Kinematics of Linear Motion: Additional Mathematics questions

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

Which expression gives the acceleration of a particle from its displacement function s?

ds/dt
d²s/dt²
∫ s dt
∫∫ s dt dt

Show answerHide answer

a = dv/dt and v = ds/dt, so a = d²s/dt². ds/dt is the velocity; the integrals go the other way.

For the teacher: Remembering, low difficulty, standard 8.2.1

Textbook p. 260

A particle moves along a straight line and motion to the right is positive. At a certain instant its velocity is negative. What does this show?

The particle is at rest.
The particle is to the left of O.
The particle is moving to the left.
The particle is slowing down.

Show answerHide answer

v < 0 means the direction of motion is left. Position to the left of O is a negative displacement, not velocity; slowing down depends on the sign of a as well.

For the teacher: Understanding, low difficulty, standard 8.1.1

Textbook p. 255

The displacement of a particle is s = 4 + 8t − t² metres at t seconds. What is its displacement when t = 10?

−16 m
16 m
4 m
84 m

Show answerHide answer

s = 4 + 80 − 100 = −16 m, which is 16 m to the left of O. 16 forgets the sign.

For the teacher: Applying, low difficulty, standard 8.1.1

Textbook p. 253

A particle has velocity v = 2t² − 5t − 3 m/s at t seconds. Its initial velocity is

3 m/s
5 m/s
−5 m/s
−3 m/s

Show answerHide answer

Initial means t = 0: v = −3 m/s. The particle starts by moving to the left.

For the teacher: Applying, low difficulty, standard 8.2.2

Textbook p. 263

A particle has displacement s = t³ − 9t² + 24t + 5 metres. Find its velocity when t = 3.

27 m/s
−18 m/s
−3 m/s
5 m/s

Show answerHide answer

v = 3t² − 18t + 24. At t = 3: 27 − 54 + 24 = −3 m/s.

For the teacher: Applying, low difficulty, standard 8.2.2

Textbook p. 263

A particle starts at O with s = t² − 6t metres. Find the total distance travelled in the first 7 seconds.

7 m
25 m
18 m
9 m

Show answerHide answer

The particle turns at t = 3 (s = −9). Distance = 9 + 9 + 7 = 25 m. 7 m is the displacement at t = 7, not the distance.

For the teacher: Applying, low difficulty, standard 8.1.2

Textbook p. 258

A particle has displacement s = t³ − 3t² − 4t metres. Find its acceleration when t = 5.

24 m/s²
41 m/s²
−6 m/s²
6 m/s²

Show answerHide answer

v = 3t² − 6t − 4, a = 6t − 6 = 24 m/s². 41 is v(5), the velocity.

For the teacher: Applying, low difficulty, standard 8.2.3

Textbook p. 265

At one instant, the acceleration of a moving particle is zero. Which statement must be true at that instant?

The particle is at rest.
The particle is at the fixed point O.
The velocity of the particle is zero at all later times.
The velocity of the particle is momentarily not changing.

Show answerHide answer

a = dv/dt = 0 means the gradient of the velocity-time graph is zero, so v is momentarily not changing (often a maximum or minimum). It does not mean v = 0.

For the teacher: Understanding, medium difficulty, standard 8.1.1

Textbook p. 257

The acceleration of a particle is a = 6 − 2t m/s² and its initial velocity is 5 m/s. Find its velocity when t = 4.

8 m/s
−3 m/s
13 m/s
29 m/s

Show answerHide answer

v = 6t − t² + c and c = 5. v(4) = 24 − 16 + 5 = 13 m/s. 8 forgets the constant c.

For the teacher: Applying, low difficulty, standard 8.3.1

Textbook p. 267

The velocity of a particle is v = t² − 4t + 3 m/s. For which values of t does the particle move to the left?

t < 1
1 < t < 3
t > 3
t < 1 or t > 3

Show answerHide answer

v = (t − 1)(t − 3) is negative between the roots, so 1 < t < 3. Outside the roots v > 0 and the particle moves right.

For the teacher: Analysing, medium difficulty, standard 8.2.2

Textbook p. 263

The velocity of a particle is v = 6t − 3t² m/s and it starts from O. Find its displacement from O when t = 2.

0 m
12 m
8 m
4 m

Show answerHide answer

s = ∫(6t − 3t²) dt = 3t² − t³ (s = 0 at t = 0). s(2) = 12 − 8 = 4 m. 0 is v(2), found by substituting into the wrong function.

For the teacher: Applying, medium difficulty, standard 8.3.2

Textbook p. 269

A particle has velocity v = t² − 4t + 7 m/s. Hafiz says: "The acceleration is zero at t = 2, so the particle is at rest at t = 2." Which judgement is correct?

Wrong. a = 0 at t = 2 but v = 3 m/s, which is the minimum velocity.
Correct. a = 0 means the particle is at rest.
Wrong. a = 0 at t = 4, not at t = 2.
Correct. v = 0 when t = 2.

Show answerHide answer

a = 2t − 4 = 0 gives t = 2. v(2) = 4 − 8 + 7 = 3, not zero. Since d²v/dt² = 2 > 0, 3 m/s is the minimum velocity.

For the teacher: Evaluating, medium difficulty, standard 8.2.3

Textbook p. 264

Section B: structured questions

32 marks

Answer all questions. Show your working. Take motion to the right as positive.

The velocity of a particle that moves along a straight line is v = t² − 8t + 7 m/s, t seconds after passing O.

[1 mark] (a)Find the velocity when t = 3.

Show answerHide answer

9 − 24 + 7 = −8 m/s [1]1

Tick each point you got.

[2 marks] (b)Find the times when the particle stops instantaneously.

Show answerHide answer

(t − 1)(t − 7) = 0 [1]1
t = 1 s, t = 7 s [1]1

Tick each point you got.

[1 mark] (c)Find the range of t when the particle moves to the left.

Show answerHide answer

1 < t < 7 [1]1

Tick each point you got.

[2 marks] (d)Find the acceleration when t = 5.

Show answerHide answer

a = dv/dt = 2t − 8 [1]1
a = 2 m/s² [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 8.2.2

Textbook p. 263

The acceleration of a particle is a = 8 − 4t m/s², t seconds after passing O. Its initial velocity is 6 m/s.

[1 mark] (a)Find the acceleration when t = 4.

Show answerHide answer

8 − 16 = −8 m/s² [1]1

Tick each point you got.

[2 marks] (b)Express the velocity v in terms of t.

Show answerHide answer

v = ∫(8 − 4t) dt = 8t − 2t² + c [1]1
t = 0, v = 6, so c = 6 and v = 8t − 2t² + 6 [1]1

Tick each point you got.

[3 marks] (c)Find the time when the velocity is a maximum, and the maximum velocity.

Show answerHide answer

a = 0, 8 − 4t = 0, t = 2 s [1]1
d²v/dt² = −4 < 0, so v is a maximum [1]1
v = 16 − 8 + 6 = 14 m/s [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 8.3.1

Textbook p. 267

Diagram 1 shows the velocity-time graph v = t² − 6t + 8 of a particle that moves along a straight line, for 0 ≤ t ≤ 5.

Graph of v = t squared minus 6t plus 8 for t from 0 to 5. It starts at 8, crosses the t-axis at t = 2 and t = 4, and has a lowest point of −1 at t = 3. — Diagram 1

[1 mark] (a)State the times when the particle is instantaneously at rest.

Show answerHide answer

t = 2 s and t = 4 s [1]1

Tick each point you got.

[1 mark] (b)State the minimum velocity.

Show answerHide answer

−1 m/s [1]1

Tick each point you got.

[2 marks] (c)Find the acceleration at t = 3 and explain why the velocity is not zero then.

Show answerHide answer

a = 2t − 6 = 0 [1]1
a = 0 means v is at its minimum (−1 m/s), not that v = 0 [1]1

Tick each point you got.

[1 mark] (d)Find the range of t when the velocity is increasing.

Show answerHide answer

a > 0, 2t − 6 > 0, t > 3 [1]1

Tick each point you got.

For the teacher: Analysing, medium difficulty, standard 8.2.3

Textbook p. 264

A particle leaves O and moves along a straight line. Its velocity is v = 3t² − 12t + 9 m/s, t seconds after leaving O.

[2 marks] (a)Find the displacement function s.

Show answerHide answer

s = ∫(3t² − 12t + 9) dt = t³ − 6t² + 9t + c [1]1
s = 0 at t = 0, so c = 0 [1]1

Tick each point you got.

[2 marks] (b)Find the times when the particle stops instantaneously and its displacement at those times.

Show answerHide answer

3(t − 1)(t − 3) = 0, t = 1 s and t = 3 s [1]1
s(1) = 4 m, s(3) = 0 m [1]1

Tick each point you got.

[2 marks] (c)Find the total distance travelled in the first 4 seconds.

Show answerHide answer

s(4) = 64 − 96 + 36 = 4 m [1]1
Distance = 4 + 4 + 4 = 12 m (0 to 4, back to 0, then to 4) [1]1

The displacement at t = 4 is only 4 m, but the particle moved 12 m in total because it turned twice.

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 8.3.2

Textbook p. 269

A particle passes the fixed point O with velocity −8 m/s. Its acceleration is a = m + nt m/s², t seconds after passing O. At t = 0 the acceleration is 6 m/s² and at t = 3 the acceleration is zero.

[2 marks] (a)Find the values of m and n.

Show answerHide answer

t = 0: m = 6 [1]1
t = 3: 6 + 3n = 0, n = −2 [1]1

Tick each point you got.

[2 marks] (b)Express the velocity v in terms of t.

Show answerHide answer

v = ∫(6 − 2t) dt = 6t − t² + c [1]1
c = −8, v = −8 + 6t − t² [1]1

Tick each point you got.

[1 mark] (c)Find the maximum velocity.

Show answerHide answer

t = 3: v = −8 + 18 − 9 = 1 m/s [1]1

Tick each point you got.

[4 marks] (d)Find the total distance travelled in the first 4 seconds.

Show answerHide answer

v = −(t − 2)(t − 4), so v = 0 at t = 2 and t = 4 [1]1
s = −8t + 3t² − t³/3 [1]1
s(2) = −20/3 m, s(4) = −16/3 m [1]1
Distance = 20/3 + (20/3 − 16/3) = 8 m [1]1

The particle moves left for 0 < t < 2, then right for 2 < t < 4.

Tick each point you got.

For the teacher: Analysing, high difficulty, standard 8.4.1

Textbook p. 274

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Section B: structured questions

Question 13

Question 14

Question 15

Question 16

Question 17