8.2 Differentiation in Kinematics of Linear Motion
In kinematics of linear motion, the movement of a particle along a straight line can be modelled through mathematical functions of time, \(t\). By applying differentiation, students can transition smoothly from a displacement function \(s = f(t)\) to a velocity function \(v = g(t)\), and subsequently to an acceleration function \(a = h(t)\).
Learning Objectives
- Identify and explain the relationship between displacement function \(s = f(t)\), velocity function \(v = g(t)\), and acceleration function \(a = h(t)\) using differentiation.
- Determine the velocity function \(v\) and acceleration function \(a\) by differentiating displacement and velocity functions with respect to time, \(t\).
- Determine and interpret the instantaneous velocity of a particle at a specific time, initial velocity, times when a particle stops instantaneously (\(v = 0\)), and time intervals when a particle moves to the right (\(v > 0\)) or to the left (\(v < 0\)).
- Determine and interpret the instantaneous acceleration of a particle at a specific time, initial acceleration, uniform or constant acceleration, conditions for maximum or minimum velocity (\(a = 0\)), and periods of deceleration (\(a < 0\)).
- Sketch and interpret graphs of displacement, velocity, and acceleration functions against time to analyse linear motion.
Kinematics Differentiation Flow and Motion Graph Linkages
Differentiating displacement gives velocity, and differentiating velocity gives acceleration. Each tier has its own key kinematic conditions.
At fixed point \(O \implies s = 0\).
Maximum or minimum \(s \implies \dfrac{ds}{dt} = 0\).
Initial velocity \(\implies t = 0\).
Instantaneously at rest \(\implies v = 0\).
Moving right \(\implies v > 0\).
Moving left \(\implies v < 0\).
Initial acceleration \(\implies t = 0\).
Maximum or minimum velocity \(\implies a = 0\).
Deceleration \(\implies a < 0\).
Short Explanation
Kinematics describes motion without considering the forces causing it.
1. Relationship between Displacement, Velocity, and Acceleration
Displacement \(s\) of a particle moving along a straight line from a fixed point \(O\) is a function of time \(t\), denoted as \(s = f(t)\). Velocity is the rate of change of displacement with respect to time \(t\): \(v = \dfrac{ds}{dt} = g(t)\). Acceleration is the rate of change of velocity with respect to time \(t\), which is also the second derivative of displacement with respect to time: \(a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2} = h(t)\).
2. Differentiation Rule Recall
If \(y = ax^n\), then \(\dfrac{dy}{dx} = anx^{n-1}\), where \(a\) is an integer and \(n\) is a constant. Sir Isaac Newton was the first person to introduce calculus and differentiation. His book Philosophiae Naturalis Principia Mathematica laid the foundation for the concept of limits in differentiation.
3. Instantaneous Velocity from Displacement Function
The gradient of the tangent to the displacement-time (\(s-t\)) curve at any specific time \(t\), given by \(v = \dfrac{ds}{dt}\). Initial velocity is found by evaluating \(v\) at \(t = 0\). For a linear displacement-time graph, the gradient of the tangent is constant at all points; hence, velocity is uniform.
4. Maximum or Minimum Displacement
Occurs when the gradient of the tangent to the \(s-t\) curve is zero: \(\dfrac{ds}{dt} = v = 0\).
5. Direction of Motion
Particle at instantaneous rest or changes direction: \(v = 0\). Particle moves to the right (positive direction): \(v > 0\). Particle moves to the left (negative direction): \(v < 0\). Total distance travelled is calculated by summing the absolute distances travelled across intervals before and after turning points (where \(v = 0\)).
6. Instantaneous Acceleration from Velocity and Displacement Functions
The gradient of the tangent to the velocity-time (\(v-t\)) curve at any specific time \(t\), given by \(a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}\). Initial acceleration is calculated by substituting \(t = 0\) into the acceleration function \(a\). For a linear velocity-time graph, the gradient \(\dfrac{dv}{dt}\) is constant throughout the motion.
7. Signs and Interpretation of Acceleration
Positive acceleration (\(a > 0\)): the gradient \(\dfrac{dv}{dt} > 0\); velocity increases with time. Negative acceleration or deceleration (\(a < 0\)): the gradient \(\dfrac{dv}{dt} < 0\); velocity decreases with time. Zero acceleration (\(a = 0\)): the gradient \(\dfrac{dv}{dt} = 0\). This occurs when the particle experiences either a maximum velocity or a minimum velocity. Crucially, \(a = 0\) does not mean that the velocity is zero.
8. Total Distance Travelled
Calculated by summing the absolute distances travelled across intervals before and after turning points (where \(v = 0\)).
| Condition | Mathematical Representation | Interpretation |
|---|---|---|
| Instantaneous rest / changes direction | \(v = 0\) | The particle momentarily stops and may reverse direction. |
| Moves to the right | \(v > 0\) | Positive direction of motion. |
| Moves to the left | \(v < 0\) | Negative direction of motion. |
| Maximum or minimum displacement | \(\dfrac{ds}{dt} = v = 0\) | The gradient of the \(s-t\) curve is zero. |
| Maximum or minimum velocity | \(\dfrac{dv}{dt} = a = 0\) | The gradient of the \(v-t\) curve is zero. |
| Deceleration | \(a < 0\) | Velocity decreases with time. |
Velocity function
\(v = \dfrac{ds}{dt}\)First derivative of displacement.
Acceleration function
\(a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}\)First derivative of velocity, second derivative of displacement.
Try to Answer First
Answer in your mind, then press "Check Answer".
A particle moves along a straight line such that its displacement from a fixed point \(O\) after \(t\) seconds is \(s = 2t^3 - 4t^2 + 2t + 1\). Determine the velocity function \(v\) and the acceleration function \(a\) of the particle in terms of \(t\).
The displacement of a particle moving along a straight line is given by \(s = 40t - 5t^2\). Find: (a) the initial velocity of the particle, (b) the time \(t\) when the displacement is maximum, (c) the maximum displacement.
The velocity of a particle is given by \(v = 8t - t^2\). (a) Find the initial acceleration of the particle. (b) Find the acceleration when the particle stops instantaneously for the second time.
Common Mistakes
- Confusing \(v = 0\) with \(a = 0\): students often mistakenly assume that when a particle is instantaneously at rest (\(v = 0\)), its acceleration must also be zero. In reality, a particle reversing direction has \(v = 0\) while maintaining a non-zero acceleration (e.g., \(a = -2\text{ ms}^{-2}\)).
- Confusing \(a = 0\) with \(v = 0\): students assume zero acceleration implies the particle is stationary. In fact, \(a = 0\) means the velocity is constant, maximum, or minimum, not necessarily zero.
- Misinterpreting the sign of velocity and displacement: assuming negative displacement (\(s < 0\)) means the particle is moving to the left. Displacement sign indicates position relative to fixed point \(O\), whereas the sign of velocity (\(v\)) indicates the direction of motion.
- Incorrect total distance calculation: students often compute total distance simply by finding displacement at the end time (\(s(t_{\text{end}})\)), neglecting direction changes where the particle turns around.
Concept Misunderstandings
Deceleration always means the particle is moving backwards.
Deceleration (\(a < 0\)) means the particle's velocity is decreasing over time. The particle can still be moving forward (to the right, \(v > 0\)) while decelerating.
Initial conditions mean when displacement or velocity equals zero.
The term "initial" strictly refers to the time \(t = 0\) at the start of motion, not \(s = 0\) or \(v = 0\).
Maximum displacement occurs when acceleration is zero.
Maximum or minimum displacement occurs when velocity is zero (\(\dfrac{ds}{dt} = v = 0\)). Maximum or minimum velocity occurs when acceleration is zero (\(\dfrac{dv}{dt} = a = 0\)).
Summary
- Velocity function: \(v = \dfrac{ds}{dt}\) (first derivative of displacement).
- Acceleration function: \(a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}\) (first derivative of velocity, second derivative of displacement).
- Initial state: substitute \(t = 0\) into the respective function to obtain initial displacement, initial velocity, or initial acceleration.
- Instantaneous rest: set \(v = 0\) to find the time(s) when the particle stops or reverses direction.
- Direction of motion: \(v > 0\) indicates motion to the right; \(v < 0\) indicates motion to the left.
- Extreme values: maximum or minimum displacement occurs when \(\dfrac{ds}{dt} = v = 0\); maximum or minimum velocity occurs when \(\dfrac{dv}{dt} = a = 0\).
- Deceleration occurs when \(a < 0\) (rate of change of velocity is negative).
Short Activity
Objective: Test your understanding of velocity, acceleration, direction of motion and key kinematic conditions.
A. Objective Quiz
1 A particle moves along a straight line such that its displacement from a fixed point \(O\) is \(s = 8 + 2t - t^2\). What is the velocity of the particle when it passes through point \(O\) again?
2 The velocity of a particle is given by \(v = t^2 - 2t - 8\). For what range of \(t\) does the particle experience deceleration?
B. Fill in the Blanks
3 Maximum or minimum displacement of a moving particle occurs when the instantaneous velocity is equal to ____ ms⁻¹.
4 A particle moving in a straight line experiences negative acceleration when its velocity decreases with time, which is known as ____.
C. Matching / Drag and Drop
Match each kinematic condition or term with its corresponding mathematical representation. On a phone, tap a term first, then tap the matching box.
