Linear motion involves the movement of an object along a straight line relative to a fixed point, known as the reference point \(O\). In this topic, kinematics explores the relationship between time \(t\) and three fundamental vector quantities: displacement \(s\), velocity \(v\), and acceleration \(a\). Understanding these quantities as functions of time allows us to determine the exact position, instantaneous rate of motion, and total distance travelled by an object at any given moment.
How the sign of displacement, velocity and acceleration relates to position on a number line and tangent slopes on \(s\)-\(t\) and \(v\)-\(t\) graphs.
With movement to the right of fixed point \(O\) defined as positive:
\(s > 0\) — particle is on the right of \(O\). \(s = 0\) — particle is at the fixed point \(O\). \(s < 0\) — particle is on the left of \(O\).
\(v > 0\) — moving to the right. \(v = 0\) — instantaneously at rest or turning. \(v < 0\) — moving to the left.
\(a > 0\) — velocity is increasing. \(a = 0\) — velocity is at a maximum or minimum. \(a < 0\) — velocity is decreasing (deceleration).
Three key functions of time describe linear motion: displacement, velocity and acceleration.
Gives the exact position relative to origin \(O\). A positive value denotes a position to the right of \(O\), and a negative value denotes a position to the left of \(O\).
Represents the slope of the displacement-time graph. A positive velocity means moving rightwards, while a negative velocity means moving leftwards. When \(v = 0\), the object is momentarily at rest.
Represents the slope of the velocity-time graph. Positive acceleration means velocity is increasing, negative acceleration means velocity is decreasing (deceleration), and zero acceleration indicates maximum or minimum velocity.
To calculate the total path length, any reversal of direction (turning point) must be taken into account by summing the absolute distances of each movement segment.
Identify turning points where instantaneous velocity \(v = 0\), or where displacement reaches a local maximum or minimum.
Tabulate values of \(s\) at boundary times and turning times.
Sketch a number line or a displacement-time graph to trace every segment of the particle's movement.
Sum the absolute lengths of all individual paths travelled.
Instantaneous velocity is the gradient of the tangent to the \(s\)-\(t\) graph
Instantaneous acceleration is the gradient of the tangent to the \(v\)-\(t\) graph
Answer in your mind, then press "Check Answer".
A particle moves along a straight line such that its displacement from a fixed point \(O\) is given by \(s = 2t^2 - 5t - 3\). Find the instantaneous displacement of the particle when \(t = 2\text{ s}\) and state its position relative to \(O\).
The velocity of a particle is given by \(v = t^2 - 8t + 7\), where \(t\) is the time in seconds after passing point \(O\). Calculate the values of \(t\) when the particle stops instantaneously.
A particle moves along a straight line with displacement \(s = t^2 - 6t\) from fixed point \(O\). Calculate the total distance travelled by the particle in the first \(7\text{ seconds}\).
Zero acceleration means the object is at rest.
Zero acceleration (\(a = 0\)) means the velocity is constant or momentarily stationary at an extreme value (maximum or minimum velocity). An object is only at rest when velocity \(v = 0\).
Negative acceleration always means an object is moving backwards.
Negative acceleration means the velocity is decreasing with time (decelerating in the positive direction). The direction of motion depends entirely on the sign of velocity \(v\), not acceleration \(a\).
Distance and displacement are identical if motion is along a straight line.
They are only equal in magnitude if the object travels in a single direction without reversing. Once it changes direction, distance accumulates positively, whereas displacements can cancel out.
Objective: Identify displacement, velocity, acceleration and total distance concepts in kinematics of linear motion.
1 A particle moves along a straight line such that its velocity is given by \(v = 3t - 12\text{ ms}^{-1}\). What is the physical state of the particle at \(t = 4\text{ s}\)?
2 The acceleration of a particle is given by \(a = 8 - 4t\text{ ms}^{-2}\). At what time is the velocity of the particle at its maximum?
3 When the velocity of an object decreases with time, the object undergoes deceleration and its acceleration value is ________.
4 On a displacement-time graph, the instantaneous velocity of a particle at time \(t\) is represented by the ________ of the tangent to the curve at that time.
Drag each kinematic condition to the correct physical interpretation. On a phone, tap a condition first, then tap the matching box.
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