Motion in a Straight Line: Displacement, Velocity and Acceleration — Answers
Answer sheet with worked solutions for the Year 13 calculus worksheet.
Success TutoringWorksheetsAnswer sheetYear 13 · Mathematics · CalculusMotion in a Straight Line: Displacement, Velocity and Acceleration — Answers
Part A: Velocity and acceleration from displacement
For a particle moving along a line with displacement x at time t: velocity v = [frac dx/dt] and acceleration a = [frac dv/dt] = [frac d[sup 2]x/dt[sup 2]]. Speed is |v|.
The particle is at rest when v = 0, moving in the positive direction when v > 0, and speeding up when v and a have the same sign (slowing down when they have opposite signs).
Worked example. x = t3 − 6t2 + 9t. Then v = 3t2 − 12t + 9 = 3(t − 1)(t − 3), so it is at rest at t = 1 and t = 3, and a = 6t − 12.
Answer: 13
- v = = 8t − 3
- At t = 2: v = 16 − 3 = 13 m/s
Answer: 8
- a = = 8 m/s[sup 2] (constant)
Answer: 2, 4
- v = 3t2 − 18t + 24 = 3(t2 − 6t + 8) = 3(t − 2)(t − 4)
- At rest when v = 0: t = 2 s and t = 4 s
Answer: 6
- a = = 6t − 18
- At t = 4: a = 24 − 18 = 6 m/s[sup 2]
Answer: False
When v and a have the same sign the speed increases, so the particle is speeding up (in the negative direction). It slows down when v and a have opposite signs.
- AMoving in the negative direction and speeding up
- BMoving in the negative direction and slowing down
- CMoving in the positive direction and slowing down
- DAt rest
Answer: Moving in the negative direction and speeding up
v < 0 means it moves in the negative direction; a has the same sign as v, so the speed is increasing.
Part B: Displacement and distance covered
Displacement is the change in position, x(end) − x(start); it can be negative. Distance covered counts the full length of the path, so you must find where the particle changes direction (where v = 0) and add the distances for each stage.
Worked example. x = t2 − 4t for 0 ≤ t ≤ 5. v = 2t − 4 = 0 at t = 2. x(0) = 0, x(2) = −4, x(5) = 5. Displacement = 5 m; distance = 4 + 9 = 13 m.
Answer: 20, 28
Positions: x(0) = 0, x(2) = 20, x(4) = 16, x(5) = 20. The particle turns at t = 2 and t = 4. Displacement = x(5) − x(0) = 20 m. Distance = 20 + |16 − 20| + |20 − 16| = 20 + 4 + 4 = 28 m.
Answer: 4
- Average velocity = = = 4 m/s
Part C: Working backwards with integration
Reverse the process: v = ∫ a dt and x = ∫ v dt. Each integration adds a constant, found from an initial condition such as the starting velocity or position.
∫ from t₁ to t₂ of v dt is the change in displacement. For the distance covered, split the integral where v changes sign and add the absolute values.
Worked example. a = 6t with v(0) = 2 and x(0) = 1. Then v = 3t2 + 2 and x = t3 + 2t + 1.
Answer: 11
- x = ∫ v dt = t3 − t2 + t + C
- x(0) = 5 gives C = 5, so x = t3 − t2 + t + 5
- x(2) = 8 − 4 + 2 + 5 = 11 m
Answer: 4,
v = 4t − t2 + C, and v(0) = 0 gives C = 0. v = t(4 − t) = 0 again at t = 4. x = 2t2 − + D with x(0) = 0, so D = 0, and x(4) = 32 − = ≈ 10.67 m.
Answer: 13
- v = 0 at t = 3, so the particle turns around then.
- ∫ from 0 to 3 of (6 − 2t) dt = [6t − t2] = 18 − 9 = 9
- ∫ from 3 to 5 of (6 − 2t) dt = (30 − 25) − (18 − 9) = −4
- Distance = 9 + 4 = 13 m (the displacement is only 9 − 4 = 5 m)
- AThe change in displacement from t = 0 to t = 4
- BThe total distance covered from t = 0 to t = 4
- CThe acceleration at t = 4
- DThe greatest speed between t = 0 and t = 4
Answer: The change in displacement from t = 0 to t = 4
Integrating velocity gives the net change in position. Sections where v < 0 subtract, so it equals the distance covered only if the particle never changes direction.
Part D: Exponential and trigonometric motion
The same rules apply to any displacement function. Exponential terms such as e−kt die away, so the particle approaches a limiting position. Sine and cosine give motion that oscillates back and forth.
Worked example. x = 2 cos t. Then v = −2 sin t and a = −2 cos t = −x. The particle is at rest when sin t = 0 (t = 0, π, 2π, …), at the ends of its path x = ±2.
Answer: 0.920, 5
v = = 2.5e−0.5t; at t = 2, v = 2.5e−1 = 0.9197… ≈ 0.920 m/s. As t grows, e−0.5t → 0, so x → 5.
Answer: 3, −6√3
v = 6 cos 2t, so v() = 6 cos() = 3. a = −12 sin 2t, so a() = −12 × = −6√3 ≈ −10.39.
Answer: True
a = −12 sin 2t = −4(3 sin 2t) = −4x. Motion where a is a negative multiple of x is called simple harmonic motion.
Answer:
- v = 6 cos 2t = 0 when 2t =
- t = [frac π/4] (at the end of its swing, x = 3)
Part E: Extended problem
A particle moves along a straight line with velocity v = t2 − 6t + 8 m/s for 0 ≤ t ≤ 5. It starts at the origin.
(a) Find the times when the particle is at rest.
(b) Find the acceleration at t = 5.
(c) Find an expression for its displacement x, and its position at t = 2, t = 4 and t = 5.
(d) Find the total distance covered in the 5 seconds.
(e) Find the greatest speed of the particle during the 5 seconds, and when it occurs.
Model answer: (a) v = (t − 2)(t − 4) = 0 at t = 2 s and t = 4 s. (b) a = 2t − 6, so a(5) = 4 m/s[sup 2]. (c) x = − 3t2 + 8t + C with x(0) = 0, so C = 0. x(2) = − 12 + 16 = [frac 20/3] ≈ 6.67 m, x(4) = − 48 + 32 = [frac 16/3] ≈ 5.33 m, x(5) = − 75 + 40 = [frac 20/3] ≈ 6.67 m. (d) It moves forward 0 → , back to , then forward to : distance = + + = [frac 28/3] ≈ 9.33 m. (e) v is a parabola with its minimum at t = 3, where v = 9 − 18 + 8 = −1 m/s. On 0 ≤ t ≤ 5, v(0) = 8 and v(5) = 3, so the greatest speed is 8 m/s at t = 0; the slowest backward motion (speed 1 m/s) is at t = 3.
Success criteria:
- (a) t = 2 and t = 4 (2 marks)
- (b) a = 2t − 6 and a(5) = 4 m/s2 (1 mark)
- (c) integrates with C = 0 and finds all three positions (3 marks)
- (d) splits the motion at the turning times to get m (3 marks)
- (e) compares v at the endpoints and the vertex to get 8 m/s at t = 0 (2 marks)
Marking tips
- Give credit for correct method even when the final answer slips — the working shows where the thinking went right.
- For open-ended tasks, use the success criteria as a checklist rather than looking for one "right" answer.
- Celebrate what went well first, then pick one thing to work on next.