Read the graph.
Find your first step.
20 questions to build confidence in trigonometric functions and kinematics. Try on paper, reveal a hint when you need it, then compare your reasoning with the worked solution.
Trigonometry · work backwards
Midline = (maximum + minimum) / 2
Amplitude = (maximum − minimum) / 2
For sine/cosine: b = 2π / period. For tangent: b = π / period.
When solving, use the graph to count intersections and check every answer against the stated interval.
1. Read the vertical features
Find (a) the maximum and minimum, (b) the amplitude and midline, and (c) the range.
Show a helpful hint
The amplitude is half the total height. The midline is halfway between the maximum and minimum.
Reveal the worked solution
Maximum 7; minimum −1. Amplitude = (7 − (−1))/2 = 4. Midline y = (7 + (−1))/2 = 3. Range: −1 ≤ y ≤ 7.
2. Build an equation and solve it
(a) Find y = a cos(bx) + d, with a,b > 0. (b) Find every x for which y = 5 on 0 ≤ x ≤ 2π, to three significant figures.
Show a helpful hint
Starting at a maximum suggests cosine. Isolate cosine, then use the graph to predict the number of intersections.
Reveal the worked solution
One cycle takes 2π, so b = 1. Hence y = 4 cos x + 3. Setting y = 5 gives cos x = 1/2. The solutions are π/3 and 5π/3, approximately 1.05 and 5.24 radians.
3. Recover the frequency
Find (a) the period, (b) a,b,d in y = a sin(bx) + d with a,b > 0, and (c) the number of complete cycles on 0 ≤ x ≤ 2π.
Show a helpful hint
Measure the distance between consecutive peaks. That distance is the period, not b.
Reveal the worked solution
Consecutive peaks are at π/4 and 5π/4, so the period is π. Amplitude 3; midline 1. Since b = 2π/π = 2, y = 3 sin(2x) + 1. There are two complete cycles.
4. Find every solution
Find all x for which y = 5/2 on 0 ≤ x ≤ 2π. Give exact answers.
Show a helpful hint
Reduce the equation to sin(2x) = 1/2. The angle 2x runs from 0 to 4π. Solve for that angle before dividing by two.
Reveal the worked solution
sin(2x) = 1/2. Thus 2x = π/6, 5π/6, 13π/6, 17π/6. Dividing every value by two gives x = π/12, 5π/12, 13π/12, 17π/12. Four intersections agree with the graph.
5. Recover a horizontal shift
Find y = a sin(x − c) + d, where a > 0 and 0 ≤ c < 2π.
Show a helpful hint
A positive sine cycle begins at an upward crossing of the midline. Find that crossing.
Reveal the worked solution
Maximum 3, minimum −1: amplitude 2 and midline 1. The upward midline crossing is at x = π/2. Therefore y = 2 sin(x − π/2) + 1.
6. Connect representations and an inequality
(a) Write y = p cos x + q. (b) Find every x for which y ≥ 2 on 0 ≤ x ≤ 2π.
Show a helpful hint
Starting at a minimum suggests negative cosine. Find the boundary intersections before deciding which interval lies above y = 2.
Reveal the worked solution
y = −2 cos x + 1. Then −2 cos x + 1 ≥ 2 gives cos x ≤ −1/2: remember to reverse the inequality when dividing by −2. Hence 2π/3 ≤ x ≤ 4π/3.
7. Recognise tangent
For y = a tan(bx), a,b > 0, find (a) a and b, (b) the period, and (c) the two displayed asymptotes.
Show a helpful hint
Tangent has period π/b. The asymptotes determine b; the labelled point then determines a.
Reveal the worked solution
The displayed branch is bounded by x = −π/2 and x = π/2, so b = 1 and the period is π. At x = π/4, tan x = 1 and y = 2, so a = 2. Equation: y = 2 tan x.
8. Extend beyond the displayed branch
Find every solution of y = 2 on −π ≤ x ≤ π.
Show a helpful hint
Start with the labelled point. Tangent repeats after one period; check backwards as well as forwards.
Reveal the worked solution
tan x = 1. Solutions repeat every π: x = π/4 + kπ. In the required interval, x = −3π/4 and π/4.
9. Build a water-depth model
Find (a) the mean depth and amplitude, (b) the period, and (c) h(t) = d + a cos(b(t − c)), where a,b > 0 and c is the first positive peak time.
Show a helpful hint
Use maximum and minimum first. Consecutive high tides give the period; the first high tide positions the cosine peak.
Reveal the worked solution
Mean depth 3 m; amplitude 2 m; period 14 − 2 = 12 hours. Thus b = 2π/12 = π/6 and c = 2. Model: h(t) = 3 + 2 cos((π/6)(t − 2)).
10. Use the model to make a decision
A boat can enter when depth is at least 4 m. Find every permitted interval during 0 ≤ t < 24. Express the boundaries as clock times.
Show a helpful hint
Solve h(t) = 4 first. Then choose the portions above that depth. Check whether the interval already begins at midnight.
Reveal the worked solution
h ≥ 4 means cos((π/6)(t − 2)) ≥ 1/2. Each peak has a permitted window extending two hours either side. Within 0 ≤ t < 24: 0 ≤ t ≤ 4 and 12 ≤ t ≤ 16. The boat can enter from 00:00–04:00 and 12:00–16:00, including the endpoints.
Kinematics · translate the words
Throughout this section, t is in seconds, s is in metres, v is in m/s and a is in m/s². Unless an exact answer is requested, round non-exact numerical answers to three significant figures.
v(t) = s′(t) · a(t) = v′(t) · speed = |v(t)|
At rest: v = 0. Displacement: final position − initial position. Distance: split at reversals and add absolute changes. On a velocity–time graph, gradient is acceleration and signed area is displacement.
1. Position, velocity and acceleration
s(t) = t³ − 6t² + 9t + 2, 0 ≤ t ≤ 5. Find (a) the initial position, (b) v(t) and a(t), and (c) velocity and speed at t = 2.
Show a helpful hint
Substitute into s for position. Differentiate for velocity. Speed is the magnitude of velocity.
Reveal the worked solution
s(0) = 2 m. v(t) = 3t² − 12t + 9; a(t) = 6t − 12. At t = 2, v = −3 m/s and speed = 3 m/s.
2. Rest and direction
v(t) = 3t² − 12t + 9, 0 ≤ t ≤ 5. Find (a) rest times, (b) intervals of positive and negative motion, and (c) whether each rest time is a reversal.
Show a helpful hint
Factorise velocity and make a sign table. A root only indicates a reversal if velocity changes sign.
Reveal the worked solution
v = 3(t − 1)(t − 3). Rest at t = 1 and 3 s. Positive motion on 0 ≤ t < 1 and 3 < t ≤ 5; negative on 1 < t < 3. The sign changes at both roots, so both are reversals.
3. Displacement versus distance
s(t) = t² − 6t + 5, 0 ≤ t ≤ 8. Find (a) displacement and (b) total distance travelled.
Show a helpful hint
Solve s′(t) = 0 before finding distance. Split at any reversal and add absolute changes in position.
Reveal the worked solution
v = 2t − 6, so reversal at t = 3. Positions: s(0) = 5, s(3) = −4, s(8) = 21. Displacement = 21 − 5 = 16 m. Distance = |−4 − 5| + |21 − (−4)| = 9 + 25 = 34 m.
4. Recover position from velocity
v(t) = 6t − 4 and s(0) = 7. Find (a) s(t), (b) s(3), and (c) displacement from t = 0 to t = 3.
Show a helpful hint
Integrate velocity and use the initial position to determine the constant.
Reveal the worked solution
s = 3t² − 4t + C. Since s(0) = 7, C = 7. Thus s(3) = 22 m. Displacement = 22 − 7 = 15 m.
5. Recover motion from acceleration
a(t) = 4t − 6, v(0) = 2 and s(0) = 5. Find (a) v(t), (b) s(t), and (c) all rest times on 0 ≤ t ≤ 3.
Show a helpful hint
Integrate twice. Each integration introduces a different constant, determined by its own initial condition.
Reveal the worked solution
v = 2t² − 6t + 2. Integrating again: s = (2/3)t³ − 3t² + 2t + 5. Solve t² − 3t + 1 = 0: t = (3 ± √5)/2, approximately 0.382 and 2.62 s. Both lie in the interval.
6. Read a velocity–time graph
Straight segments join (0,0), (4,8), (7,8), (11,−8). Find (a) acceleration on each segment, (b) the reversal time, and (c) displacement and total distance over 11 s.
Show a helpful hint
Gradient gives acceleration; signed area gives displacement. Split the final segment at the time-axis crossing for distance.
Reveal the worked solution
Accelerations: 2 m/s² on 0 < t < 4; 0 on 4 < t < 7; −4 m/s² on 7 < t < 11. Acceleration is not defined at the sharp corners. Reversal at t = 9. Areas: +16, +24, +8, −8 m. Displacement = 40 m; total distance = 56 m.
7. Speeding up or slowing down
v(t) = t² − 6t + 8, 0 ≤ t ≤ 6. Find the open intervals on which the particle is (a) speeding up and (b) slowing down.
Show a helpful hint
Find signs of both velocity and acceleration. Matching signs mean increasing speed. Split at zeros of either function.
Reveal the worked solution
v = (t − 2)(t − 4); a = 2t − 6. Speeding up: 2 < t < 3 and 4 < t < 6. Slowing down: 0 < t < 2 and 3 < t < 4. At t = 3 acceleration is zero; at t = 2 and 4 speed is zero.
8. Find an unknown parameter
s(t) = t³ − kt² + 12t. The particle is at rest at t = 2. Find (a) k, (b) a(2), and (c) whether t = 2 is a reversal.
Show a helpful hint
Translate “at rest” into s′(2) = 0. A zero velocity does not necessarily mean a change of direction.
Reveal the worked solution
v = 3t² − 2kt + 12. At t = 2: 24 − 4k = 0, so k = 6. Then a = 6t − 12 and a(2) = 0. Velocity becomes 3(t − 2)², positive on both sides: the particle stops momentarily without reversing.
9. Two particles meeting
sA(t) = t² + 2t and sB(t) = 12 − t, t ≥ 0. Find (a) when they meet, (b) the meeting position, and (c) both velocities at that time.
Show a helpful hint
Meeting means equal positions at the same time. Solve that equation before differentiating the position functions separately.
Reveal the worked solution
t² + 3t − 12 = 0 gives t = (−3 + √57)/2 ≈ 2.27 s; the other root is negative. Meeting position = 12 − t ≈ 9.73 m. vA = 2t + 2 = √57 − 1 ≈ 6.55 m/s; vB = −1 m/s.
10. Combine calculus and trigonometry
s(t) = 4 sin(πt/6), 0 ≤ t ≤ 12. Find (a) v(t) and a(t), (b) every rest time, (c) maximum speed, and (d) total distance.
Show a helpful hint
Apply the chain rule, including π/6. For distance, evaluate position at endpoints and all reversal times.
Reveal the worked solution
v = (2π/3) cos(πt/6); a = −(π²/9) sin(πt/6). Rest at t = 3 and 9 s. Maximum speed = 2π/3 ≈ 2.09 m/s. Positions at 0,3,9,12 are 0,4,−4,0: distance = 4 + 8 + 4 = 16 m.