5.9

The Calculus Motion Hierarchy ($s \to v \to a$)

  • Displacement $s(t)$: Position of the particle relative to a fixed origin $O$ at time $t$ (metres, $\text{m}$). Can be positive, zero, or negative.
  • Velocity $v(t)$: Rate of change of displacement with respect to time: $$v(t) = s'(t) = \frac{ds}{dt} \quad (\text{m s}^{-1})$$ Sign indicates direction of motion: $v > 0$ moves in positive direction (right/up); $v < 0$ moves in negative direction (left/down).
  • Acceleration $a(t)$: Rate of change of velocity with respect to time: $$a(t) = v'(t) = s''(t) = \frac{d^2s}{dt^2} \quad (\text{m s}^{-2})$$
  • Speed vs. Velocity: $\text{Speed} = |v(t)|$ (scalar magnitude, strictly non-negative).

Example 1

Evaluating Kinematics Derivatives

The position of a particle moving in a straight line is given by $s(t) = 2t^3 - 9t^2 + 12t - 4$ metres for $t \ge 0$.
(a) Find expressions for velocity $v(t)$ and acceleration $a(t)$.
(b) Find the initial position, initial velocity, and initial acceleration of the particle.
(c) Find the times when the particle passes through the origin.

  1. (a) Differentiate: $$v(t) = s'(t) = 6t^2 - 18t + 12 \quad (\text{m s}^{-1})$$ $$a(t) = v'(t) = 12t - 18 \quad (\text{m s}^{-2})$$
  2. (b) Initial values ($t = 0$): $$s(0) = 2(0) - 9(0) + 12(0) - 4 = \mathbf{-4\text{ m}}$$ $$v(0) = 6(0) - 18(0) + 12 = \mathbf{12\text{ m s}^{-1}}$$ $$a(0) = 12(0) - 18 = \mathbf{-18\text{ m s}^{-2}}$$
  3. (c) At origin ($s(t) = 0$): $$2t^3 - 9t^2 + 12t - 4 = 0$$ Test integer factors: $t = 1 \implies 2(1) - 9 + 12 - 4 = 1 \ne 0$; $t = 2 \implies 2(8) - 9(4) + 12(2) - 4 = 16 - 36 + 24 - 4 = 0$.
    Factor $(t - 2)$: $(t - 2)(2t^2 - 5t + 2) = (t - 2)(2t - 1)(t - 2) = (t - 2)^2(2t - 1) = 0$.
    The particle passes through the origin at $\mathbf{t = 0.5\text{ s}}$ and reaches/touches origin at $\mathbf{t = 2\text{ s}}$.

Answers: (a) $v = 6t^2 - 18t + 12$, $a = 12t - 18$; (b) $s(0) = -4\text{ m}$, $v(0) = 12\text{ m s}^{-1}$, $a(0) = -18\text{ m s}^{-2}$; (c) $t = 0.5\text{ s}, 2\text{ s}$.

5.9

Instantaneous Rest & Direction Changes

  • Instantaneous Rest: A particle is at rest when its velocity is zero: $$v(t) = 0$$
  • Reversal of Direction: A particle changes its direction of motion if and only if $v(t)$ crosses through zero with a change of sign (e.g. from positive to negative, or negative to positive).
  • Momentarily Stationary vs. Turning Point: If $v(t) = 0$ but $v(t)$ does not change sign (e.g. $v(t) = (t - 2)^2 \ge 0$), the particle pauses momentarily but continues moving in the same direction.

Example 2

Identifying Direction Reversals

A toy car moves along a track with displacement $s(t) = t^3 - 6t^2 + 9t + 2$ metres for $t \ge 0$.
(a) Find the times when the car is instantaneously at rest.
(b) Determine the intervals of time when the car is moving in the negative direction.
(c) State the position of the car each time it changes direction.

  1. (a) Velocity: $$v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t - 1)(t - 3)$$ Set $v(t) = 0 \implies \mathbf{t = 1\text{ s} \quad \text{and} \quad t = 3\text{ s}}$.
  2. (b) Negative direction ($v(t) < 0$): $$3(t - 1)(t - 3) < 0 \implies \mathbf{1 < t < 3\text{ s}}$$
  3. (c) Positions at turning points: $$\text{At } t = 1: \quad s(1) = (1)^3 - 6(1)^2 + 9(1) + 2 = 1 - 6 + 9 + 2 = \mathbf{6\text{ m}}$$ $$\text{At } t = 3: \quad s(3) = (3)^3 - 6(3)^2 + 9(3) + 2 = 27 - 54 + 27 + 2 = \mathbf{2\text{ m}}$$

Answers: (a) $t = 1\text{ s}, 3\text{ s}$; (b) $1 < t < 3\text{ s}$; (c) Positions are $6\text{ m}$ (at $t=1$) and $2\text{ m}$ (at $t=3$).

5.9

Displacement vs. Total Distance Travelled

  • Displacement over Interval $[t_1, t_2]$: Net change in position: $$\Delta s = s(t_2) - s(t_1)$$
  • Total Distance Travelled: The cumulative path length traversed. Must account for reversals:
    1. Find all times in $[t_1, t_2]$ where $v(t) = 0$ (turnaround times $c_1, c_2, \dots$).
    2. Calculate position at start, at each turnaround time, and at the end: $s(t_1), s(c_1), s(c_2), s(t_2)$.
    3. Sum the absolute displacements of each leg: $$\text{Total Distance} = |s(c_1) - s(t_1)| + |s(c_2) - s(c_1)| + |s(t_2) - s(c_2)|$$
Displacement vs Total Distance Travelled Shows a particle starting at position s(0)=2, moving forward to turnaround at s(1)=6, then reversing backward to s(3)=2, illustrating total distance 8 meters vs zero displacement. 0 (Origin) s = 2 s = 6 Leg 1: t = 0 → 1 (Dist = 4 m) Start (t=0) Turnaround (t=1, v=0) Leg 2: t = 1 → 3 (Dist = 4 m) End (t=3) Displacement = s(3) − s(0) = 0 m | Total Distance = 4 + 4 = 8 m
Figure 5.6: A direction change means total distance ($4\text{ m} + 4\text{ m} = 8\text{ m}$) exceeds net displacement ($0\text{ m}$).

Example 3

Calculating Total Distance Travelled

A particle moves with displacement $s(t) = t^3 - 6t^2 + 9t + 2$ metres. Find the total distance travelled between $t = 0$ and $t = 4$.

  1. Identify turnaround points ($v(t) = 0$): $$v(t) = 3(t - 1)(t - 3) = 0 \implies t = 1\text{ s} \quad \text{and} \quad t = 3\text{ s}$$ Both $t = 1$ and $t = 3$ lie inside the interval $[0, 4]$.
  2. Calculate position at each milestone: $$s(0) = 0 - 0 + 0 + 2 = 2\text{ m}$$ $$s(1) = 1 - 6 + 9 + 2 = 6\text{ m}$$ $$s(3) = 27 - 54 + 27 + 2 = 2\text{ m}$$ $$s(4) = 4^3 - 6(4^2) + 9(4) + 2 = 64 - 96 + 36 + 2 = 6\text{ m}$$
  3. Sum segment distances: $$\text{Leg 1 } [0, 1]: \quad |s(1) - s(0)| = |6 - 2| = 4\text{ m}$$ $$\text{Leg 2 } [1, 3]: \quad |s(3) - s(1)| = |2 - 6| = 4\text{ m}$$ $$\text{Leg 3 } [3, 4]: \quad |s(4) - s(3)| = |6 - 2| = 4\text{ m}$$ $$\text{Total Distance} = 4 + 4 + 4 = \mathbf{12\text{ m}}$$

Answer: Total distance travelled is $12\text{ metres}$ (Net displacement is $s(4) - s(0) = 4\text{ m}$).

5.9

Speeding Up vs. Slowing Down (Signs of $v$ & $a$)

  • Core IB Physical Principle: $$\text{Speeding Up: } v(t) \text{ and } a(t) \text{ have the } \mathbf{SAME\ SIGN} \quad (v > 0, a > 0 \text{ OR } v < 0, a < 0)$$ $$\text{Slowing Down: } v(t) \text{ and } a(t) \text{ have } \mathbf{OPPOSITE\ SIGNS} \quad (v > 0, a < 0 \text{ OR } v < 0, a > 0)$$
  • Intuition: When acceleration acts in the same direction as velocity, the magnitude of velocity ($|v|$) increases. When acceleration opposes velocity, the particle is braking/decelerating.
  • Common Misconception: Negative acceleration ($a < 0$) does NOT automatically mean slowing down! If $v < 0$ and $a < 0$, the particle is moving to the left and accelerating to the left, so it is speeding up.

Example 4

Determining Intervals of Speeding Up and Slowing Down

A particle moves with displacement $s(t) = \frac{1}{3}t^3 - 3t^2 + 8t$ for $t \ge 0$. Find the intervals of time when the particle is speeding up and when it is slowing down.

  1. Find $v(t)$ and $a(t)$: $$v(t) = s'(t) = t^2 - 6t + 8 = (t - 2)(t - 4)$$ $$a(t) = v'(t) = 2t - 6 = 2(t - 3)$$
  2. Find transition roots: $$v(t) = 0 \implies t = 2, 4$$ $$a(t) = 0 \implies t = 3$$ Test intervals: $[0, 2)$, $(2, 3)$, $(3, 4)$, $(4, \infty)$.
  3. Create Sign Chart:
    • On $0 ≤ t < 2$: $v > 0$ and $a < 0 \implies$ Opposite signs $\implies$ Slowing down
    • On $2 < t < 3$: $v < 0$ and $a < 0 \implies$ Same signs $\implies$ Speeding up
    • On $3 < t < 4$: $v < 0$ and $a > 0 \implies$ Opposite signs $\implies$ Slowing down
    • On $t > 4$: $v > 0$ and $a > 0 \implies$ Same signs $\implies$ Speeding up

Answers: Speeding up on $(2, 3) \cup (4, \infty)$; Slowing down on $[0, 2) \cup (3, 4)$.

5.9

Maximum & Minimum Velocity

  • Optimizing Velocity: Since acceleration is the derivative of velocity ($a = v'$), the stationary values of velocity occur when: $$a(t) = 0$$
  • Classification of Maximum Velocity: $$\text{If } a(t) = 0 \text{ and } a'(t) < 0 \implies \mathbf{Maximum\ Velocity}$$ $$\text{If } a(t) = 0 \text{ and } a'(t) > 0 \implies \mathbf{Minimum\ Velocity}$$

Example 5

Finding Maximum Velocity

The velocity of an electric drone is modeled by $v(t) = 18t - 3t^2$ for $0 \le t \le 6$ ($\text{m s}^{-1}$).
(a) Find the time at which the drone attains its maximum velocity.
(b) Calculate the maximum velocity.
(c) Find the total displacement over the 6-second flight.

  1. (a) Optimize velocity ($a(t) = 0$): $$a(t) = v'(t) = 18 - 6t = 0 \implies 6t = 18 \implies \mathbf{t = 3\text{ s}}$$ Check second derivative of velocity: $v''(t) = -6 < 0 \implies$ confirms a maximum.
  2. (b) Maximum velocity: $$v(3) = 18(3) - 3(3^2) = 54 - 27 = \mathbf{27\text{ m s}^{-1}}$$
  3. (c) Displacement ($s(t) = \int v(t) dt$ or anti-differentiation): $$s(t) = 9t^2 - t^3$$ $$s(6) = 9(6^2) - 6^3 = 9(36) - 216 = 324 - 216 = \mathbf{108\text{ m}}$$

Answers: (a) $t = 3\text{ s}$; (b) $v_{\max} = 27\text{ m s}^{-1}$; (c) Displacement $= 108\text{ m}$.

Tier 1: Core Motion Differentiation (Questions 1–4)

  1. 1. Instantaneous Velocity and Acceleration: A particle has displacement $s(t) = 4t^3 - 5t^2 + 6t - 2$ metres. Find its velocity and acceleration at $t = 2\text{ seconds}$.

    Show solution

    $v(t) = s'(t) = 12t^2 - 10t + 6$
    At $t = 2$: $v(2) = 12(4) - 10(2) + 6 = 48 - 20 + 6 = \mathbf{34\text{ m s}^{-1}}$.

    $a(t) = v'(t) = 24t - 10$
    At $t = 2$: $a(2) = 24(2) - 10 = 48 - 10 = \mathbf{38\text{ m s}^{-2}}$.

  2. 2. Instantaneous Rest: A model rocket ascends with displacement $s(t) = 60t - 5t^2$ metres until it reaches its highest point. Find the time taken to reach maximum height, and calculate this height.

    Show solution

    At maximum height, $v(t) = 0$:
    $$v(t) = 60 - 10t = 0 \implies 10t = 60 \implies \mathbf{t = 6\text{ s}}$$ Maximum height: $$s(6) = 60(6) - 5(6^2) = 360 - 180 = \mathbf{180\text{ m}}$$

  3. 3. Initial Acceleration: The velocity of a particle is given by $v(t) = 10(1 - e^{-2t})\text{ m s}^{-1}$. Find its acceleration at $t = 0$ and describe what happens to the velocity as $t \to \infty$.

    Show solution

    $v(t) = 10 - 10e^{-2t}$
    $a(t) = v'(t) = -10(-2e^{-2t}) = 20e^{-2t}$
    At $t = 0$: $a(0) = 20e^0 = \mathbf{20\text{ m s}^{-2}}$.

    As $t \to \infty$, $e^{-2t} \to 0$, so $v(t) \to \mathbf{10\text{ m s}^{-1}}$ (terminal velocity).

  4. 4. Negative Velocity: A particle moves with position $s(t) = t^2 - 8t + 12$. Find the interval of time when the particle is moving to the left.

    Show solution

    Moving to the left corresponds to negative velocity ($v(t) < 0$):
    $$v(t) = 2t - 8 < 0 \implies 2t < 8 \implies \mathbf{0 \le t < 4\text{ s}}$$

Tier 2: Distance & Speed Modeling (Questions 5–7)

  1. 5. Total Distance Travelled: A particle moves along a straight line such that its displacement is $s(t) = 2t^3 - 15t^2 + 24t$ metres for $0 \le t \le 5$. Calculate the total distance travelled by the particle.

    Show solution

    Step 1: Find turning points ($v(t) = 0$): $$v(t) = 6t^2 - 30t + 24 = 6(t^2 - 5t + 4) = 6(t - 1)(t - 4) = 0 \implies t = 1\text{ s}, 4\text{ s}$$ Both lie in $[0, 5]$.
    Step 2: Evaluate positions: $$s(0) = 0\text{ m}$$ $$s(1) = 2(1) - 15(1) + 24(1) = 11\text{ m}$$ $$s(4) = 2(64) - 15(16) + 24(4) = 128 - 240 + 96 = -16\text{ m}$$ $$s(5) = 2(125) - 15(25) + 24(5) = 250 - 375 + 120 = -5\text{ m}$$ Step 3: Sum leg distances: $$\text{Leg 1 } [0, 1]: \quad |11 - 0| = 11\text{ m}$$ $$\text{Leg 2 } [1, 4]: \quad |-16 - 11| = 27\text{ m}$$ $$\text{Leg 3 } [4, 5]: \quad |-5 - (-16)| = 11\text{ m}$$ $$\text{Total Distance} = 11 + 27 + 11 = \mathbf{49\text{ metres}}$$

  2. 6. Speeding Up vs. Slowing Down: For the position function $s(t) = t^3 - 3t^2 - 9t + 5$ ($t \ge 0$), determine all intervals of time where the particle is speeding up.

    Show solution

    Step 1: Find $v(t)$ and $a(t)$: $$v(t) = 3t^2 - 6t - 9 = 3(t^2 - 2t - 3) = 3(t - 3)(t + 1)$$ Since $t \ge 0$, $v(t) = 0$ at $t = 3$.
    $$a(t) = 6t - 6 = 6(t - 1) = 0 \implies t = 1$$ Step 2: Check intervals:

    • On $[0, 1)$: $v < 0$, $a < 0 \implies$ Same sign $\implies$ Speeding up
    • On $(1, 3)$: $v < 0$, $a > 0 \implies$ Opposite sign $\implies$ Slowing down
    • On $(3, \infty)$: $v > 0$, $a > 0 \implies$ Same sign $\implies$ Speeding up
    The particle is speeding up on $\mathbf{[0, 1) \cup (3, \infty)}$.

  3. 7. Acceleration when at Rest: A particle moves with displacement $s(t) = 3t^4 - 16t^3 + 18t^2$ for $t \ge 0$. Find the acceleration of the particle each time it is momentarily at rest (excluding $t = 0$).

    Show solution

    $$v(t) = 12t^3 - 48t^2 + 36t = 12t(t^2 - 4t + 3) = 12t(t - 1)(t - 3)$$ At rest for $t > 0$ at $t = 1\text{ s}$ and $t = 3\text{ s}$.
    $$a(t) = 36t^2 - 96t + 36$$ $$\text{At } t = 1: \quad a(1) = 36(1) - 96(1) + 36 = \mathbf{-24\text{ m s}^{-2}}$$ $$\text{At } t = 3: \quad a(3) = 36(9) - 96(3) + 36 = 324 - 288 + 36 = \mathbf{72\text{ m s}^{-2}}$$

Tier 3: Advanced Examination Contexts (Questions 8–10)

  1. 8. Trigonometric Motion: A buoy oscillates vertically in the ocean with displacement $s(t) = 4\cos(2t)$ metres at time $t$ seconds.
    (a) Find expressions for velocity and acceleration.
    (b) Show that $a(t) = -4s(t)$.
    (c) State the maximum speed of the buoy and the positions where this maximum speed occurs.

    Show solution

    (a) By Chain Rule: $$v(t) = s'(t) = 4(-2\sin(2t)) = \mathbf{-8\sin(2t)\text{ m s}^{-1}}$$ $$a(t) = v'(t) = -8(2\cos(2t)) = \mathbf{-16\cos(2t)\text{ m s}^{-2}}$$ (b) $a(t) = -16\cos(2t) = -4(4\cos(2t)) = \mathbf{-4s(t)}$. (Simple Harmonic Motion).

    (c) Speed $= |-8\sin(2t)|$. Maximum speed is $\mathbf{8\text{ m s}^{-1}}$, which occurs when $|\sin(2t)| = 1$.
    When $|\sin(2t)| = 1$, $\cos(2t) = 0$, so $s(t) = 4(0) = \mathbf{0\text{ m}}$ (at the equilibrium mean position).

  2. 9. Maximum Velocity Determination: A maglev train's position during acceleration is modeled by $s(t) = 5t^2 - \frac{1}{6}t^3$ for $0 \le t \le 30$.
    (a) Find the velocity of the train as a function of $t$.
    (b) Find the time at which the train achieves its maximum velocity.
    (c) Calculate this maximum velocity in $\text{m s}^{-1}$ and convert to $\text{km h}^{-1}$.

    Show solution

    (a) $v(t) = s'(t) = 10t - \frac{1}{2}t^2\text{ m s}^{-1}$.
    (b) Maximum velocity occurs when $a(t) = v'(t) = 0$: $$a(t) = 10 - t = 0 \implies \mathbf{t = 10\text{ s}}$$ $a'(t) = -1 < 0 \implies$ maximum confirmed.

    (c) Maximum velocity: $$v(10) = 10(10) - \frac{1}{2}(10^2) = 100 - 50 = \mathbf{50\text{ m s}^{-1}}$$ In $\text{km h}^{-1}$: $50 \times 3.6 = \mathbf{180\text{ km h}^{-1}}$.

  3. 10. Multi-Condition Parameters: A particle moves with displacement $s(t) = a t^3 + b t^2 + c t$. It is initially at rest at the origin ($t = 0$). When $t = 2$, its displacement is $20\text{ m}$ and its acceleration is $6\text{ m s}^{-2}$. Find the values of $a, b,$ and $c$.

    Show solution

    Condition 1 ($t = 0, s(0) = 0, v(0) = 0$): $$s(0) = 0 \implies \text{satisfied}$$ $$v(t) = 3at^2 + 2bt + c \implies v(0) = c = 0 \implies \mathbf{c = 0}$$ So $s(t) = at^3 + bt^2$ and $v(t) = 3at^2 + 2bt$, $a(t) = 6at + 2b$.

    Condition 2 ($s(2) = 20$): $$a(2^3) + b(2^2) = 20 \implies 8a + 4b = 20 \implies 2a + b = 5 \quad \text{--- (1)}$$ Condition 3 ($a(2) = 6$): $$6a(2) + 2b = 6 \implies 12a + 2b = 6 \implies 6a + b = 3 \quad \text{--- (2)}$$ Subtract (1) from (2): $$(6a + b) - (2a + b) = 3 - 5 \implies 4a = -2 \implies \mathbf{a = -0.5}$$ Substitute into (1): $$2(-0.5) + b = 5 \implies -1 + b = 5 \implies \mathbf{b = 6}$$ $$\mathbf{a = -0.5, \quad b = 6, \quad c = 0}$$

Velocity Definition

Memorize
$$v(t) = \frac{ds}{dt} = s'(t)$$

Instantaneous velocity as the first derivative of displacement with respect to time.

Acceleration Definition

Memorize
$$a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} = s''(t)$$

Instantaneous acceleration as the derivative of velocity / second derivative of position.

Speed vs Velocity

Memorize
$$\text{Speed} = |v(t)|$$

Speed is the non-negative scalar magnitude of velocity without direction sign.

Instantaneous Rest

Memorize
$$v(t) = 0$$

Occurs at turning points where the particle reverses its physical direction.

Speeding Up / Slowing Down

Memorize
$$v \cdot a > 0 \implies \text{Speeding Up}, \quad v \cdot a < 0 \implies \text{Slowing Down}$$

Same signs mean speeding up; opposite signs mean braking / decelerating.

Total Distance Travelled

Memorize
$$\text{Dist} = \sum |\Delta s_{\text{segments}}| = \int_{t_1}^{t_2} |v(t)| dt$$

Sum of distances between turnaround times where $v(t) = 0$.

Critical IB Examination Pitfalls

  • Confusing displacement with total distance: Displacement is simply $s(t_2) - s(t_1)$. Total distance requires finding all roots of $v(t) = 0$ in the interval and summing absolute leg lengths.
  • Assuming negative acceleration always means slowing down: If $v < 0$ and $a < 0$, the particle is accelerating in the negative direction, so its speed $|v|$ is increasing (speeding up!).
  • Misinterpreting physical phrases: "Initially" $\implies t = 0$; "At the origin" $\implies s(t) = 0$; "At rest" $\implies v(t) = 0$; "Maximum speed" $\implies a(t) = 0$.
  • Forgetting units in final answers: Displacement in $\text{m}$, velocity in $\text{m s}^{-1}$, acceleration in $\text{m s}^{-2}$.

Section A: Diagnostic Concept Check

Instant feedback questions on kinematic derivatives, turning points, and speed conditions.

Score: 0 / 4

1. If a particle has velocity $v(t) = -6\text{ m s}^{-1}$ and acceleration $a(t) = -3\text{ m s}^{-2}$, what is happening to the particle?

Correct! Since $v(t) < 0$, the particle moves in the negative direction. Because $v$ and $a$ share the same negative sign ($v \cdot a > 0$), the speed $|v|$ is increasing (speeding up).

2. The position of a particle is $s(t) = t^3 - 3t$. At what positive time is the particle momentarily at rest?

Correct! $v(t) = s'(t) = 3t^2 - 3 = 0 \implies 3(t^2 - 1) = 0 \implies t = 1\text{ s}$.

3. A particle starts at $s(0) = 4\text{ m}$, travels to $s(2) = 10\text{ m}$ (where it turns around), and finishes at $s(5) = 1\text{ m}$. What is its total distance travelled?

Correct! Leg 1: $|10 - 4| = 6\text{ m}$. Leg 2: $|1 - 10| = 9\text{ m}$. Total distance $= 6 + 9 = 15\text{ m}$ (Displacement is $1 - 4 = -3\text{ m}$).

4. When a particle achieves its maximum velocity in a given interval, what condition must hold?

Correct! Since $a(t) = v'(t)$, the critical points of velocity occur where its derivative is zero: $a(t) = 0$.

Section B: Examination Practice

Multi-step questions with markscheme breakdowns.

Paper 1 (Non-Calculator) [6 marks]

A particle moves along a horizontal line with displacement $s(t) = 2t^3 - 9t^2 + 12t + 1$ metres for $t \ge 0$.

(a) Find the times when the particle is instantaneously at rest. [2 marks]
(b) Find the total distance travelled by the particle between $t = 0$ and $t = 3$. [4 marks]
View Markscheme
(a)
M1 for finding velocity: $v(t) = 6t^2 - 18t + 12 = 6(t - 1)(t - 2)$
Set $v(t) = 0 \implies \mathbf{t = 1\text{ s} \quad \text{and} \quad t = 2\text{ s}}$ A1

(b)
M1 for evaluating positions at milestone times:
$s(0) = 1\text{ m}$
$s(1) = 2(1) - 9(1) + 12(1) + 1 = 6\text{ m}$ A1
$s(2) = 2(8) - 9(4) + 12(2) + 1 = 16 - 36 + 24 + 1 = 5\text{ m}$ A1
$s(3) = 2(27) - 9(9) + 12(3) + 1 = 54 - 81 + 36 + 1 = 10\text{ m}$
M1 for summing segment distances:
$\text{Dist} = |6 - 1| + |5 - 6| + |10 - 5| = 5 + 1 + 5 = \mathbf{11\text{ metres}}$ A1
Paper 2 (Calculator Allowed) [6 marks]

A drone takes off vertically with displacement $s(t) = 15t^2 - t^3$ metres for $0 \le t \le 15$.

(a) Find the maximum velocity of the drone, and the time at which this occurs. [3 marks]
(b) Determine the time interval during which the drone is slowing down. [3 marks]
View Markscheme
(a)
$v(t) = 30t - 3t^2$
M1 for setting $a(t) = 0$: $a(t) = 30 - 6t = 0 \implies \mathbf{t = 5\text{ s}}$ A1
$v(5) = 30(5) - 3(25) = 150 - 75 = \mathbf{75\text{ m s}^{-1}}$ A1

(b)
M1 for finding roots: $v(t) = 3t(10 - t) = 0 \implies t = 10\text{ s}$
$a(t) = 30 - 6t = 0 \implies t = 5\text{ s}$
M1 for testing signs of $v$ and $a$:
On $0 < t < 5$: $v > 0, a > 0$ (speeding up)
On $5 < t < 10$: $v > 0, a < 0$ (slowing down)
On $10 < t < 15$: $v < 0, a < 0$ (speeding up)
The drone is slowing down on $\mathbf{5 < t < 10\text{ seconds}}$ A1

Calculus Unit

Next Lessons in Calculus

5.1, 5.3 Limits and First Derivatives Rates of change, first principles limits, and power rule for polynomials. 5.2, 5.7, 5.8 Increasing, Decreasing & Stationary Points Monotonicity tests, second derivatives, local extrema, and inflection points. 5.4, 5.6 Tangents, Normals & Rules Product, quotient, and chain rules with tangent and normal line equations. 5.9 Kinematics with Derivatives Motion in a line: displacement, velocity, acceleration, and total distance. 5.5, 5.10, 5.11 Integration & Area Anti-differentiation, definite integrals, and area enclosed between curves.