WAEC Further Mathematics 2022 past questions and answers

50 questions. Try each one, then open the answer to check yourself and read the explanation.

Question 1

Binary Operations

A binary operation Δ is defined on the set of real numbers, R, by x Δ y = √(x + y − xy/4), where x, y ∈ R. Find the value of 4 Δ 3.

  1. A.16
  2. B.8
  3. C.4
  4. D.2
Show the answer and explanation

Answer: D (2)

Explanation

4 Δ 3 = √(4 + 3 − 12/4) = √4 = 2.

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Question 2

Surds

Simplify: ((3√6 + √54)/(√5(3√5)))−1

  1. A.5√3/6
  2. B.3√15/6
  3. C.5√6/12
  4. D.5√3/12
Show the answer and explanation

Answer: C (5√6/12)

Explanation

√54 = 3√6, so the fraction is 6√6/15 = 2√6/5. Its reciprocal is 5/(2√6) = 5√6/12.

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Question 3

Indices and Logarithms

If log10(3x − 1) + log104 = log10(9x + 2), find the value of x.

  1. A.1/3
  2. B.1
  3. C.2
  4. D.3
Show the answer and explanation

Answer: C (2)

Explanation

log 4(3x − 1) = log(9x + 2), so 12x − 4 = 9x + 2 and x = 2.

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Question 4

Indices and Logarithms

Simplify: (9 × 3n+1 − 3n+2)/(3n+1 − 3n)

  1. A.3
  2. B.9
  3. C.27
  4. D.81
Show the answer and explanation

Answer: B (9)

Explanation

Take out 3ⁿ: top = 3ⁿ(27 − 9) = 18 × 3ⁿ, bottom = 3ⁿ(3 − 1) = 2 × 3ⁿ. The answer is 9.

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Question 5

Sets and Logic

Consider the following statements:
x: All wrestlers are strong
y: Some wrestlers are not weightlifters
Which of the following is a valid conclusion?

  1. A.All strong wrestlers are weightlifters
  2. B.Some strong wrestlers are not weightlifters
  3. C.Some weak wrestlers are weightlifters
  4. D.All weightlifters are wrestlers
Show the answer and explanation

Answer: B (Some strong wrestlers are not weightlifters)

Explanation

The wrestlers who are not weightlifters are still wrestlers, so they are strong. Hence some strong wrestlers are not weightlifters.

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Question 6

Functions

The functions f: x → 2x2 + 3x − 7 and g: x → 5x2 + 7x − 6 are defined on the set of real numbers, R. Find the values of x for which 3f(x) = g(x).

  1. A.x = −3 or −5
  2. B.x = −3 or 5
  3. C.x = 3 or −5
  4. D.x = 3 or 5
Show the answer and explanation

Answer: C (x = 3 or −5)

Explanation

6x² + 9x − 21 = 5x² + 7x − 6 gives x² + 2x − 15 = 0, (x + 5)(x − 3) = 0, so x = 3 or −5.

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Question 7

Trigonometry

Express 4π/5 radians in degrees.

  1. A.288°
  2. B.200°
  3. C.144°
  4. D.120°
Show the answer and explanation

Answer: C (144°)

Explanation

π radians = 180°, so 4π/5 = 4 × 36° = 144°.

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Question 8

Polynomials

Given that (8x + m)/(x2 − 3x − 4) ≡ 5/(x + 1) + 3/(x − 4), find the value of m.

  1. A.23
  2. B.17
  3. C.−17
  4. D.−23
Show the answer and explanation

Answer: C (−17)

Explanation

8x + m = 5(x − 4) + 3(x + 1) = 8x − 17, so m = −17.

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Question 9

Differentiation

If x2 + y2 − 2x − 6y + 5 = 0, evaluate dy/dx when x = 3 and y = 2.

  1. A.−2
  2. B.2
  3. C.4
  4. D.−4
Show the answer and explanation

Answer: B (2)

Explanation

2x + 2y dy/dx − 2 − 6 dy/dx = 0, so dy/dx = (2 − 2x)/(2y − 6) = (2 − 6)/(4 − 6) = 2.

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Question 10

Integration

Evaluate ∫01 x2(x3 + 2)3 dx.

  1. A.56/12
  2. B.65/12
  3. C.12
  4. D.65
Show the answer and explanation

Answer: B (65/12)

Explanation

Let u = x³ + 2, du = 3x² dx: ∫ = (x³ + 2)⁴/12 from 0 to 1 = (81 − 16)/12 = 65/12.

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Question 13

Permutations and Combinations

Evaluate: 4P2 + 4C2 − 4P3.

  1. A.18
  2. B.6
  3. C.−6
  4. D.−18
Show the answer and explanation

Answer: C (−6)

Explanation

⁴P₂ = 12, ⁴C₂ = 6, ⁴P₃ = 24. 12 + 6 − 24 = −6.

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Question 14

Binomial Theorem

Find the coefficient of x2 in the binomial expansion of (x + 2/x2)5.

  1. A.10
  2. B.40
  3. C.32
  4. D.80
Show the answer and explanation

Answer: A (10)

Explanation

The general term ⁵Cᵣ x⁵⁻ʳ(2/x²)ʳ has power 5 − 3r. For x², r = 1: ⁵C₁ × 2 = 10.

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Question 15

Sets and Logic

Given that P = {x : x is a multiple of 5}, Q = {x : x is a multiple of 3} and R = {x : x is an odd number} are subsets of μ = {x : 20 ≤ x ≤ 35}, find (P ∪ Q) ∩ R.

  1. A.{20, 21, 25, 30, 33}
  2. B.{21, 25, 27, 33, 35}
  3. C.{20, 21, 25, 27, 33, 35}
  4. D.{21, 25, 27, 30, 33, 35}
Show the answer and explanation

Answer: B ({21, 25, 27, 33, 35})

Explanation

P = {20, 25, 30, 35}, Q = {21, 24, 27, 30, 33}. Their union's odd members are 21, 25, 27, 33, 35.

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Question 16

Dynamics

A particle moving with a velocity of 5 ms−1 accelerates at 2 ms−2. Find the distance it covers in 4 seconds.

  1. A.16 m
  2. B.26 m
  3. C.36 m
  4. D.46 m
Show the answer and explanation

Answer: C (36 m)

Explanation

s = ut + ½at² = 5(4) + ½(2)(16) = 20 + 16 = 36 m.

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Question 17

Sequences and Series

If Un = kn2 + pn, U1 = −1, U5 = 15, find the values of k and p.

  1. A.k = −1, p = 2
  2. B.k = −1, p = −2
  3. C.k = 1, p = −2
  4. D.k = 1, p = 2
Show the answer and explanation

Answer: C (k = 1, p = −2)

Explanation

k + p = −1 and 25k + 5p = 15 (so 5k + p = 3). Subtracting: 4k = 4, k = 1 and p = −2.

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Question 18

Permutations and Combinations

In how many ways can six persons be paired?

  1. A.5
  2. B.10
  3. C.15
  4. D.25
Show the answer and explanation

Answer: C (15)

Explanation

Choosing a pair of 2 from 6 people: ⁶C₂ = 15 (this is also the number of ways to split 6 people into 3 pairs: 6!/(2³ × 3!) = 15).

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Question 19

Indices and Logarithms

Solve: 32x−2 − 28(3x−2) + 3 = 0.

  1. A.x = −2 or x = 1
  2. B.x = 0 or x = −3
  3. C.x = 2 or x = 1
  4. D.x = 0 or x = 3
Show the answer and explanation

Answer: D (x = 0 or x = 3)

Explanation

Let p = 3ˣ: p²/9 − 28p/9 + 3 = 0, so p² − 28p + 27 = 0, (p − 1)(p − 27) = 0. 3ˣ = 1 or 27 gives x = 0 or 3.

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Question 20

Vectors

Given that P = (−4, −5) and Q = (2, 3), express PQ in the form (k, θ), where k is the magnitude and θ the bearing.

  1. A.(10 units, 063°)
  2. B.(9 units, 049°)
  3. C.(10 units, 037°)
  4. D.(9 units, 027°)
Show the answer and explanation

Answer: C ((10 units, 037°))

Explanation

PQ = (6, 8), so k = √(36 + 64) = 10. The bearing is measured from north: tan θ = 6/8, θ ≈ 037°.

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Question 21

Vectors

If PQ = −2i + 5j and RQ = −i − 7j, find PR.

  1. A.−3i + 12j
  2. B.−3i − 12j
  3. C.−i + 12j
  4. D.i − 12j
Show the answer and explanation

Answer: C (−i + 12j)

Explanation

PR = PQ + QR = PQ − RQ = (−2i + 5j) − (−i − 7j) = −i + 12j.

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Question 22

Probability

The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.
Distance (km): 3, 4, 5, 6, 7, 8
Frequency: 5, 4, x, 9, 2x, 1
If a hunter is selected at random, find the probability that the hunter covered at least 6 km.

  1. A.3/5
  2. B.2/5
  3. C.3/8
  4. D.9/40
Show the answer and explanation

Answer: A (3/5)

Explanation

19 + 3x = 40 gives x = 7. Hunters covering 6 km or more: 9 + 14 + 1 = 24. P = 24/40 = 3/5.

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Question 23

Statistics

The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.
Distance (km): 3, 4, 5, 6, 7, 8
Frequency: 5, 4, x, 9, 2x, 1
Find the mode of the distribution

  1. A.5
  2. B.6
  3. C.7
  4. D.8
Show the answer and explanation

Answer: C (7)

Explanation

x = 7, so 7 km has frequency 2x = 14, the highest. Mode = 7.

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Question 24

Functions

If g(x) = √(1 − x2), find the domain of g(x).

  1. A.x < −1 or x > 1
  2. B.x ≤ −1 or x ≥ 1
  3. C.−1 ≤ x ≤ 1
  4. D.−1 < x < 1
Show the answer and explanation

Answer: C (−1 ≤ x ≤ 1)

Explanation

We need 1 − x² ≥ 0, i.e. x² ≤ 1, so −1 ≤ x ≤ 1.

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Question 25

Binomial Theorem

Find the coefficient of x3y2 in the binomial expansion of (x − 2y)5.

  1. A.−80
  2. B.−10
  3. C.40
  4. D.80
Show the answer and explanation

Answer: C (40)

Explanation

The term is ⁵C₂x³(−2y)² = 10 × 4x³y² = 40x³y².

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Question 26

Sequences and Series

The first, second and third terms of an exponential sequence (G.P) are (x − 4), (x + 2) and (3x + 1) respectively. Find the values of x.

  1. A.−1/2, 8
  2. B.1/2, −8
  3. C.−1/2, −8
  4. D.1/2, 8
Show the answer and explanation

Answer: A (−1/2, 8)

Explanation

(x + 2)² = (x − 4)(3x + 1) gives x² + 4x + 4 = 3x² − 11x − 4, so 2x² − 15x − 8 = 0, (2x + 1)(x − 8) = 0: x = −½ or 8.

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Question 27

Dynamics

A body of mass 18 kg moving with velocity 4 ms−1 collides with another body of mass 6 kg moving in the opposite direction with velocity 10 ms−1. If they stick together after collision, find their common velocity.

  1. A.−½ ms−1
  2. B.⅓ ms−1
  3. C.2 ms−1
  4. D.3 ms−1
Show the answer and explanation

Answer: A (−½ ms−1)

Explanation

Taking the 6 kg body's direction as positive: 18(−4) + 6(10) = 24v, so v = −12/24 = −½ ms⁻¹ (0.5 ms⁻¹ in the direction the 18 kg body was moving).

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Question 28

Statistics

The mean heights of three groups of students consisting of 20, 16 and 14 students each are 1.67 m, 1.50 m and 1.40 m respectively. Find the mean height of all the students.

  1. A.1.63 m
  2. B.1.54 m
  3. C.1.52 m
  4. D.1.42 m
Show the answer and explanation

Answer: B (1.54 m)

Explanation

Total height = 20(1.67) + 16(1.50) + 14(1.40) = 33.4 + 24 + 19.6 = 77; 77 ÷ 50 = 1.54 m.

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Question 29

Coordinate Geometry

Find, correct to the nearest degree, the acute angle formed by the lines y = 2x + 5 and 2y = x − 6.

  1. A.76°
  2. B.53°
  3. C.37°
  4. D.14°
Show the answer and explanation

Answer: C (37°)

Explanation

Gradients 2 and ½. tan θ = (2 − ½)/(1 + 2 × ½) = 0.75, so θ ≈ 37°.

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Question 30

Trigonometry

Solve: 4sin2θ + 1 = 2, where 0° < θ < 180°.

  1. A.60°, 120°
  2. B.30°, 150°
  3. C.30°, 120°
  4. D.60°, 150°
Show the answer and explanation

Answer: B (30°, 150°)

Explanation

sin²θ = ¼, so sin θ = ½ (sin θ is positive in this range): θ = 30° or 150°.

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Question 32

Probability

The probability that a student will graduate from a college is 0.4. If 3 students are selected from the college, what is the probability that at least one student will graduate?

  1. A.0.06
  2. B.0.22
  3. C.0.78
  4. D.0.80
Show the answer and explanation

Answer: C (0.78)

Explanation

P(none graduate) = 0.6³ = 0.216, so P(at least one) = 1 − 0.216 = 0.784 ≈ 0.78.

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Question 33

Coordinate Geometry

The equation of a circle is given as 2x2 + 2y2 − x − 3y − 41 = 0. Find the coordinates of its centre.

  1. A.(−¼, ¾)
  2. B.(¼, ¾)
  3. C.(−½, 3/2)
  4. D.(−½, −3/2)
Show the answer and explanation

Answer: B ((¼, ¾))

Explanation

Divide by 2: x² + y² − ½x − (3/2)y − 41/2 = 0. The centre is (¼, ¾).

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Question 34

Integration

The gradient of a function at any point (x, y) is 2x − 6. If the function passes through (1, 2), find the function.

  1. A.y = x2 − 6x − 5
  2. B.y = x2 − 6x + 5
  3. C.y = x2 − 6x − 3
  4. D.y = x2 − 6x + 7
Show the answer and explanation

Answer: D (y = x2 − 6x + 7)

Explanation

y = x² − 6x + c. At (1, 2): 1 − 6 + c = 2, so c = 7 and y = x² − 6x + 7.

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Question 35

Dynamics

A particle of mass 3 kg moving along a straight line under the action of a F N, covers a line distance, d, at time, t, such that d = t2 + 3t. Find the magnitude of F at time t.

  1. A.0 N
  2. B.2 N
  3. C.3(2t + 3) N
  4. D.6 N
Show the answer and explanation

Answer: D (6 N)

Explanation

v = 2t + 3 and a = 2 ms⁻², so F = ma = 3 × 2 = 6 N.

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Question 36

Polynomials

If α and β are the roots of x2 + mx − n = 0, where m and n are constants, form the equation whose roots are 1/α and 1/β.

  1. A.mnx2 − n2x − m = 0
  2. B.mx2 − nx + 1 = 0
  3. C.nx2 − mx + 1 = 0
  4. D.nx2 − mx − 1 = 0
Show the answer and explanation

Answer: D (nx2 − mx − 1 = 0)

Explanation

α + β = −m, αβ = −n. New sum = (α + β)/αβ = m/n, new product = 1/αβ = −1/n. So x² − (m/n)x − 1/n = 0, i.e. nx² − mx − 1 = 0.

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Question 37

Statics

A particle is acted upon by forces F = (10 N, 060°), P = (15 N, 120°) and Q = (12 N, 200°). Express the force that will keep the particle in equilibrium in the form xi + yj, where x and y are scalars.

  1. A.17.55i + 13.78j
  2. B.17.55i − 13.78j
  3. C.−17.55i + 13.78j
  4. D.−17.55i − 13.78j
Show the answer and explanation

Answer: C (−17.55i + 13.78j)

Explanation

With i east and j north, a force (F, θ) is F sin θ i + F cos θ j. The resultant is 17.55i − 13.78j, so the equilibrant is −17.55i + 13.78j.

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Question 38

Differentiation

Evaluate: limx→−2 (x3 + 8)/(x + 2)

  1. A.8
  2. B.2
  3. C.4
  4. D.12
Show the answer and explanation

Answer: D (12)

Explanation

x³ + 8 = (x + 2)(x² − 2x + 4), so the limit is (−2)² − 2(−2) + 4 = 12.

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Question 39

Functions

If f(x − 1) = x3 + 3x2 + 4x − 5, find f(2)

  1. A.61
  2. B.25
  3. C.20
  4. D.13
Show the answer and explanation

Answer: A (61)

Explanation

x − 1 = 2 when x = 3: f(2) = 27 + 27 + 12 − 5 = 61.

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Question 40

Coordinate Geometry

The length of the line joining points (x, 4) and (−x, 3) is 7 units. Find the value of x.

  1. A.4√3
  2. B.2√6
  3. C.3√2
  4. D.2√3
Show the answer and explanation

Answer: D (2√3)

Explanation

(2x)² + 1² = 49, so 4x² = 48, x² = 12 and x = 2√3 (taking the positive value).

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Theory question 101

Binary Operations

A binary operation * is defined on the set T = {−2, −1, 1, 2} by p * q = p2 + 2pq − q2, where p, q ∈ T.
(a) Copy and complete the table (row p, column q; blanks to be filled):
Row −2: q = −2: __, q = −1: 7, q = 1: __, q = 2: −8
Row −1: q = −2: __, q = −1: 2, q = 1: −2, q = 2: __
Row 1: q = −2: −7, q = −1: __, q = 1: __, q = 2: 1
Row 2: q = −2: __, q = −1: −1, q = 1: __, q = 2: __
(b) Using the table in 1(a), find the value of p such that (−2 * p) * 2 = −7.

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Model answer

(a) Complete table (columns q = −2, −1, 1, 2):
Row −2: 8, 7, −1, −8
Row −1: 1, 2, −2, −7
Row 1: −7, −2, 2, 1
Row 2: −8, −1, 7, 8
(e.g. −2 * −2 = 4 + 8 − 4 = 8; 1 * −1 = 1 − 2 − 1 = −2.)
(b) From the column q = 2, a * 2 = −7 only when a = −1. So −2 * p = −1, and row −2 gives −1 at q = 1. Hence p = 1.

Explanation

Substitute each pair carefully, keeping track of the minus signs.

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Theory question 102

Indices and Logarithms

Solve 2(2y+1) − 5(2y) + 2 = 0

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Model answer

Let p = 2ʸ: 2p² − 5p + 2 = 0, (2p − 1)(p − 2) = 0.
p = ½: 2ʸ = 2⁻¹, y = −1. p = 2: y = 1.
y = −1 or y = 1.

Explanation

2^(2y + 1) = 2 × (2ʸ)².

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Theory question 104

Trigonometry

Solve 3cos 2x − sin x = 0 for 0° ≤ x ≤ 360°.

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Model answer

cos 2x = 1 − 2sin²x: 3 − 6sin²x − sin x = 0, i.e. 6sin²x + sin x − 3 = 0.
sin x = (−1 ± √73)/12 = 0.6287 or −0.7953.
sin x = 0.6287: x = 38.95°, 141.05°.
sin x = −0.7953: x = 232.69°, 307.31°.
x ≈ 39.0°, 141.0°, 232.7°, 307.3°.

Explanation

Write cos 2x in terms of sin x to get a quadratic in sin x.

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Theory question 105

Probability

The probability that Abiola will be late to office on a given day is 2/5. In a given working week of six days, find, correct to four significant figures, the probability that he will: (a) be late for only 3 days; (b) not be late in the week; (c) be late throughout the six days.

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Model answer

Binomial: n = 6, p = 0.4, q = 0.6.
(a) ⁶C₃(0.4)³(0.6)³ = 20 × 0.064 × 0.216 = 0.2765.
(b) (0.6)⁶ = 0.04666.
(c) (0.4)⁶ = 0.004096.

Explanation

Each day is an independent trial with the same probability.

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Theory question 106

Statistics

The table shows the scores obtained by a group of artistes in Vocal (X) and Instrumental (Y) music competion.
Vocal (X): 63, 69, 72, 59, 82, 91, 95, 68
Instrumental (Y): 58, 61, 67, 51, 53, 79, 92, 57
Calculate the Spearman's rank correlation coefficient between the scores.

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Model answer

Ranks (highest = 1): X: 7, 5, 4, 8, 3, 2, 1, 6; Y: 5, 4, 3, 8, 7, 2, 1, 6.
d = 2, 1, 1, 0, −4, 0, 0, 0; Σd² = 22.
ρ = 1 − 6Σd²/(n(n² − 1)) = 1 − 132/504 ≈ 0.738.

Explanation

Rank each set separately, then use the differences in rank.

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Theory question 107

Statics

A body of mass 18 kg is suspended by an inextensible string from a rigid support and is pulled by a horizontal force F until the angle of inclination of the string to the vertical is 35°. If the system is in equilibrium, calculate the: (a) value of F; (b) tension in the string. [Take g = 10ms−2]

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Model answer

Resolve: vertically T cos 35° = 180; horizontally T sin 35° = F.
(a) F = 180 tan 35° ≈ 126.0 N.
(b) T = 180/cos 35° ≈ 219.7 N.

Explanation

Resolve the tension into vertical and horizontal parts.

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Theory question 108

Vectors

Given that p = (8 N, 030°) and q = (9 N, 150°), find, in component form, the unit vector along (p − q).

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Model answer

With i east and j north: p = 8 sin 30° i + 8 cos 30° j = 4i + 6.928j; q = 9 sin 150° i + 9 cos 150° j = 4.5i − 7.794j.
p − q = −0.5i + 14.722j; |p − q| = √(0.25 + 216.74) ≈ 14.731.
Unit vector = (−0.5i + 14.722j)/14.731 ≈ −0.034i + 0.999j.

Explanation

Convert bearings to components, subtract, then divide by the magnitude.

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Theory question 109

Sequences and Series

Given that nC4, nC5 and nC6 are the first 3 terms of a linear sequence (A. P.), find the: (a) values of n; (b) common differences of the sequence.

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Model answer

(a) 2 ⁿC₅ = ⁿC₄ + ⁿC₆. Divide by ⁿC₄: 2(n − 4)/5 = 1 + (n − 4)(n − 5)/30.
12(n − 4) = 30 + (n − 4)(n − 5), so n² − 21n + 98 = 0, (n − 7)(n − 14) = 0: n = 7 or 14.
(b) n = 7: terms 35, 21, 7, d = −14. n = 14: terms 1001, 2002, 3003, d = 1001.

Explanation

In an A.P. the middle term is the average of its neighbours.

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Theory question 110

Differentiation

A solid rectangular block has a base which measures 3x cm by 2x cm. The height of the block is y cm and its volume is 72 cm3. (a) Express y in terms of x. (b) Find: (i) an expression for the total surface area of the block in terms of x only; (ii) the value of x for which the total surface area has a stationary value.

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Model answer

(a) 6x²y = 72, so y = 12/x².
(b)(i) A = 2(6x²) + 2(3xy) + 2(2xy) = 12x² + 10xy = 12x² + 120/x.
(ii) dA/dx = 24x − 120/x² = 0 gives x³ = 5, x = ∛5 ≈ 1.71 cm.

Explanation

Eliminate y using the volume before differentiating.

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Theory question 111

Binomial Theorem

(a) Find the binomial expansion of (1 + 2x)7 and (1 − 2x)7.
(b) Using the result in 11(a), find, correct to three decimal places, the value of (1.2)7 − (0.8)7.

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Model answer

(a) (1 + 2x)⁷ = 1 + 14x + 84x² + 280x³ + 560x⁴ + 672x⁵ + 448x⁶ + 128x⁷.
(1 − 2x)⁷ = 1 − 14x + 84x² − 280x³ + 560x⁴ − 672x⁵ + 448x⁶ − 128x⁷.
(b) Put x = 0.1: the difference = 2(14x + 280x³ + 672x⁵ + 128x⁷) = 2(1.4 + 0.28 + 0.00672 + 0.0000128) ≈ 3.373.

Explanation

Subtracting the two expansions cancels the even powers.

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Theory question 112

Probability

A basket contains 12 fruits: orange, apple and avocado pear, all of the same size. The number of oranges, apples and avocado pear forms three consecutive integers. Two fruits are drawn one after the other without replacement. Calculate the probability that: (a) the first is an orange and the second is an avocado pear; (b) both are of the same fruit; (c) at least one is an apple.

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Model answer

n + (n + 1) + (n + 2) = 12 gives 3 oranges, 4 apples, 5 avocado pears.
(a) 3/12 × 5/11 = 15/132 = 5/44.
(b) (3×2 + 4×3 + 5×4)/(12×11) = 38/132 = 19/66.
(c) 1 − P(no apple) = 1 − (8/12)(7/11) = 1 − 56/132 = 19/33.

Explanation

Without replacement, the second draw is out of 11.

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Theory question 114

Dynamics

(a) A particle initially at rest moves in a straight line with an acceleration of (10t − 4t2) ms−2. Find the: (i) velocity of the particle after t seconds; (ii) average acceleration of the particle during the 4th second.
(b) A load of mass 120 kg is placed on a lift. Calculate the reaction between the floor of the lift and the load when the lift moves upwards: (i) at a constant velocity; (ii) with an acceleration of 3 ms−2. [Take g = 10ms−2]

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Model answer

(a)(i) v = ∫(10t − 4t²) dt = 5t² − (4/3)t³ (v = 0 at t = 0).
(ii) The 4th second is from t = 3 to t = 4: v(3) = 45 − 36 = 9; v(4) = 80 − 256/3 = −16/3.
Average acceleration = (v(4) − v(3))/1 = −16/3 − 9 = −43/3 ≈ −14.3 ms⁻².
(b)(i) R = mg = 1200 N. (ii) R − 1200 = 120 × 3, R = 1560 N.

Explanation

The 4th second runs from t = 3 to t = 4.

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Theory question 115

Vectors

The vectors 6i + 8j and 8i − 6j are parallel to OP and OQ respectively. If the magnitude of OP and OQ are 80 units and 120 units respectively, express: (a) OP and OQ in terms of i and j; (b) |PQ|, in the form c√k, where c and k are constants.

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Model answer

(a) |6i + 8j| = 10, so OP = 8(6i + 8j) = 48i + 64j. |8i − 6j| = 10, so OQ = 12(8i − 6j) = 96i − 72j.
(b) PQ = OQ − OP = 48i − 136j; |PQ| = √(2304 + 18496) = √20800 = 40√13.

Explanation

Scale each direction vector to the required length.

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