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 \(a = h(t)\).
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 \(\implies a < 0\).
Kinematics describes motion without considering the forces causing it.
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: \(a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2} = h(t)\).
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.
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.
Occurs when the gradient of the tangent to the \(s-t\) curve is zero: \(\dfrac{ds}{dt} = v = 0\).
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 \(v = 0\)).
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.
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.
Calculated by summing the absolute distances travelled across intervals before and after turning points (where \(v = 0\)).
Velocity function
First derivative of displacement.
Acceleration function
First derivative of velocity, second derivative of displacement.
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.
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 (\(\dfrac{dv}{dt} = a = 0\)).
Objective: Test your understanding of velocity, acceleration, direction of motion and key kinematic conditions.
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?
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 ____.
Match each kinematic condition or term with its corresponding mathematical representation. On a phone, tap a term first, then tap the matching box.
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