IB Mathematics: Applications and Interpretation Quiz: 2d Vectors
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2d VectorsQuestion 1 of 20

Two particles start at the origin at the same time. Particle 1 has velocity v⃗1=(512)\vec{v}_1 = \begin{pmatrix} 5 \\ 12 \end{pmatrix} m/s and Particle 2 has velocity v⃗2=(8−3)\vec{v}_2 = \begin{pmatrix} 8 \\ -3 \end{pmatrix} m/s. What is the distance between the two particles after 4 seconds?

15.3 m
61.2 m
84.1 m
116 m
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: 2d Vectors

Practice 2d Vectors 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.

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This quiz focuses on 2d Vectors, 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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Question 1

Two particles start at the origin at the same time. Particle 1 has velocity v⃗1=(512)\vec{v}_1 = \begin{pmatrix} 5 \\ 12 \end{pmatrix} m/s and Particle 2 has velocity v⃗2=(8−3)\vec{v}_2 = \begin{pmatrix} 8 \\ -3 \end{pmatrix} m/s. What is the distance between the two particles after 4 seconds?

  1. 15.3 m
  2. 61.2 m (correct answer)
  3. 84.1 m
  4. 116 m
Explanation: First, find the relative velocity of Particle 2 with respect to Particle 1: v⃗rel=v⃗2−v⃗1=(8−3)−(512)=(3−15) m/s\vec{v}_{rel} = \vec{v}_2 - \vec{v}_1 = \begin{pmatrix} 8 \\ -3 \end{pmatrix} - \begin{pmatrix} 5 \\ 12 \end{pmatrix} = \begin{pmatrix} 3 \\ -15 \end{pmatrix} \text{ m/s} The displacement vector between them after 4 seconds is d⃗=t×v⃗rel\vec{d} = t \times \vec{v}_{rel}: d⃗=4(3−15)=(12−60) m\vec{d} = 4 \begin{pmatrix} 3 \\ -15 \end{pmatrix} = \begin{pmatrix} 12 \\ -60 \end{pmatrix} \text{ m} The distance between them is the magnitude of this displacement vector: Distance=∣d⃗∣=122+(−60)2=144+3600=3744≈61.19 m\text{Distance} = |\vec{d}| = \sqrt{12^2 + (-60)^2} = \sqrt{144 + 3600} = \sqrt{3744} \approx 61.19\text{ m}

Question 2

Two monitoring stations, A and B, have position vectors a⃗=(512)\vec{a} = \begin{pmatrix} 5 \\ 12 \end{pmatrix} and b⃗=(134)\vec{b} = \begin{pmatrix} 13 \\ 4 \end{pmatrix} relative to a central command post. An emergency beacon is activated at a point C, which is the midpoint of the line segment AB. What is the position vector of the beacon, c⃗\vec{c}?

  1. (4−4)\begin{pmatrix} 4 \\ -4 \end{pmatrix}
  2. (8−8)\begin{pmatrix} 8 \\ -8 \end{pmatrix}
  3. (98)\begin{pmatrix} 9 \\ 8 \end{pmatrix} (correct answer)
  4. (1816)\begin{pmatrix} 18 \\ 16 \end{pmatrix}
Explanation: The position vector of the midpoint of a line segment is the average of the position vectors of its endpoints. c⃗=12(a⃗+b⃗)\vec{c} = \frac{1}{2}(\vec{a} + \vec{b}) $$\vec{c} = \frac{1}{2}\left( \begin{pmatrix} 5 \ 12 \end{pmatrix} + \begin{pmatrix} 13 \ 4 \end{pmatrix} \right) = \frac{1}{2} \begin{pmatrix} 5+13 \ 12+4 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 18 \ 16 \end{pmatrix} = \begin{pmatrix} 9 \ 8 \end{pmatrix}

Question 3

A particle starts at the origin and is displaced by vector a⃗=(10−4)\vec{a} = \begin{pmatrix} 10 \\ -4 \end{pmatrix}. It then undergoes a second displacement, b⃗\vec{b}, which is parallel to a⃗\vec{a} and has a magnitude of 2.5 times the magnitude of a⃗\vec{a}. What is the final position vector of the particle?

  1. (25−10)\begin{pmatrix} 25 \\ -10 \end{pmatrix}
  2. (35−14)\begin{pmatrix} 35 \\ -14 \end{pmatrix} (correct answer)
  3. (35.40)\begin{pmatrix} 35.4 \\ 0 \end{pmatrix}
  4. (45.4−18.1)\begin{pmatrix} 45.4 \\ -18.1 \end{pmatrix}
Explanation: The second displacement vector b⃗\vec{b} is parallel to a⃗\vec{a} and its magnitude is 2.5 times that of a⃗\vec{a}. This means b⃗=2.5a⃗\vec{b} = 2.5\vec{a}. b⃗=2.5(10−4)=(25−10)\vec{b} = 2.5 \begin{pmatrix} 10 \\ -4 \end{pmatrix} = \begin{pmatrix} 25 \\ -10 \end{pmatrix} The final position vector is the sum of the two displacements: $$\text{Final Position} = \vec{a} + \vec{b} = \begin{pmatrix} 10 \ -4 \end{pmatrix} + \begin{pmatrix} 25 \ -10 \end{pmatrix} = \begin{pmatrix} 35 \ -14 \end{pmatrix}

Question 4

A weather balloon is tracked from a ground station. At 10:00 am its position vector is r⃗1=(28)\vec{r}_1 = \begin{pmatrix} 2 \\ 8 \end{pmatrix} km. At 10:30 am its position vector is r⃗2=(104)\vec{r}_2 = \begin{pmatrix} 10 \\ 4 \end{pmatrix} km.

What is the magnitude of the balloon's average velocity in km/h during this time interval?

  1. 8.94 km/h
  2. 12.0 km/h
  3. 17.9 km/h (correct answer)
  4. 24.0 km/h
Explanation: First, find the displacement vector from time 1 to time 2. Δr⃗=r⃗2−r⃗1=(104)−(28)=(8−4) km\Delta\vec{r} = \vec{r}_2 - \vec{r}_1 = \begin{pmatrix} 10 \\ 4 \end{pmatrix} - \begin{pmatrix} 2 \\ 8 \end{pmatrix} = \begin{pmatrix} 8 \\ -4 \end{pmatrix}\text{ km} The time interval is 30 minutes, which is 0.5 hours. The average velocity vector is the displacement divided by the time interval. v⃗avg=Δr⃗Δt=10.5(8−4)=(16−8) km/h\vec{v}_{avg} = \frac{\Delta\vec{r}}{\Delta t} = \frac{1}{0.5} \begin{pmatrix} 8 \\ -4 \end{pmatrix} = \begin{pmatrix} 16 \\ -8 \end{pmatrix}\text{ km/h} The magnitude of this average velocity vector is: ∣v⃗avg∣=162+(−8)2=256+64=320≈17.89 km/h|\vec{v}_{avg}| = \sqrt{16^2 + (-8)^2} = \sqrt{256 + 64} = \sqrt{320} \approx 17.89\text{ km/h}

Question 5

On a map, a starting point is at the origin O. A treasure hunter follows the displacement vector a⃗=(4020)\vec{a} = \begin{pmatrix} 40 \\ 20 \end{pmatrix} metres to reach point A, then follows the displacement vector b⃗=(−1060)\vec{b} = \begin{pmatrix} -10 \\ 60 \end{pmatrix} metres to reach point B.

The treasure is buried at the midpoint of the line segment AB. What is the position vector of the treasure relative to the starting point O?

  1. (1540)\begin{pmatrix} 15 \\ 40 \end{pmatrix}
  2. (3080)\begin{pmatrix} 30 \\ 80 \end{pmatrix}
  3. (3550)\begin{pmatrix} 35 \\ 50 \end{pmatrix} (correct answer)
  4. (70100)\begin{pmatrix} 70 \\ 100 \end{pmatrix}
Explanation: First, we need the position vectors of points A and B. The position vector of A is OA⃗=a⃗=(4020)\vec{OA} = \vec{a} = \begin{pmatrix} 40 \\ 20 \end{pmatrix}. The position vector of B is OB⃗=OA⃗+AB⃗=a⃗+b⃗\vec{OB} = \vec{OA} + \vec{AB} = \vec{a} + \vec{b}. OB⃗=(4020)+(−1060)=(3080)\vec{OB} = \begin{pmatrix} 40 \\ 20 \end{pmatrix} + \begin{pmatrix} -10 \\ 60 \end{pmatrix} = \begin{pmatrix} 30 \\ 80 \end{pmatrix} The treasure T is at the midpoint of AB, so its position vector OT⃗\vec{OT} is the average of the position vectors of A and B. $$\vec{OT} = \frac{1}{2}(\vec{OA} + \vec{OB}) = \frac{1}{2}\left( \begin{pmatrix} 40 \ 20 \end{pmatrix} + \begin{pmatrix} 30 \ 80 \end{pmatrix} \right) = \frac{1}{2}\begin{pmatrix} 70 \ 100 \end{pmatrix} = \begin{pmatrix} 35 \ 50 \end{pmatrix}

Question 6

A submarine travels from its base on two legs. The first displacement is d⃗1=(80−120)\vec{d}_1 = \begin{pmatrix} 80 \\ -120 \end{pmatrix} km, and the second is d⃗2=(−50180)\vec{d}_2 = \begin{pmatrix} -50 \\ 180 \end{pmatrix} km. What displacement vector d⃗3\vec{d}_3 must the submarine follow to return directly to its base?

  1. (−130−60)\begin{pmatrix} -130 \\ -60 \end{pmatrix}
  2. (−30−60)\begin{pmatrix} -30 \\ -60 \end{pmatrix} (correct answer)
  3. (3060)\begin{pmatrix} 30 \\ 60 \end{pmatrix}
  4. (130−300)\begin{pmatrix} 130 \\ -300 \end{pmatrix}
Explanation: To return to the base, the sum of all displacement vectors must be the zero vector: d⃗1+d⃗2+d⃗3=(00)\vec{d}_1 + \vec{d}_2 + \vec{d}_3 = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. First, find the current displacement from the base: d⃗1+d⃗2=(80−120)+(−50180)=(3060)\vec{d}_1 + \vec{d}_2 = \begin{pmatrix} 80 \\ -120 \end{pmatrix} + \begin{pmatrix} -50 \\ 180 \end{pmatrix} = \begin{pmatrix} 30 \\ 60 \end{pmatrix} The return vector d⃗3\vec{d}_3 must be the negative of this resultant vector: $$\vec{d}_3 = -(\vec{d}_1 + \vec{d}_2) = -\begin{pmatrix} 30 \ 60 \end{pmatrix} = \begin{pmatrix} -30 \ -60 \end{pmatrix}

Question 7

A triangular park has vertices at coordinates P(2, 6), Q(8, 3), and R(5, 10). A person walks from P to Q, and then from Q to R. Which vector represents the person's total displacement from their starting point P?

  1. (−63)\begin{pmatrix} -6 \\ 3 \end{pmatrix}
  2. (34)\begin{pmatrix} 3 \\ 4 \end{pmatrix} (correct answer)
  3. (3−3)\begin{pmatrix} 3 \\ -3 \end{pmatrix}
  4. (6−3)\begin{pmatrix} 6 \\ -3 \end{pmatrix}
Explanation: The total displacement is the sum of the individual displacement vectors, PQ⃗+QR⃗\vec{PQ} + \vec{QR}. By the triangle rule of vector addition, this sum is equal to the vector PR⃗\vec{PR}. We can calculate PR⃗\vec{PR} directly: PR⃗=OR⃗−OP⃗=(510)−(26)=(34)\vec{PR} = \vec{OR} - \vec{OP} = \begin{pmatrix} 5 \\ 10 \end{pmatrix} - \begin{pmatrix} 2 \\ 6 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} Alternatively, one could calculate each vector and add them: PQ⃗=(8−23−6)=(6−3)\vec{PQ} = \begin{pmatrix} 8-2 \\ 3-6 \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \end{pmatrix} and QR⃗=(5−810−3)=(−37)\vec{QR} = \begin{pmatrix} 5-8 \\ 10-3 \end{pmatrix} = \begin{pmatrix} -3 \\ 7 \end{pmatrix}. Then, PQ⃗+QR⃗=(6−3−3+7)=(34)\vec{PQ} + \vec{QR} = \begin{pmatrix} 6-3 \\ -3+7 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}.

Question 8

In a flight simulator, the velocity of the wind is given by the vector w⃗=(−2015)\vec{w} = \begin{pmatrix} -20 \\ 15 \end{pmatrix} km/h. The airplane's velocity relative to the air is v⃗a=(8000)\vec{v}_a = \begin{pmatrix} 800 \\ 0 \end{pmatrix} km/h. The components represent East and North respectively.

What is the ground speed of the airplane, correct to the nearest integer?

  1. 775 km/h
  2. 780 km/h (correct answer)
  3. 800 km/h
  4. 825 km/h
Explanation: The airplane's velocity relative to the ground, v⃗g\vec{v}_g, is the sum of its velocity relative to the air and the velocity of the wind: v⃗g=v⃗a+w⃗=(8000)+(−2015)=(78015)\vec{v}_g = \vec{v}_a + \vec{w} = \begin{pmatrix} 800 \\ 0 \end{pmatrix} + \begin{pmatrix} -20 \\ 15 \end{pmatrix} = \begin{pmatrix} 780 \\ 15 \end{pmatrix} The ground speed is the magnitude of this resultant vector: ∣v⃗g∣=7802+152=608400+225=608625≈780.14 km/h|\vec{v}_g| = \sqrt{780^2 + 15^2} = \sqrt{608400 + 225} = \sqrt{608625} \approx 780.14\text{ km/h} To the nearest integer, the speed is 780 km/h.

Question 9

An airplane flies from city P to city Q. The displacement vector for this journey is PQ⃗=(450−120)\vec{PQ} = \begin{pmatrix} 450 \\ -120 \end{pmatrix} km. The flight takes 1.5 hours. What is the velocity vector of the airplane?

  1. (300−80)\begin{pmatrix} 300 \\ -80 \end{pmatrix} (correct answer)
  2. (451.5−121.5)\begin{pmatrix} 451.5 \\ -121.5 \end{pmatrix}
  3. (675−180)\begin{pmatrix} 675 \\ -180 \end{pmatrix}
  4. 468 km/h\text{468 km/h}
Explanation: Velocity is displacement divided by time. The velocity vector is found by multiplying the displacement vector by the scalar 1time\frac{1}{\text{time}}. $$\vec{v} = \frac{1}{1.5} \vec{PQ} = \frac{1}{1.5} \begin{pmatrix} 450 \ -120 \end{pmatrix} = \begin{pmatrix} 450/1.5 \ -120/1.5 \end{pmatrix} = \begin{pmatrix} 300 \ -80 \end{pmatrix}

Question 10

The displacement vector from point A to point B is AB⃗=(−74)\vec{AB} = \begin{pmatrix} -7 \\ 4 \end{pmatrix}. The coordinates of point B are (1, -3). What are the coordinates of point A?

  1. (-6, 1)
  2. (-8, 7)
  3. (6, -1)
  4. (8, -7) (correct answer)
Explanation: Let the coordinates of A be (x,y)(x, y). The position vectors are OA⃗=(xy)\vec{OA} = \begin{pmatrix} x \\ y \end{pmatrix} and OB⃗=(1−3)\vec{OB} = \begin{pmatrix} 1 \\ -3 \end{pmatrix}. We know that AB⃗=OB⃗−OA⃗\vec{AB} = \vec{OB} - \vec{OA}. We need to solve for OA⃗\vec{OA}. OA⃗=OB⃗−AB⃗\vec{OA} = \vec{OB} - \vec{AB} $$\vec{OA} = \begin{pmatrix} 1 \ -3 \end{pmatrix} - \begin{pmatrix} -7 \ 4 \end{pmatrix} = \begin{pmatrix} 1 - (-7) \ -3 - 4 \end{pmatrix} = \begin{pmatrix} 8 \ -7 \end{pmatrix}

Question 11

Two tugboats are pulling a large ship. The forces are represented by vectors F⃗1=(6090)\vec{F}_1 = \begin{pmatrix} 60 \\ 90 \end{pmatrix} kN and F⃗2=(p−50)\vec{F}_2 = \begin{pmatrix} p \\ -50 \end{pmatrix} kN. The resultant force has a magnitude of 100 kN. Given that the horizontal component of the resultant force is positive, find the value of pp, correct to one decimal place.

  1. 31.7 (correct answer)
  2. 91.7
  3. 151.7
  4. 211.7
Explanation: First, find the resultant force vector by adding the two force vectors: F⃗R=F⃗1+F⃗2=(6090)+(p−50)=(60+p40)\vec{F}_R = \vec{F}_1 + \vec{F}_2 = \begin{pmatrix} 60 \\ 90 \end{pmatrix} + \begin{pmatrix} p \\ -50 \end{pmatrix} = \begin{pmatrix} 60+p \\ 40 \end{pmatrix} The magnitude of this resultant force is given as 100 kN. ∣F⃗R∣=(60+p)2+402=100|\vec{F}_R| = \sqrt{(60+p)^2 + 40^2} = 100 Square both sides: (60+p)2+1600=10000(60+p)^2 + 1600 = 10000 (60+p)2=8400(60+p)^2 = 8400 60+p=±8400≈±91.6560+p = \pm\sqrt{8400} \approx \pm 91.65 The question states the horizontal component, 60+p60+p, is positive, so we take the positive root. 60+p=91.6560+p = 91.65 p=91.65−60=31.65≈31.7p = 91.65 - 60 = 31.65 \approx 31.7

Question 12

A drone flies along a displacement vector d⃗=(6−8)\vec{d} = \begin{pmatrix} 6 \\ -8 \end{pmatrix} km from its base. It then flies back towards its base, covering exactly half of the return journey. What is the drone's final position vector relative to its base?

  1. (3−4)\begin{pmatrix} 3 \\ -4 \end{pmatrix} (correct answer)
  2. (−34)\begin{pmatrix} -3 \\ 4 \end{pmatrix}
  3. (9−12)\begin{pmatrix} 9 \\ -12 \end{pmatrix}
  4. (12−16)\begin{pmatrix} 12 \\ -16 \end{pmatrix}
Explanation: The initial displacement is d⃗=(6−8)\vec{d} = \begin{pmatrix} 6 \\ -8 \end{pmatrix}. The vector for the full return journey would be −d⃗=(−68)-\vec{d} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}. The drone covers half of this return journey, so its second displacement is 12(−d⃗)=(−34)\frac{1}{2}(-\vec{d}) = \begin{pmatrix} -3 \\ 4 \end{pmatrix}. The final position is the sum of the two displacements: Final Position=(6−8)+(−34)=(3−4)\text{Final Position} = \begin{pmatrix} 6 \\ -8 \end{pmatrix} + \begin{pmatrix} -3 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix} This is equivalent to finding the midpoint of the initial displacement vector, 12d⃗\frac{1}{2}\vec{d}.

Question 13

A submarine travels from its base on two legs. The first displacement is d⃗1=(80−120)\vec{d}_1 = \begin{pmatrix} 80 \\ -120 \end{pmatrix} km, and the second is d⃗2=(−50180)\vec{d}_2 = \begin{pmatrix} -50 \\ 180 \end{pmatrix} km. What displacement vector d⃗3\vec{d}_3 must the submarine follow to return directly to its base?

  1. (−130−60)\begin{pmatrix} -130 \\ -60 \end{pmatrix}
  2. (−30−60)\begin{pmatrix} -30 \\ -60 \end{pmatrix} (correct answer)
  3. (3060)\begin{pmatrix} 30 \\ 60 \end{pmatrix}
  4. (130−300)\begin{pmatrix} 130 \\ -300 \end{pmatrix}
Explanation: To return to the base, the sum of all displacement vectors must be the zero vector: d⃗1+d⃗2+d⃗3=(00)\vec{d}_1 + \vec{d}_2 + \vec{d}_3 = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. First, find the current displacement from the base: d⃗1+d⃗2=(80−120)+(−50180)=(3060)\vec{d}_1 + \vec{d}_2 = \begin{pmatrix} 80 \\ -120 \end{pmatrix} + \begin{pmatrix} -50 \\ 180 \end{pmatrix} = \begin{pmatrix} 30 \\ 60 \end{pmatrix} The return vector d⃗3\vec{d}_3 must be the negative of this resultant vector: $$\vec{d}_3 = -(\vec{d}_1 + \vec{d}_2) = -\begin{pmatrix} 30 \ 60 \end{pmatrix} = \begin{pmatrix} -30 \ -60 \end{pmatrix}

Question 14

A delivery robot starts at a warehouse located at the origin. It travels to point A with position vector OA⃗=(12−5)\vec{OA} = \begin{pmatrix} 12 \\ -5 \end{pmatrix} km. It then travels to point B, such that the displacement from A to B is AB⃗=(−820)\vec{AB} = \begin{pmatrix} -8 \\ 20 \end{pmatrix} km. What is the magnitude of the robot's final displacement vector from the warehouse, OB⃗\vec{OB}?

  1. 13.0 km
  2. 15.5 km (correct answer)
  3. 21.5 km
  4. 34.5 km
Explanation: The final displacement vector from the warehouse (origin O) to point B is the sum of the two displacement vectors: OB⃗=OA⃗+AB⃗\vec{OB} = \vec{OA} + \vec{AB}. OB⃗=(12−5)+(−820)=(12−8−5+20)=(415)\vec{OB} = \begin{pmatrix} 12 \\ -5 \end{pmatrix} + \begin{pmatrix} -8 \\ 20 \end{pmatrix} = \begin{pmatrix} 12-8 \\ -5+20 \end{pmatrix} = \begin{pmatrix} 4 \\ 15 \end{pmatrix} The magnitude of this vector is ∣OB⃗∣=42+152=16+225=241≈15.52|\vec{OB}| = \sqrt{4^2 + 15^2} = \sqrt{16 + 225} = \sqrt{241} \approx 15.52 km.

Question 15

A triangular park has vertices at coordinates P(2, 6), Q(8, 3), and R(5, 10). A person walks from P to Q, and then from Q to R. Which vector represents the person's total displacement from their starting point P?

  1. (−63)\begin{pmatrix} -6 \\ 3 \end{pmatrix}
  2. (34)\begin{pmatrix} 3 \\ 4 \end{pmatrix} (correct answer)
  3. (3−3)\begin{pmatrix} 3 \\ -3 \end{pmatrix}
  4. (6−3)\begin{pmatrix} 6 \\ -3 \end{pmatrix}
Explanation: The total displacement is the sum of the individual displacement vectors, PQ⃗+QR⃗\vec{PQ} + \vec{QR}. By the triangle rule of vector addition, this sum is equal to the vector PR⃗\vec{PR}. We can calculate PR⃗\vec{PR} directly: PR⃗=OR⃗−OP⃗=(510)−(26)=(34)\vec{PR} = \vec{OR} - \vec{OP} = \begin{pmatrix} 5 \\ 10 \end{pmatrix} - \begin{pmatrix} 2 \\ 6 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} Alternatively, one could calculate each vector and add them: PQ⃗=(8−23−6)=(6−3)\vec{PQ} = \begin{pmatrix} 8-2 \\ 3-6 \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \end{pmatrix} and QR⃗=(5−810−3)=(−37)\vec{QR} = \begin{pmatrix} 5-8 \\ 10-3 \end{pmatrix} = \begin{pmatrix} -3 \\ 7 \end{pmatrix}. Then, PQ⃗+QR⃗=(6−3−3+7)=(34)\vec{PQ} + \vec{QR} = \begin{pmatrix} 6-3 \\ -3+7 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}.

Question 16

In a flight simulator, the velocity of the wind is given by the vector w⃗=(−2015)\vec{w} = \begin{pmatrix} -20 \\ 15 \end{pmatrix} km/h. The airplane's velocity relative to the air is v⃗a=(8000)\vec{v}_a = \begin{pmatrix} 800 \\ 0 \end{pmatrix} km/h. The components represent East and North respectively.

What is the ground speed of the airplane, correct to the nearest integer?

  1. 775 km/h
  2. 780 km/h (correct answer)
  3. 800 km/h
  4. 825 km/h
Explanation: The airplane's velocity relative to the ground, v⃗g\vec{v}_g, is the sum of its velocity relative to the air and the velocity of the wind: v⃗g=v⃗a+w⃗=(8000)+(−2015)=(78015)\vec{v}_g = \vec{v}_a + \vec{w} = \begin{pmatrix} 800 \\ 0 \end{pmatrix} + \begin{pmatrix} -20 \\ 15 \end{pmatrix} = \begin{pmatrix} 780 \\ 15 \end{pmatrix} The ground speed is the magnitude of this resultant vector: ∣v⃗g∣=7802+152=608400+225=608625≈780.14 km/h|\vec{v}_g| = \sqrt{780^2 + 15^2} = \sqrt{608400 + 225} = \sqrt{608625} \approx 780.14\text{ km/h} To the nearest integer, the speed is 780 km/h.

Question 17

The velocity of a ship is represented by vector u⃗=(9−12)\vec{u} = \begin{pmatrix} 9 \\ -12 \end{pmatrix} km/h. A speedboat is moving parallel to the ship, with its velocity represented by vector v⃗=(x8)\vec{v} = \begin{pmatrix} x \\ 8 \end{pmatrix} km/h. Find the value of xx.

  1. -6.0 (correct answer)
  2. -6.75
  3. 6.0
  4. 10.7
Explanation: Since the vectors are parallel, one must be a scalar multiple of the other. Let v⃗=ku⃗\vec{v} = k\vec{u}. (x8)=k(9−12)\begin{pmatrix} x \\ 8 \end{pmatrix} = k \begin{pmatrix} 9 \\ -12 \end{pmatrix} From the y-components, we can find the scalar kk: 8=k(−12)  ⟹  k=−812=−238 = k(-12) \implies k = -\frac{8}{12} = -\frac{2}{3} Now, use this scalar to find xx from the x-components: x=k(9)=(−23)×9=−6x = k(9) = \left(-\frac{2}{3}\right) \times 9 = -6

Question 18

An airplane flies from city P to city Q. The displacement vector for this journey is PQ⃗=(450−120)\vec{PQ} = \begin{pmatrix} 450 \\ -120 \end{pmatrix} km. The flight takes 1.5 hours. What is the velocity vector of the airplane?

  1. (300−80)\begin{pmatrix} 300 \\ -80 \end{pmatrix} (correct answer)
  2. (451.5−121.5)\begin{pmatrix} 451.5 \\ -121.5 \end{pmatrix}
  3. (675−180)\begin{pmatrix} 675 \\ -180 \end{pmatrix}
  4. 468 km/h\text{468 km/h}
Explanation: Velocity is displacement divided by time. The velocity vector is found by multiplying the displacement vector by the scalar 1time\frac{1}{\text{time}}. $$\vec{v} = \frac{1}{1.5} \vec{PQ} = \frac{1}{1.5} \begin{pmatrix} 450 \ -120 \end{pmatrix} = \begin{pmatrix} 450/1.5 \ -120/1.5 \end{pmatrix} = \begin{pmatrix} 300 \ -80 \end{pmatrix}

Question 19

A food manufacturer is creating a new snack mix. The nutritional content (in grams per serving) for two ingredients, nuts and dried fruit, is given by vectors where the components represent (protein, carbohydrates). For nuts, the vector is n⃗=(58)\vec{n} = \begin{pmatrix} 5 \\ 8 \end{pmatrix}. For dried fruit, the vector is f⃗=(125)\vec{f} = \begin{pmatrix} 1 \\ 25 \end{pmatrix}.

A new mix is created using 2 servings of nuts and 1.5 servings of dried fruit. What is the total nutritional vector for this new mix?

  1. (633)\begin{pmatrix} 6 \\ 33 \end{pmatrix}
  2. (7.545.5)\begin{pmatrix} 7.5 \\ 45.5 \end{pmatrix}
  3. (11.553.5)\begin{pmatrix} 11.5 \\ 53.5 \end{pmatrix} (correct answer)
  4. (1582.5)\begin{pmatrix} 15 \\ 82.5 \end{pmatrix}
Explanation: The total nutritional vector is a linear combination of the ingredient vectors: T⃗=2n⃗+1.5f⃗\vec{T} = 2\vec{n} + 1.5\vec{f}. T⃗=2(58)+1.5(125)\vec{T} = 2 \begin{pmatrix} 5 \\ 8 \end{pmatrix} + 1.5 \begin{pmatrix} 1 \\ 25 \end{pmatrix} $$\vec{T} = \begin{pmatrix} 10 \ 16 \end{pmatrix} + \begin{pmatrix} 1.5 \ 37.5 \end{pmatrix} = \begin{pmatrix} 10+1.5 \ 16+37.5 \end{pmatrix} = \begin{pmatrix} 11.5 \ 53.5 \end{pmatrix}

Question 20

Two tugboats are pulling a large ship. The forces are represented by vectors F⃗1=(6090)\vec{F}_1 = \begin{pmatrix} 60 \\ 90 \end{pmatrix} kN and F⃗2=(p−50)\vec{F}_2 = \begin{pmatrix} p \\ -50 \end{pmatrix} kN. The resultant force has a magnitude of 100 kN. Given that the horizontal component of the resultant force is positive, find the value of pp, correct to one decimal place.

  1. 31.7 (correct answer)
  2. 91.7
  3. 151.7
  4. 211.7
Explanation: First, find the resultant force vector by adding the two force vectors: F⃗R=F⃗1+F⃗2=(6090)+(p−50)=(60+p40)\vec{F}_R = \vec{F}_1 + \vec{F}_2 = \begin{pmatrix} 60 \\ 90 \end{pmatrix} + \begin{pmatrix} p \\ -50 \end{pmatrix} = \begin{pmatrix} 60+p \\ 40 \end{pmatrix} The magnitude of this resultant force is given as 100 kN. ∣F⃗R∣=(60+p)2+402=100|\vec{F}_R| = \sqrt{(60+p)^2 + 40^2} = 100 Square both sides: (60+p)2+1600=10000(60+p)^2 + 1600 = 10000 (60+p)2=8400(60+p)^2 = 8400 60+p=±8400≈±91.6560+p = \pm\sqrt{8400} \approx \pm 91.65 The question states the horizontal component, 60+p60+p, is positive, so we take the positive root. 60+p=91.6560+p = 91.65 p=91.65−60=31.65≈31.7p = 91.65 - 60 = 31.65 \approx 31.7