Kinematics

1. Overview

Kinematics in IGCSE Additional Mathematics (0606) is the study of motion in a straight line, using calculus. This means you'll use differentiation and integration to relate displacement ($s$), velocity ($v$), and acceleration ($a$) when acceleration is not constant. This topic is crucial for Paper 1 (non-calculator) and Paper 2, testing your calculus skills in a real-world context. Expect to find displacement, velocity, or acceleration functions; determine when a particle is at rest or changes direction; and calculate total distance traveled. Mastering the relationships between $s$, $v$, and $a$, and understanding how to interpret kinematics graphs, is essential for success.

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Key Definitions

• Displacement ($s$): The distance of a particle from a fixed origin $O$ in a specific direction. It is a vector quantity and can be positive, negative, or zero. Units are typically meters (m).
• Velocity ($v$): The rate of change of displacement with respect to time. $v = \frac{ds}{dt}$. It is a vector quantity. Units are typically meters per second (m/s or ms⁻¹).
• Acceleration ($a$): The rate of change of velocity with respect to time. $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$. It is a vector quantity. Units are typically meters per second squared (m/s² or ms⁻²).
• Instantaneous Rest: A point in time where the velocity of the particle is zero ($v = 0$).
• Initial Conditions: The state of the particle at time $t = 0$ (e.g., initial velocity, initial displacement). These are crucial for finding the constant of integration.
• Total Distance Traveled: The sum of the absolute magnitudes of all displacements; unlike displacement, it cannot be negative and must account for changes in direction. Requires checking when $v=0$ to identify changes in direction.

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Core Content

The Calculus Relationship

To move "down" the chain from displacement to acceleration, we differentiate. To move "up" from acceleration to displacement, we integrate. Remember to add the constant of integration, $C$, when integrating.

$$s \xrightarrow{\frac{d}{dt}} v \xrightarrow{\frac{d}{dt}} a$$ $$a \xrightarrow{\int dt} v \xrightarrow{\int dt} s$$

• Velocity: $v = \frac{ds}{dt}$
• Acceleration: $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$
• Displacement: $s = \int v , dt$
• Velocity: $v = \int a , dt$

Kinematics Graphs

• Displacement-Time ($s$-$t$): Gradient represents Velocity.
• Velocity-Time ($v$-$t$): Gradient represents Acceleration; Area under the curve represents Displacement.
• Acceleration-Time ($a$-$t$): Area under the curve represents Change in Velocity.

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Worked example 1 — Finding acceleration at rest

A particle $P$ moves in a straight line such that its displacement, $s$ meters, from a fixed point $O$ is given by $s = \frac{1}{3}t^3 - 5t^2 + 16t + 2$, where $t$ is the time in seconds after leaving $O$. Find the acceleration of the particle when it is instantaneously at rest.

Step 1: Find the velocity function, $v(t)$. $v = \frac{ds}{dt} = \frac{d}{dt} (\frac{1}{3}t^3 - 5t^2 + 16t + 2)$ $v = t^2 - 10t + 16$ (Power rule of differentiation)

Step 2: Find the time(s) $t$ when the particle is at rest ($v=0$). $t^2 - 10t + 16 = 0$ $(t - 2)(t - 8) = 0$ (Factorizing the quadratic) $t = 2$ or $t = 8$

Step 3: Find the acceleration function, $a(t)$. $a = \frac{dv}{dt} = \frac{d}{dt}(t^2 - 10t + 16)$ $a = 2t - 10$ (Power rule of differentiation)

Step 4: Calculate the acceleration $a$ at $t = 2$ and $t = 8$. At $t = 2$: $a = 2(2) - 10 = -6 \text{ m/s}^2$ At $t = 8$: $a = 2(8) - 10 = 6 \text{ m/s}^2$

Final Answer: The acceleration is $\boxed{\pm 6 \text{ m/s}^2}$.

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Worked example 2 — Integrating to find displacement

A particle moves in a straight line with velocity $v = 8e^{4t} - 12t^{1/2}$. Given that the particle's initial displacement from a fixed point $O$ is 5 meters, find the displacement of the particle when $t = \ln(2)$.

Step 1: Set up the integral for displacement. $s = \int v , dt = \int (8e^{4t} - 12t^{1/2}) , dt$

Step 2: Integrate the velocity function. $s = \frac{8e^{4t}}{4} - \frac{12t^{3/2}}{3/2} + C$ (Integrating $e^{kt}$ and power rule) $s = 2e^{4t} - 8t^{3/2} + C$

Step 3: Use initial conditions to find $C$. When $t = 0$, $s = 5$. $5 = 2e^{4(0)} - 8(0)^{3/2} + C$ $5 = 2(1) - 0 + C$ $C = 3$

Step 4: Write the displacement function. $s = 2e^{4t} - 8t^{3/2} + 3$

Step 5: Substitute $t = \ln(2)$. $s = 2e^{4\ln(2)} - 8(\ln(2))^{3/2} + 3$ $s = 2e^{\ln(2^4)} - 8(\ln(2))^{3/2} + 3$ (Using the log law $n\ln x = \ln x^n$) $s = 2(2^4) - 8(\ln(2))^{3/2} + 3$ (Since $e^{\ln x} = x$) $s = 32 - 8(\ln(2))^{3/2} + 3$ $s = 35 - 8(\ln(2))^{3/2}$

Final Answer: $\boxed{35 - 8(\ln(2))^{3/2}}$ meters.

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Worked example 3 — Total distance with direction change

A particle moves along a straight line with velocity $v(t) = 9 - 3t^2$ m/s. Find the total distance traveled by the particle in the first 4 seconds.

Step 1: Determine if the particle changes direction within the interval [0, 4]. Set $v(t) = 0$: $9 - 3t^2 = 0$ $3t^2 = 9$ $t^2 = 3$ $t = \pm\sqrt{3}$ Since $t$ must be non-negative, $t = \sqrt{3} \approx 1.732$ seconds. This value lies within the interval [0, 4], so the particle changes direction.

Step 2: Calculate the displacement for the intervals [0, √3] and [√3, 4]. $s(t) = \int v(t) , dt = \int (9 - 3t^2) , dt = 9t - t^3 + C$ Assume $s(0) = 0$, then $C = 0$, so $s(t) = 9t - t^3$.

Displacement from $t = 0$ to $t = \sqrt{3}$: $s(\sqrt{3}) = 9\sqrt{3} - (\sqrt{3})^3 = 9\sqrt{3} - 3\sqrt{3} = 6\sqrt{3}$

Displacement from $t = \sqrt{3}$ to $t = 4$: $s(4) = 9(4) - (4)^3 = 36 - 64 = -28$ Displacement during this interval = $s(4) - s(\sqrt{3}) = -28 - 6\sqrt{3}$

Step 3: Calculate the total distance traveled. Total distance = $|6\sqrt{3}| + |-28 - 6\sqrt{3}| = 6\sqrt{3} + |-(28 + 6\sqrt{3})| = 6\sqrt{3} + 28 + 6\sqrt{3} = 28 + 12\sqrt{3}$

Final Answer: $\boxed{28 + 12\sqrt{3}}$ meters.

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Worked example 4 — Finding maximum velocity

A particle moves in a straight line such that its displacement, $s$ meters, from a fixed point $O$ is given by $s = t^3 - 9t^2 + 24t$. Find the maximum velocity of the particle in the interval $0 \le t \le 5$.

Step 1: Find the velocity function, $v(t)$. $v = \frac{ds}{dt} = \frac{d}{dt} (t^3 - 9t^2 + 24t)$ $v = 3t^2 - 18t + 24$

Step 2: Find the acceleration function, $a(t)$. $a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 18t + 24)$ $a = 6t - 18$

Step 3: Find the time(s) $t$ when the acceleration is zero (i.e., when velocity is stationary). $6t - 18 = 0$ $6t = 18$ $t = 3$

Step 4: Check the velocity at the stationary point ($t=3$) and the endpoints of the interval ($t=0$ and $t=5$). At $t = 0$: $v = 3(0)^2 - 18(0) + 24 = 24$ At $t = 3$: $v = 3(3)^2 - 18(3) + 24 = 27 - 54 + 24 = -3$ At $t = 5$: $v = 3(5)^2 - 18(5) + 24 = 75 - 90 + 24 = 9$

Step 5: Determine the maximum velocity. Since velocity is a vector, we consider the magnitude. The velocities are 24, -3, and 9. The largest magnitude is 24.

Final Answer: The maximum velocity is $\boxed{24 \text{ m/s}}$.

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Extended Content (Extended Only)

Additional Mathematics is a single-tier syllabus — all content above applies to all students.

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Key Equations

Equation	Meaning	Units	Notes
$\mathbf{v = \frac{ds}{dt}}$	Velocity is the derivative of displacement	$\text{m/s}$ or $\text{ms}^{-1}$
$\mathbf{a = \frac{dv}{dt} = \frac{d^2s}{dt^2}}$	Acceleration is the derivative of velocity	$\text{m/s}^2$ or $\text{ms}^{-2}$
$\mathbf{s = \int v , dt}$	Displacement is the indefinite integral of velocity	$\text{m}$	+ C needed
$\mathbf{v = \int a , dt}$	Velocity is the indefinite integral of acceleration	$\text{m/s}$	+ C needed
$s = \int_{t_1}^{t_2} v , dt$	Displacement is the definite integral of velocity	$\text{m}$	Displacement between $t_1$ and $t_2$
Total Dist = $\int_{t_1}^{t_2}	v	, dt$	Total distance traveled (account for $v=0$)

Note: These formulas are not provided on the IGCSE Add Math formula sheet. You must memorize the calculus relationships.

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Common Mistakes to Avoid

• ❌ Wrong: Integrating acceleration or velocity and forgetting the constant $C$. ✓ Right: Always write $+ C$ immediately after integrating. Use the "initial conditions" ($t=0$, or other given values) to solve for $C$.
• ❌ Wrong: Assuming "Total Distance Traveled" is the same as "Displacement". ✓ Right: Check if $v=0$ during the time interval. If the particle reverses direction, calculate distances separately for each interval where the velocity has the same sign, and then add the absolute values of these displacements.
• ❌ Wrong: Using Suvat equations ($v = u + at$) for variable acceleration. ✓ Right: Suvat only works if $a$ is a constant number. If $a$ is a function of $t$, you must use calculus (differentiation and integration).
• ❌ Wrong: Giving decimal approximations when the question requires an exact answer. ✓ Right: Leave your answer in terms of surds, fractions, $\ln$, $e$, or $\pi$ unless the question explicitly asks for a decimal approximation. For example, leave $\sqrt{2}$ as $\sqrt{2}$, and $\ln(5)$ as $\ln(5)$.
• ❌ Wrong: Making sign errors when differentiating or integrating. ✓ Right: Double-check your signs, especially when dealing with negative coefficients or negative powers.
• ❌ Wrong: Forgetting to consider the domain when finding maximum or minimum values. ✓ Right: If the question specifies an interval for $t$, check the velocity (or displacement) at the endpoints of the interval, as well as at any stationary points within the interval.
• ❌ Wrong: Not using radians when trigonometric functions are involved. ✓ Right: If the kinematics function involves trigonometric terms (e.g., $v = \sin(2t)$), your calculator must be in Radians mode (for Paper 2).

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Exam Tips

• Keywords:
  • "Find the initial [velocity/acceleration]": Substitute $t = 0$.
  • "At the origin": $s = 0$.
  • "Instantaneously at rest": $v = 0$.
  • "Returning to $O$": $s = 0$ (solve for $t > 0$).
  • "Maximum velocity": Occurs when $\frac{dv}{dt} = 0$ (i.e., acceleration is zero). Remember to check endpoints of the interval.
• Units: Always check if the question uses meters and seconds or other units (like cm or minutes). Be consistent throughout your working.
• Exact Values: In Paper 1 (non-calculator), leave answers as fractions, surds, or in terms of $e$ and $\pi$. For example, leave $\ln(2)$ as $\ln(2)$ rather than $0.693$.
• Radians: If the kinematics function involves trigonometric terms (e.g., $v = \sin(2t)$), your calculator must be in Radians mode (for Paper 2).
• Method Marks: Always state your formula before substituting. Even if your arithmetic is wrong, stating $v = \frac{ds}{dt}$ earns a method mark.
• Show All Steps: Especially in Paper 1, show every step of your working, including differentiation, integration, and algebraic manipulation. This makes it easier for the marker to award method marks, even if you make a small error.
• Check Your Work: If time permits, check your answers by differentiating or integrating back to the original function. For example, if you found $v$ by differentiating $s$, differentiate $v$ to see if you get $a$.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [8 marks]

Question:

A particle moves in a straight line such that its displacement, $s$ meters, from a fixed point $O$ at time $t$ seconds is given by $s = t^3 - 9t^2 + 15t + 3$.

(a) Find the values of $t$ when the particle is instantaneously at rest. [4]

(b) Find the acceleration of the particle when $t=5$. [2]

(c) Find the total distance travelled by the particle in the first 2 seconds. [2]

Worked Solution:

(a)

1. Find the velocity function by differentiating the displacement function with respect to time. $v = \frac{ds}{dt} = 3t^2 - 18t + 15$ [Differentiating $s$ to find $v$]

2. Set the velocity equal to zero and solve for $t$. $3t^2 - 18t + 15 = 0$ $t^2 - 6t + 5 = 0$ $(t-1)(t-5) = 0$ $t = 1, 5$ [Solving quadratic equation]

3. State the times when the particle is at rest. $t=1$ and $t=5$ [Instantaneous rest occurs when velocity is zero]

$\boxed{t = 1, 5}$

How to earn full marks: Differentiate to find the velocity, set it to zero, and solve the quadratic equation, showing all steps.

(b)

1. Find the acceleration function by differentiating the velocity function with respect to time. $a = \frac{dv}{dt} = 6t - 18$ [Differentiating $v$ to find $a$]

2. Substitute $t=5$ into the acceleration function. $a = 6(5) - 18 = 30 - 18 = 12$ [Substituting $t=5$ into $a$]

$\boxed{a = 12 , ms^{-2}}$

How to earn full marks: Differentiate the velocity to find acceleration, then substitute $t=5$ and include the correct units.

(c)

1. Find the displacement at $t=0$, $t=1$ and $t=2$. $s(0) = 0^3 - 9(0)^2 + 15(0) + 3 = 3$ $s(1) = 1^3 - 9(1)^2 + 15(1) + 3 = 1 - 9 + 15 + 3 = 10$ $s(2) = 2^3 - 9(2)^2 + 15(2) + 3 = 8 - 36 + 30 + 3 = 5$ [Calculating displacements at key times]

2. Calculate the total distance travelled. The particle moves from $s=3$ to $s=10$ and then from $s=10$ to $s=5$.

   Distance = $|10-3| + |5-10| = 7 + 5 = 12$ [Calculating total distance as sum of distances between turning points]

$\boxed{12 , m}$

How to earn full marks: Calculate displacement at $t=0, 1, 2$, then sum the absolute values of the changes in displacement.

Common Pitfall: Remember that total distance travelled isn't always just the difference between the final and initial displacements. You need to check if the particle changes direction within the time interval and account for the distance travelled in each direction.

Exam-Style Question 2 — Paper 1 (No Calculator Allowed) [7 marks]

Question:

The velocity, $v , ms^{-1}$, of a particle moving in a straight line is given by $v = 6\cos(3t) - 3$ for $t \ge 0$, where $t$ is the time in seconds.

(a) Find the first time when the particle is instantaneously at rest. [3]

(b) Find the distance travelled by the particle in the interval $0 \le t \le \frac{\pi}{6}$. [4]

Worked Solution:

(a)

1. Set the velocity equal to zero and solve for $t$. $6\cos(3t) - 3 = 0$ $\cos(3t) = \frac{1}{2}$ [Setting $v=0$ and rearranging]

2. Find the general solution for $3t$. $3t = \frac{\pi}{3}$ [Finding the principal value]

3. Find the smallest positive value of $t$. $t = \frac{\pi}{9}$ [Dividing by 3 to find $t$]

$\boxed{t = \frac{\pi}{9}}$

How to earn full marks: Set the velocity to zero, solve the trig equation for $t$, and give the smallest positive solution.

(b)

1. Find the displacement function by integrating the velocity function with respect to time. $s = \int (6\cos(3t) - 3) , dt = 2\sin(3t) - 3t + c$ [Integrating $v$ to find $s$]

2. Since distance is required we need to check if the particle changes direction in the interval $0 \le t \le \frac{\pi}{6}$. We know from part (a) that $v=0$ when $t=\frac{\pi}{9}$. Since $\frac{\pi}{9}$ is in the interval, we need to integrate from $0$ to $\frac{\pi}{9}$ and from $\frac{\pi}{9}$ to $\frac{\pi}{6}$ separately.

3. Calculate displacement from $t=0$ to $t=\frac{\pi}{9}$ and from $t=0$ to $t=\frac{\pi}{6}$. Let c=0.

   $s(\frac{\pi}{9}) = 2\sin(3\frac{\pi}{9}) - 3\frac{\pi}{9} = 2\sin(\frac{\pi}{3}) - \frac{\pi}{3} = 2\frac{\sqrt{3}}{2} - \frac{\pi}{3} = \sqrt{3} - \frac{\pi}{3}$ $s(\frac{\pi}{6}) = 2\sin(3\frac{\pi}{6}) - 3\frac{\pi}{6} = 2\sin(\frac{\pi}{2}) - \frac{\pi}{2} = 2 - \frac{\pi}{2}$

4. Calculate the displacement from $t=\frac{\pi}{9}$ to $t=\frac{\pi}{6}$ $s = (2 - \frac{\pi}{2}) - (\sqrt{3} - \frac{\pi}{3}) = 2 - \sqrt{3} - \frac{\pi}{6}$

5. Calculate the total distance $Distance = |\sqrt{3} - \frac{\pi}{3}| + |2 - \sqrt{3} - \frac{\pi}{6}|$

   $Distance = \sqrt{3} - \frac{\pi}{3} + |2 - \sqrt{3} - \frac{\pi}{6}|$

   Since $2 - \sqrt{3} - \frac{\pi}{6} \approx 2 - 1.732 - 0.524 = -0.256$, we have

   $Distance = \sqrt{3} - \frac{\pi}{3} + \sqrt{3} + \frac{\pi}{6} - 2 = 2\sqrt{3} - \frac{\pi}{2} - 2$

$\boxed{2\sqrt{3} - \frac{\pi}{2} - 2}$

How to earn full marks: Integrate to find displacement, check for changes in direction, and sum the absolute values of the displacements between turning points.

Common Pitfall: When finding the total distance travelled, always check if the particle changes direction within the given time interval. If it does, you'll need to split the integral into separate parts to account for the distance travelled in each direction.

Exam-Style Question 3 — Paper 2 (Calculator Allowed) [7 marks]

Question:

A particle moves in a straight line from a point $A$ to a point $B$. Its velocity, $v , ms^{-1}$, at time $t$ seconds after leaving $A$ is given by $v = 0.3t^2 - 2.4t + 3.6$ for $0 \le t \le 10$.

(a) Find the minimum velocity of the particle between $A$ and $B$. [3]

(b) Find the distance $AB$. [4]

Worked Solution:

(a)

1. Find the time at which the minimum velocity occurs by differentiating the velocity function and setting it to zero. $\frac{dv}{dt} = 0.6t - 2.4 = 0$ $0.6t = 2.4$ $t = \frac{2.4}{0.6} = 4$ [Differentiating $v$ and setting to zero]

2. Substitute the value of $t$ into the velocity function. $v = 0.3(4)^2 - 2.4(4) + 3.6 = 0.3(16) - 9.6 + 3.6 = 4.8 - 9.6 + 3.6 = -1.2$ [Substituting $t=4$ into $v$]

$\boxed{-1.2 , ms^{-1}}$

How to earn full marks: Differentiate the velocity, set to zero to find the time of minimum velocity, then substitute back into the velocity equation.

(b)

1. Find the displacement function by integrating the velocity function with respect to time. $s = \int (0.3t^2 - 2.4t + 3.6) , dt = \frac{0.3}{3}t^3 - \frac{2.4}{2}t^2 + 3.6t + c = \frac{1}{10}t^3 - 1.2t^2 + 3.6t + c$ [Integrating $v$ to find $s$]

2. Assuming that $A$ is at $t=0$ and $B$ is at $t=10$, we want to find the displacement between $t=0$ and $t=10$. Let $c=0$.

3. Substitute $t=0$ and $t=10$ into the displacement function. $s(0) = 0$ $s(10) = \frac{1}{10}(10)^3 - 1.2(10)^2 + 3.6(10) = \frac{1000}{10} - 120 + 36 = 100 - 120 + 36 = 16$

4. Calculate the distance $AB$. $AB = 16$

$\boxed{16 , m}$

How to earn full marks: Integrate the velocity to find displacement, then evaluate the displacement at $t=0$ and $t=10$, and subtract.

Common Pitfall: Don't forget to include the constant of integration when finding the displacement function. However, in this case, since we're finding the distance between two points, the constant cancels out.

Exam-Style Question 4 — Paper 2 (Calculator Allowed) [9 marks]

Question:

A particle $P$ moves in a straight line such that its acceleration, $a , ms^{-2}$, at time $t$ seconds after passing through a fixed point $O$ is given by $a = 8e^{-0.4t}$. The particle has a velocity of $3 , ms^{-1}$ at $O$.

(a) Find the velocity of $P$ when $t = 3$. [4]

(b) Find an expression for the displacement of $P$ from $O$ at time $t$. [3]

(c) Find the distance of $P$ from $O$ as $t$ becomes very large. [2]

Worked Solution:

(a)

1. Find the velocity function by integrating the acceleration function with respect to time. $v = \int 8e^{-0.4t} , dt = -20e^{-0.4t} + c$ [Integrating $a$ to find $v$]

2. Use the initial condition $v = 3$ when $t = 0$ to find the constant of integration $c$. $3 = -20e^{-0.4(0)} + c$ $3 = -20(1) + c$ $c = 23$ [Using initial conditions to find $c$]

3. Write the velocity function. $v = -20e^{-0.4t} + 23$

4. Substitute $t = 3$ into the velocity function. $v = -20e^{-0.4(3)} + 23 = -20e^{-1.2} + 23 = -20(0.3012) + 23 = -6.024 + 23 = 16.976$

$\boxed{17.0 , ms^{-1}}$ (3 s.f.)

How to earn full marks: Integrate acceleration to find velocity, use the initial condition to find $c$, then substitute $t=3$.

(b)

1. Find the displacement function by integrating the velocity function with respect to time. $s = \int (-20e^{-0.4t} + 23) , dt = 50e^{-0.4t} + 23t + k$ [Integrating $v$ to find $s$]

2. Since $P$ passes through $O$ at $t=0$, we have $s=0$ when $t=0$. Use this to find the constant of integration $k$. $0 = 50e^{-0.4(0)} + 23(0) + k$ $0 = 50(1) + 0 + k$ $k = -50$ [Using initial conditions to find $k$]

3. Write the displacement function. $s = 50e^{-0.4t} + 23t - 50$

$\boxed{s = 50e^{-0.4t} + 23t - 50}$

How to earn full marks: Integrate the velocity to find displacement, use the initial condition ($s=0$ at $t=0$) to find $k$, and state the final expression.

(c)

1. As $t$ becomes very large, $e^{-0.4t}$ approaches 0. $\lim_{t \to \infty} s = \lim_{t \to \infty} (50e^{-0.4t} + 23t - 50) = 0 + \infty - 50 = \infty$

2. However, the question implies a finite distance, so we consider the difference between the displacement at large $t$ and the initial displacement. As $t$ gets very large, the $23t$ term dominates, and the $50e^{-0.4t}$ term approaches zero. We are looking for the limiting value of $s$.

   As $t\to\infty$, $s \approx 23t - 50$. However, since we are interested in the distance of $P$ from $O$ as $t$ becomes very large, the effect of the constant term $-50$ becomes negligible.

   Therefore, displacement approaches infinity. The particle travels infinitely far from O. The question is poorly worded, as it implies a finite distance.

$\boxed{\infty}$

How to earn full marks: Recognize that as $t$ approaches infinity, the exponential term goes to zero, and the displacement increases without bound.

Common Pitfall: When dealing with limits as $t$ approaches infinity, remember that exponential terms with negative exponents will approach zero. This can simplify the expression and help you determine the overall behavior of the function. Also, always include the constant of integration and use initial conditions to find its value.