IB · Mathematics: Applications & Interpretation SL & HL · Geometry & trigonometry

Vectors & kinematics: IB Mathematics: Applications & Interpretation questions with answers

Free Vectors & kinematics questions for IB Mathematics: Applications & Interpretation (SL & HL), HL only, part of Geometry & trigonometry. Each question has the answer and the marking points an examiner would look for.

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Vectors & kinematics practice questions

Question 1Structured · 14 marks

An aircraft’s position is (x, y, z) km, where x and y are displacements east and north of an airport and z is the height. At 13:00 it is at (40, 20, 6) and moves with constant velocity −200−100−15 km h−1. Let t be the time in hours after 13:00.

(a)

Write down a vector equation for the position r of the aircraft.

[2]
(b)

Verify that the aircraft would pass directly over the airport, and find its height at that time.

[4]
(c)

Find the time when the aircraft is 4 km above the ground, and its distance from the airport then.

[4]
(d)

The aircraft now adjusts its descent so that it lands at (0, 0, 0), keeping the same horizontal velocity −200−100. Find the new vertical component of velocity.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)r = 40206 + t−200−100−15
position + t × velocity (M1)
r = 40206 + t−200−100−15 A1
2
(b)t = 0.200; 3.00 km
x = 0 ⇒ t = 0.200 A1
y = 0 at the same t R1
substituting into z (M1)
3.00 km A1
4
(c)t = 0.133 h; 15.4 km
6 − 15t = 4 (M1)
t = 0.133 A1
position and distance (M1)
15.4 km A1
4
(d)−60.0 km h−1
time to reach the airport horizontally (M1)
0.0667 h (A1)
−4 ÷ time (M1)
−60.0 km h−1 A1
4
Question 2Structured · 6 marks

A boat leaves point P with position vector −2−3 km and travels with constant velocity 53 km h−1. A lighthouse is at L(16, 2.8).

(a)

Find the speed of the boat.

[2]
(b)

Find the time at which the boat is due north of the lighthouse, and the distance between them at that time.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)5.83 km h−1
|53| (M1)
5.83 km h−1 A1
2
(b)t = 3.60 h; 5.00 km
−2 + 5t = 16 (M1)
t = 3.60 A1
position (16, 7.80) (A1)
5.00 km A1
4
Question 3Structured · 4 marks

A boat’s position vector t hours after noon is r = −14 + t−1−1 km.

(a)

Find the speed of the boat.

[2]
(b)

Find the position of the boat at 2 pm.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)1.41 km h−1
magnitude of velocity (M1)
1.41 A1
2
(b)(−3, 2)
substitute t = 2 (M1)
(−3, 2) A1
2
Question 4Structured · 14 marks

An aircraft’s position is (x, y, z) km, where x and y are displacements east and north of an airport and z is the height. At 13:00 it is at (30, 10, 5) and moves with constant velocity −150−50−15 km h−1. Let t be the time in hours after 13:00.

(a)

Write down a vector equation for the position r of the aircraft.

[2]
(b)

Verify that the aircraft would pass directly over the airport, and find its height at that time.

[4]
(c)

Find the time when the aircraft is 4 km above the ground, and its distance from the airport then.

[4]
(d)

The aircraft now adjusts its descent so that it lands at (0, 0, 0), keeping the same horizontal velocity −150−50. Find the new vertical component of velocity.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)r = 30105 + t−150−50−15
position + t × velocity (M1)
r = 30105 + t−150−50−15 A1
2
(b)t = 0.200; 2.00 km
x = 0 ⇒ t = 0.200 A1
y = 0 at the same t R1
substituting into z (M1)
2.00 km A1
4
(c)t = 0.0667 h; 21.5 km
5 − 15t = 4 (M1)
t = 0.0667 A1
position and distance (M1)
21.5 km A1
4
(d)−30.0 km h−1
time to reach the airport horizontally (M1)
0.133 h (A1)
−4 ÷ time (M1)
−30.0 km h−1 A1
4
Question 5Structured · 6 marks

A boat leaves point P with position vector −75 km and travels with constant velocity 63 km h−1. A lighthouse is at L(12, 18.5).

(a)

Find the speed of the boat.

[2]
(b)

Find the time at which the boat is due south of the lighthouse, and the distance between them at that time.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)6.71 km h−1
|63| (M1)
6.71 km h−1 A1
2
(b)t = 3.17 h; 4.00 km
−7 + 6t = 12 (M1)
t = 3.17 A1
position (12, 14.5) (A1)
4.00 km A1
4
Question 6Structured · 4 marks

A boat’s position vector t hours after noon is r = 01 + t−22 km.

(a)

Find the speed of the boat.

[2]
(b)

Find the position of the boat at 4 pm.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)2.83 km h−1
magnitude of velocity (M1)
2.83 A1
2
(b)(−8, 9)
substitute t = 4 (M1)
(−8, 9) A1
2

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in