Motion of a Particle in a Straight Line
Questions on the motion of a particle in a straight line ask you to differentiate a displacement equation to find velocity and acceleration. It is the final calculus skill in the Edexcel IGCSE Maths Graphs unit, and the marks come quickly once the chain from s to v to a is automatic. This page shows you how displacement, velocity and acceleration are connected, walks through clear worked examples, and then lets you drill the skill with auto-marked practice questions below.
Motion of a Particle: Displacement, Velocity and Acceleration
In these questions a particle (or a ball, drone or car) moves along a straight line, and its displacement \(s\) from a fixed point O is given as an equation in time \(t\). Displacement is measured from O, so a positive value means one direction along the line and a negative value means the other. Everything on this page comes from a single idea: velocity and acceleration are rates of change, so each one is found by differentiating.
displacement \(s\) → differentiate → velocity \( v = \dfrac{\mathrm{d}s}{\mathrm{d}t} \) → differentiate → acceleration \( a = \dfrac{\mathrm{d}v}{\mathrm{d}t} \)
How to Answer a Motion Question
- Write down the displacement equation \(s\) in terms of \(t\), expanding any brackets first.
- Differentiate once to get the velocity \(v\): multiply each term by its power of \(t\), then reduce the power by 1. Constants disappear.
- Differentiate again if the question needs the acceleration \(a\).
- Then answer the actual question: substitute the given time into \(v\) or \(a\), or set \( v = 0 \) to find when the particle is at rest or at its greatest displacement.
Keep the units straight: with \(s\) in metres and \(t\) in seconds, velocity is in m/s and acceleration is in m/s2.
Worked Examples
💡 Example 1: From s to v to a
The displacement, \(s\) metres, of a particle after \(t\) seconds is \( s = 2t^{3} - 5t^{2} + 4t + 7 \). Find expressions for the particle's velocity and acceleration.
Differentiate \(s\) once to find the velocity:
\[ v = 6t^{2} - 10t + 4 \]Differentiate \(v\) once more to find the acceleration:
\[ a = 12t - 10 \]What's happening?
Term by term: multiply by the power, then knock the power down by 1. So \(2t^{3}\) becomes \(6t^{2}\), \(-5t^{2}\) becomes \(-10t\), \(4t\) becomes \(4\), and the constant \(7\) vanishes.
\(v\) is in m/s; differentiating again gives \(a\) in m/s2.
💡 Example 2: Evaluate at a given time
A particle moves so that \( s = t^{3} - 4t^{2} + 9t \), with \(s\) in metres and \(t\) in seconds. Find its velocity and acceleration when \( t = 2 \).
Differentiate first, substitute last:
\[ v = 3t^{2} - 8t + 9 \]When \( t = 2 \):
\[ \begin{array}{rcl} v &=& 3(2)^{2} - 8(2) + 9 \\ &=& 12 - 16 + 9 \\ &=& 5 \end{array} \]So the velocity is 5 m/s. Differentiate again:
\[ a = 6t - 8 \] \[ \begin{array}{rcl} a &=& 6(2) - 8 \\ &=& 4 \end{array} \]The acceleration is 4 m/s2.
What's happening?
The most common error is substituting \( t = 2 \) into \(s\) itself: that gives the position, not the velocity.
Always differentiate to the quantity you need, and only then substitute the time.
💡 Example 3: Greatest height, where v = 0
A ball is thrown straight up so that its height, \(s\) metres, after \(t\) seconds is \( s = 18t - 3t^{2} \). Find the greatest height the ball reaches.
Differentiate to find the velocity:
\[ v = 18 - 6t \]At the greatest height the ball is momentarily at rest, so set \(v\) equal to zero:
\[ \begin{array}{rcl} 18 - 6t &=& 0 \\ t &=& 3 \end{array} \]Substitute \( t = 3 \) back into \(s\):
\[ \begin{array}{rcl} s_{\max} &=& 18(3) - 3(3)^{2} \\ &=& 54 - 27 \\ &=& 27 \end{array} \]The greatest height is 27 m.
What's happening?
This is a stationary point in disguise: the top of the flight is the maximum of the displacement-time curve, and the gradient there is zero.
Setting \( v = 0 \) finds when it happens; substituting that time into \(s\) finds how high.
Displacement against time
Velocity against time
The two graphs tell one story: where the displacement curve turns (left), the velocity line crosses zero (right).
💡 Example 4: At rest twice
A particle moves in a straight line so that its displacement from O is \( s = t^{3} - 6t^{2} + 9t + 5 \) metres after \(t\) seconds. Find the times at which the particle is at rest.
Differentiate to find the velocity:
\[ v = 3t^{2} - 12t + 9 \]At rest means the velocity is zero, so solve:
\[ \begin{array}{rcl} 3t^{2} - 12t + 9 &=& 0 \\ 3(t - 1)(t - 3) &=& 0 \end{array} \]So the particle is at rest at \( t = 1 \) second and again at \( t = 3 \) seconds.
What's happening?
A cubic displacement gives a quadratic velocity, so there can be up to two at-rest times: factorise and solve.
If one of the solutions is negative, reject it: the motion only makes sense for \( t \ge 0 \).
🔑 Key Points
- Velocity is the rate of change of displacement: \( v = \dfrac{\mathrm{d}s}{\mathrm{d}t} \). It is the gradient of the displacement-time graph.
- Acceleration is the rate of change of velocity: \( a = \dfrac{\mathrm{d}v}{\mathrm{d}t} \). It is the gradient of the velocity-time graph.
- Differentiate term by term with the power rule; a constant term differentiates to zero.
- At rest, and at the greatest displacement or height, the velocity is zero. Maximum velocity happens where the acceleration is zero.
⚠️ Pitfalls
- Substituting the time into \(s\) when the question asks for velocity: differentiate first, substitute last.
- Mixing up units: velocity is m/s, acceleration is m/s2.
- Reading "at rest" as \( s = 0 \). At rest means the velocity is zero, not the displacement.
- Keeping a negative time from a quadratic: if solving gives a negative \(t\), reject it.
- Dropping the sign: a negative velocity or acceleration is a meaningful answer, not a mistake.