer C (-0.847) might result from only considering the polynomial term's contribution. Answer D (28.2) appears to confuse signs or mix up derivative calculations entirely. Study tip: Always remember the kinematic hierarchy: position → velocity (first derivative) → acceleration (second derivative). Practice chain rule applications with trigonometric functions, as they appear frequently in oscillatory motion problems.

Question 7

An object is launched upwards. Its acceleration is modelled by a(t)=−9.8−0.1ta(t) = -9.8 - 0.1t m/s2^2, for t≥0t \ge 0. The initial velocity is 50 m/s and the initial height is 2 m.

What is the maximum height, to the nearest metre, reached by the object?

  1. 127 m (correct answer)
  2. 125 m
  3. 130 m
  4. 247 m
Explanation: When you see motion problems with given acceleration functions, you need to work backwards through calculus to find velocity, then position, then determine when maximum height occurs. Starting with a(t)=−9.8−0.1ta(t) = -9.8 - 0.1t, integrate to find velocity: v(t)=−9.8t−0.05t2+Cv(t) = -9.8t - 0.05t^2 + C. Using the initial velocity of 50 m/s, you get v(t)=50−9.8t−0.05t2v(t) = 50 - 9.8t - 0.05t^2. Integrate again for position: s(t)=50t−4.9t2−0.05t33+Cs(t) = 50t - 4.9t^2 - \frac{0.05t^3}{3} + C. With initial height 2 m: s(t)=2+50t−4.9t2−t360s(t) = 2 + 50t - 4.9t^2 - \frac{t^3}{60}. Maximum height occurs when velocity equals zero. Setting 50−9.8t−0.05t2=050 - 9.8t - 0.05t^2 = 0 and solving this quadratic equation gives t≈4.69t ≈ 4.69 seconds (taking the positive root since we want the upward journey). Substituting back: s(4.69)=2+50(4.69)−4.9(4.69)2−(4.69)360≈127s(4.69) = 2 + 50(4.69) - 4.9(4.69)^2 - \frac{(4.69)^3}{60} ≈ 127 m. A) 127 m is correct - this accounts for the changing acceleration properly. B) 125 m likely results from computational errors in the quadratic solution or rounding too early. C) 130 m might come from ignoring the −0.1t-0.1t term in acceleration, treating it as constant −9.8-9.8 m/s². D) 247 m suggests a major conceptual error, possibly confusing the time calculation or using incorrect integration. Remember: in variable acceleration problems, always integrate step-by-step from acceleration to velocity to position, and find extremes by setting velocity to zero.

Question 8

The velocity of a rocket, vv in m/s, is modelled by the function v(t)=−0.1t3+2t2+5tv(t) = -0.1t^3 + 2t^2 + 5t for 0≤t≤200 \le t \le 20 seconds after launch.

What is the acceleration of the rocket at the time its velocity is at a maximum?

  1. 0 m/s2^2 (correct answer)
  2. 14.5 m/s2^2
  3. 18.3 m/s2^2
  4. 187 m/s2^2
Explanation: The velocity is at a maximum when its rate of change is zero. The rate of change of velocity is acceleration, a(t)=v′(t)a(t) = v'(t). Therefore, at the time of maximum velocity, the acceleration must be 0. To verify, we can find the time of maximum velocity by setting a(t)=v′(t)=−0.3t2+4t+5=0a(t) = v'(t) = -0.3t^2 + 4t + 5 = 0. Using a solver, this occurs at t≈14.5t \approx 14.5 s. The question asks for the acceleration at this time, which is by definition 0 m/s2^2.

Question 9

A particle moves in a straight line with acceleration a(t)=cos⁡(0.5t)a(t) = \cos(0.5t) m/s2^2 for t≥0t \ge 0. At t=0t=0, its velocity is v=3v=3 m/s and its displacement is s=1s=1 m.

Find the displacement of the particle at t=πt=\pi seconds.

  1. 5.00 m
  2. 9.42 m
  3. 14.4 m (correct answer)
  4. 17.4 m
Explanation: First, find velocity by integrating acceleration: v(t)=∫cos⁡(0.5t) dt=2sin⁡(0.5t)+Cv(t) = \int \cos(0.5t) \,dt = 2\sin(0.5t) + C. Given v(0)=3v(0)=3, 3=2sin⁡(0)+C⇒C=33 = 2\sin(0) + C \Rightarrow C=3. So, v(t)=2sin⁡(0.5t)+3v(t) = 2\sin(0.5t) + 3. Next, find displacement by integrating velocity: s(t)=∫(2sin⁡(0.5t)+3) dt=−4cos⁡(0.5t)+3t+Ds(t) = \int (2\sin(0.5t) + 3) \,dt = -4\cos(0.5t) + 3t + D. Given s(0)=1s(0)=1, 1=−4cos⁡(0)+0+D⇒1=−4+D⇒D=51 = -4\cos(0) + 0 + D \Rightarrow 1 = -4 + D \Rightarrow D=5. So, s(t)=−4cos⁡(0.5t)+3t+5s(t) = -4\cos(0.5t) + 3t + 5. At t=πt=\pi, s(π)=−4cos⁡(0.5π)+3π+5=−4(0)+3π+5=3π+5≈14.42s(\pi) = -4\cos(0.5\pi) + 3\pi + 5 = -4(0) + 3\pi + 5 = 3\pi + 5 \approx 14.42 m.

Question 10

The rate of consumption of oil in a country is modelled by R(t)=1.2e0.05tR(t) = 1.2e^{0.05t} million barrels per year, where tt is the number of years after the start of 2020.

What is the total amount of oil, in millions of barrels, predicted to be consumed from the start of 2025 to the start of 2030?

  1. 0.437
  2. 8.75 (correct answer)
  3. 9.89
  4. 15.6
Explanation: The total amount consumed is the definite integral of the rate function R(t)R(t). The interval is from the start of 2025 (t=5t=5) to the start of 2030 (t=10t=10). We need to calculate ∫5101.2e0.05t dt\int_5^{10} 1.2e^{0.05t} \,dt. Using a calculator, this integral is approximately 8.75. Analytically, the integral is [1.20.05e0.05t]510=24[e0.5−e0.25]≈8.75[\frac{1.2}{0.05}e^{0.05t}]_5^{10} = 24[e^{0.5} - e^{0.25}] \approx 8.75 million barrels.

Question 11

The displacement of a car from its starting point is given by s(t)=t3−3t2+5ts(t) = t^3 - 3t^2 + 5t metres for t≥0t \ge 0 seconds.

Find the time t>0t>0 at which the car's instantaneous velocity is equal to its average velocity over the interval [0,6][0, 6].

  1. 1.00 s
  2. 3.00 s
  3. 23.0 s
  4. 3.65 s (correct answer)
Explanation: This question tests the relationship between instantaneous velocity (derivative) and average velocity over an interval - a fundamental concept in calculus applications. To find when instantaneous velocity equals average velocity, you need both values. First, find the instantaneous velocity by taking the derivative: v(t)=s′(t)=3t2−6t+5v(t) = s'(t) = 3t^2 - 6t + 5. Next, calculate the average velocity over [0,6][0, 6] using s(6)−s(0)6−0\frac{s(6) - s(0)}{6 - 0}. You get s(6)=63−3(62)+5(6)=216−108+30=138s(6) = 6^3 - 3(6^2) + 5(6) = 216 - 108 + 30 = 138 and s(0)=0s(0) = 0, so average velocity is 1386=23\frac{138}{6} = 23 m/s. Now set instantaneous velocity equal to average velocity: 3t2−6t+5=233t^2 - 6t + 5 = 23. Rearranging gives 3t2−6t−18=03t^2 - 6t - 18 = 0, or t2−2t−6=0t^2 - 2t - 6 = 0. Using the quadratic formula: t=2±4+242=2±282=1±7t = \frac{2 \pm \sqrt{4 + 24}}{2} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7}. Since t>0t > 0, you want t=1+7≈3.65t = 1 + \sqrt{7} \approx 3.65 seconds. Choice A (1.00 s) likely comes from taking 1−71 - \sqrt{7} and using its absolute value. Choice B (3.00 s) might result from approximating 7≈2\sqrt{7} \approx 2. Choice C (23.0 s) confuses the time with the average velocity value itself. Strategy tip: When solving "instantaneous equals average" problems, always find the derivative first, then calculate the average rate of change, and finally solve the resulting equation. Watch for negative solutions that need to be rejected based on the domain.

Question 12

The rate of change of a company's profit is modelled by P′(t)=5cos⁡(0.2t)+10P'(t) = 5\cos(0.2t) + 10, in thousands of dollars per month, where tt is the number of months from the start of the year. The company's total profit at the start of the year (t=0t=0) was $20,000.

What is the company's total profit, to the nearest thousand dollars, after 6 months?

  1. $32,000
  2. $83,000
  3. $103,000 (correct answer)
  4. $120,000
Explanation: The total profit after 6 months is the initial profit plus the accumulated profit over the 6 months. Accumulated profit = ∫06P′(t) dt=∫06(5cos⁡(0.2t)+10) dt\int_0^6 P'(t) \,dt = \int_0^6 (5\cos(0.2t) + 10) \,dt. We can solve this analytically or with a calculator. Analytically: [5sin⁡(0.2t)0.2+10t]06=[25sin⁡(0.2t)+10t]06=(25sin⁡(1.2)+60)−0≈23.30+60=83.30[\frac{5\sin(0.2t)}{0.2} + 10t]_0^6 = [25\sin(0.2t) + 10t]_0^6 = (25\sin(1.2) + 60) - 0 \approx 23.30 + 60 = 83.30. This is in thousands of dollars. The total profit is P(6)=P(0)+Accumulated Profit=20+83.30=103.30P(6) = P(0) + \text{Accumulated Profit} = 20 + 83.30 = 103.30 thousand dollars. To the nearest thousand, this is $103,000.

Question 13

An object is launched upwards. Its acceleration is modelled by a(t)=−9.8−0.1ta(t) = -9.8 - 0.1t m/s2^2, for t≥0t \ge 0. The initial velocity is 50 m/s and the initial height is 2 m.

What is the maximum height, to the nearest metre, reached by the object?

  1. 127 m (correct answer)
  2. 125 m
  3. 130 m
  4. 247 m
Explanation: When you see motion problems with given acceleration functions, you need to work backwards through calculus to find velocity, then position, then determine when maximum height occurs. Starting with a(t)=−9.8−0.1ta(t) = -9.8 - 0.1t, integrate to find velocity: v(t)=−9.8t−0.05t2+Cv(t) = -9.8t - 0.05t^2 + C. Using the initial velocity of 50 m/s, you get v(t)=50−9.8t−0.05t2v(t) = 50 - 9.8t - 0.05t^2. Integrate again for position: s(t)=50t−4.9t2−0.05t33+Cs(t) = 50t - 4.9t^2 - \frac{0.05t^3}{3} + C. With initial height 2 m: s(t)=2+50t−4.9t2−t360s(t) = 2 + 50t - 4.9t^2 - \frac{t^3}{60}. Maximum height occurs when velocity equals zero. Setting 50−9.8t−0.05t2=050 - 9.8t - 0.05t^2 = 0 and solving this quadratic equation gives t≈4.69t ≈ 4.69 seconds (taking the positive root since we want the upward journey). Substituting back: s(4.69)=2+50(4.69)−4.9(4.69)2−(4.69)360≈127s(4.69) = 2 + 50(4.69) - 4.9(4.69)^2 - \frac{(4.69)^3}{60} ≈ 127 m. A) 127 m is correct - this accounts for the changing acceleration properly. B) 125 m likely results from computational errors in the quadratic solution or rounding too early. C) 130 m might come from ignoring the −0.1t-0.1t term in acceleration, treating it as constant −9.8-9.8 m/s². D) 247 m suggests a major conceptual error, possibly confusing the time calculation or using incorrect integration. Remember: in variable acceleration problems, always integrate step-by-step from acceleration to velocity to position, and find extremes by setting velocity to zero.

Question 14

The rate of change of a fish population in a lake is given by P′(t)=100(e−0.1t−0.5)P'(t) = 100(e^{-0.1t} - 0.5) fish per month, where tt is time in months.

What is the net change in the fish population, to the nearest fish, during the first 10 months?

  1. -13
  2. 132 (correct answer)
  3. 175
  4. 184
Explanation: The net change in the population is the definite integral of the rate of change from t=0t=0 to t=10t=10. We need to calculate ∫010100(e−0.1t−0.5) dt\int_0^{10} 100(e^{-0.1t} - 0.5) \,dt. Using a GDC, the value of the integral is approximately 132.1. Analytically, the integral is 100[−e−0.1t0.1−0.5t]010=100[−10e−0.1t−0.5t]010=100[(−10e−1−5)−(−10e0−0)]=100[−10/e−5+10]=100(5−10/e)≈132.1100[-\frac{e^{-0.1t}}{0.1} - 0.5t]_0^{10} = 100[-10e^{-0.1t} - 0.5t]_0^{10} = 100[(-10e^{-1} - 5) - (-10e^0 - 0)] = 100[-10/e - 5 + 10] = 100(5-10/e) \approx 132.1. To the nearest fish, the net change is 132.

Question 15

The height hh in metres of a roller coaster car is modelled by h(t)=10sin⁡(0.5t)−0.2t2+20h(t) = 10\sin(0.5t) - 0.2t^2 + 20, where tt is the time in seconds.

What is the acceleration of the roller coaster car at t=3t=3 seconds, to three significant figures?

  1. -2.89 m/s2^2 (correct answer)
  2. -10.4 m/s2^2
  3. -0.847 m/s2^2
  4. 28.2 m/s2^2
Explanation: When you encounter a physics problem asking for acceleration given a position function, remember that acceleration is the second derivative of position with respect to time. This connects calculus directly to kinematics. Given the height function h(t)=10sin⁡(0.5t)−0.2t2+20h(t) = 10\sin(0.5t) - 0.2t^2 + 20, you need to find the second derivative and evaluate it at t=3t = 3. First, find the velocity by taking the first derivative: h′(t)=10⋅0.5cos⁡(0.5t)−0.4t=5cos⁡(0.5t)−0.4th'(t) = 10 \cdot 0.5 \cos(0.5t) - 0.4t = 5\cos(0.5t) - 0.4t Then find acceleration by taking the second derivative: h′′(t)=5⋅(−0.5)sin⁡(0.5t)−0.4=−2.5sin⁡(0.5t)−0.4h''(t) = 5 \cdot (-0.5)\sin(0.5t) - 0.4 = -2.5\sin(0.5t) - 0.4 At t=3t = 3: h′′(3)=−2.5sin⁡(0.5⋅3)−0.4=−2.5sin⁡(1.5)−0.4h''(3) = -2.5\sin(0.5 \cdot 3) - 0.4 = -2.5\sin(1.5) - 0.4 Since sin⁡(1.5)≈0.9975\sin(1.5) \approx 0.9975: h′′(3)=−2.5(0.9975)−0.4=−2.494−0.4=−2.894h''(3) = -2.5(0.9975) - 0.4 = -2.494 - 0.4 = -2.894 To three significant figures, this is −2.89-2.89 m/s², which is answer A. Answer B (-10.4) likely comes from incorrectly using the coefficient 10 instead of properly applying the chain rule. Answer C (-0.847) might result from only considering the polynomial term's contribution. Answer D (28.2) appears to confuse signs or mix up derivative calculations entirely. Study tip: Always remember the kinematic hierarchy: position → velocity (first derivative) → acceleration (second derivative). Practice chain rule applications with trigonometric functions, as they appear frequently in oscillatory motion problems.

Question 16

The velocity of a rocket, vv in m/s, is modelled by the function v(t)=−0.1t3+2t2+5tv(t) = -0.1t^3 + 2t^2 + 5t for 0≤t≤200 \le t \le 20 seconds after launch.

What is the acceleration of the rocket at the time its velocity is at a maximum?

  1. 0 m/s2^2 (correct answer)
  2. 14.5 m/s2^2
  3. 18.3 m/s2^2
  4. 187 m/s2^2
Explanation: The velocity is at a maximum when its rate of change is zero. The rate of change of velocity is acceleration, a(t)=v′(t)a(t) = v'(t). Therefore, at the time of maximum velocity, the acceleration must be 0. To verify, we can find the time of maximum velocity by setting a(t)=v′(t)=−0.3t2+4t+5=0a(t) = v'(t) = -0.3t^2 + 4t + 5 = 0. Using a solver, this occurs at t≈14.5t \approx 14.5 s. The question asks for the acceleration at this time, which is by definition 0 m/s2^2.

Question 17

The volume of water in a tank, VV in litres, is given by V(t)=5000(1−0.01t)3V(t) = 5000(1 - 0.01t)^3, where tt is the time in hours.

Find the rate, in litres per hour, at which water is leaking from the tank after 10 hours.

  1. 50.0
  2. 121.5 (correct answer)
  3. 3645.0
  4. 12150.0
Explanation: The rate of change of volume is the derivative, V′(t)V'(t). Using the chain rule: V′(t)=5000⋅3(1−0.01t)2⋅(−0.01)=−150(1−0.01t)2V'(t) = 5000 \cdot 3(1 - 0.01t)^2 \cdot (-0.01) = -150(1 - 0.01t)^2. At t=10t=10, the rate of change is V′(10)=−150(1−0.01⋅10)2=−150(0.9)2=−150(0.81)=−121.5V'(10) = -150(1 - 0.01 \cdot 10)^2 = -150(0.9)^2 = -150(0.81) = -121.5 litres/hour. Since the question asks for the rate of leaking, it implies a positive value, which is 121.5 litres/hour.

Question 18

A particle moves in a straight line with acceleration a(t)=cos⁡(0.5t)a(t) = \cos(0.5t) m/s2^2 for t≥0t \ge 0. At t=0t=0, its velocity is v=3v=3 m/s and its displacement is s=1s=1 m.

Find the displacement of the particle at t=πt=\pi seconds.

  1. 5.00 m
  2. 9.42 m
  3. 14.4 m (correct answer)
  4. 17.4 m
Explanation: First, find velocity by integrating acceleration: v(t)=∫cos⁡(0.5t) dt=2sin⁡(0.5t)+Cv(t) = \int \cos(0.5t) \,dt = 2\sin(0.5t) + C. Given v(0)=3v(0)=3, 3=2sin⁡(0)+C⇒C=33 = 2\sin(0) + C \Rightarrow C=3. So, v(t)=2sin⁡(0.5t)+3v(t) = 2\sin(0.5t) + 3. Next, find displacement by integrating velocity: s(t)=∫(2sin⁡(0.5t)+3) dt=−4cos⁡(0.5t)+3t+Ds(t) = \int (2\sin(0.5t) + 3) \,dt = -4\cos(0.5t) + 3t + D. Given s(0)=1s(0)=1, 1=−4cos⁡(0)+0+D⇒1=−4+D⇒D=51 = -4\cos(0) + 0 + D \Rightarrow 1 = -4 + D \Rightarrow D=5. So, s(t)=−4cos⁡(0.5t)+3t+5s(t) = -4\cos(0.5t) + 3t + 5. At t=πt=\pi, s(π)=−4cos⁡(0.5π)+3π+5=−4(0)+3π+5=3π+5≈14.42s(\pi) = -4\cos(0.5\pi) + 3\pi + 5 = -4(0) + 3\pi + 5 = 3\pi + 5 \approx 14.42 m.

Question 19

The displacement of a particle from an origin is given by s(t)=t3−9t2+24t−1s(t) = t^3 - 9t^2 + 24t - 1 for t≥0t \ge 0.

For what values of tt is the particle moving in the positive direction?

  1. 2<t<42 < t < 4
  2. t=2t=2 and t=4t=4
  3. 0≤t<20 \le t < 2 or t>4t > 4 (correct answer)
  4. t<2t < 2 or t>4t > 4
Explanation: The particle moves in the positive direction when its velocity, v(t)v(t), is positive. First, find the velocity function by differentiating the displacement function: v(t)=s′(t)=3t2−18t+24v(t) = s'(t) = 3t^2 - 18t + 24. To find where v(t)>0v(t) > 0, first find the roots of v(t)=0v(t)=0: 3(t2−6t+8)=0⇒3(t−2)(t−4)=03(t^2 - 6t + 8) = 0 \Rightarrow 3(t-2)(t-4) = 0. The roots are t=2t=2 and t=4t=4. Since the velocity function is a parabola opening upwards, it is positive outside its roots. Therefore, v(t)>0v(t) > 0 for t<2t < 2 and t>4t > 4. Considering the domain t≥0t \ge 0, the particle moves in the positive direction for 0≤t<20 \le t < 2 or t>4t > 4.

Question 20

The displacement of a car from its starting point is given by s(t)=t3−3t2+5ts(t) = t^3 - 3t^2 + 5t metres for t≥0t \ge 0 seconds.

Find the time t>0t>0 at which the car's instantaneous velocity is equal to its average velocity over the interval [0,6][0, 6].

  1. 1.00 s
  2. 3.00 s
  3. 23.0 s
  4. 3.65 s (correct answer)
Explanation: This question tests the relationship between instantaneous velocity (derivative) and average velocity over an interval - a fundamental concept in calculus applications. To find when instantaneous velocity equals average velocity, you need both values. First, find the instantaneous velocity by taking the derivative: v(t)=s′(t)=3t2−6t+5v(t) = s'(t) = 3t^2 - 6t + 5. Next, calculate the average velocity over [0,6][0, 6] using s(6)−s(0)6−0\frac{s(6) - s(0)}{6 - 0}. You get s(6)=63−3(62)+5(6)=216−108+30=138s(6) = 6^3 - 3(6^2) + 5(6) = 216 - 108 + 30 = 138 and s(0)=0s(0) = 0, so average velocity is 1386=23\frac{138}{6} = 23 m/s. Now set instantaneous velocity equal to average velocity: 3t2−6t+5=233t^2 - 6t + 5 = 23. Rearranging gives 3t2−6t−18=03t^2 - 6t - 18 = 0, or t2−2t−6=0t^2 - 2t - 6 = 0. Using the quadratic formula: t=2±4+242=2±282=1±7t = \frac{2 \pm \sqrt{4 + 24}}{2} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7}. Since t>0t > 0, you want t=1+7≈3.65t = 1 + \sqrt{7} \approx 3.65 seconds. Choice A (1.00 s) likely comes from taking 1−71 - \sqrt{7} and using its absolute value. Choice B (3.00 s) might result from approximating 7≈2\sqrt{7} \approx 2. Choice C (23.0 s) confuses the time with the average velocity value itself. Strategy tip: When solving "instantaneous equals average" problems, always find the derivative first, then calculate the average rate of change, and finally solve the resulting equation. Watch for negative solutions that need to be rejected based on the domain.