Displacement and velocity carry a sign; distance and speed do not. Differentiating displacement with respect to time gives velocity.
A particle can travel ten metres and end up back where it started. Displacement says zero, distance says ten, and knowing which one a question wants is half the marks.
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Recall, your gut answer first
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Recall, your gut answer first
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A particle moves $6$ m right, then $4$ m left. What is its displacement from the start? What total distance did it travel? Why are they different?
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Sign is the whole difference
Work through the core explanation before applying it.
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Sign is the whole difference
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Displacement $x$ is position relative to the origin and carries a sign; distance is how far was travelled and never decreases. Velocity $v = \dfrac{dx}{dt}$ carries a sign; speed is $|v|$.
$v = \dfrac{dx}{dt} = \dot{x}$ speed $= |v|$ at rest when $v = 0$
Negative velocity is not slowing down
It means moving in the negative direction. A particle with $v = -8$ is moving faster than one with $v = 3$; its speed is $8$.
At rest means $v = 0$
Solve $\dot{x} = 0$ to find when the particle stops. Those times are also where the direction can change, so they split a distance calculation into stages.
Distance needs the turning points
To find total distance, find when $v = 0$, compute the displacement on each stage, and add the magnitudes. Subtracting endpoints gives displacement, not distance.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
Displacement is signed position; distance travelled is never negative.
Velocity is the derivative of displacement with respect to time, $v = \dfrac{dx}{dt} = \dot{x}$.
Speed is the magnitude of velocity, $|v|$.
A particle is at rest when $v = 0$.
Understand
Concepts
Why displacement and distance differ once the particle turns around.
Why a negative velocity describes direction rather than slowness.
Why total distance must be computed stage by stage between times when $v = 0$.
Can do
Skills
Differentiate a displacement function to find velocity.
Find when a particle is at rest, and where it is at that moment.
Distinguish displacement from distance travelled in a worded problem.
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Key terms
DisplacementSigned position relative to the origin. Like this: $x = -3$ means $3$ m on the negative side of the origin.
Distance travelledThe total path length, always positive. Like this: $6$ m right then $4$ m left is a distance of $10$ m but a displacement of $2$ m.
VelocityThe rate of change of displacement with time, carrying a sign. Like this: $x = t^2$ gives $v = 2t$.
SpeedThe magnitude of velocity, with no direction. Like this: $v = -8$ means a speed of $8$.
Dot notationA dot above a variable means its derivative with respect to time. Like this: $\dot{x}$ means $\dfrac{dx}{dt}$.
At restThe moment when velocity is zero. Like this: $v = 2t - 6$ is zero at $t = 3$, so the particle is at rest then.
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Displacement against distance
Work through the core explanation before applying it.
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Displacement against distance
core concept
Displacement answers where the particle is, measured from the origin with a sign. Distance travelled answers how far it has moved altogether, and never decreases.
A particle that goes $6$ m in the positive direction then $4$ m back has displacement $2$ m and has travelled $10$ m.
If the particle never turns around, the two agree in magnitude. They differ only once the direction changes, which is why finding the turning times matters.
Read the question word. "How far from the origin" is displacement. "How far has it travelled" is distance. The two have different answers whenever the particle has reversed.
Quick check: a particle moves $5$ m right then $9$ m left. What is its displacement?
Displacement is signed position from the origin; distance travelled is total path length and never decreases. They differ as soon as the particle reverses, so $6$ m out and $4$ m back is a displacement of $2$ m and a distance of $10$ m.
Pause, copy both definitions with the $6$ m and $4$ m example, into your book.
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Velocity as a derivative
core concept
We just saw the difference between position and path. That raises a question: how fast is the position changing? This card answers it → differentiate displacement with respect to time.
Velocity is the rate of change of displacement: $v = \dfrac{dx}{dt}$, often written $\dot{x}$ in motion problems.
If $x = t^2 - 6t + 5$ then $v = 2t - 6$. At $t = 1$, $v = -4$, so the particle is moving in the negative direction at a speed of $4$.
Speed is $|v|$. A velocity of $-4$ and a velocity of $4$ describe the same speed in opposite directions.
The dot is not decoration. $\dot{x}$ specifically means differentiation with respect to time, and it appears in exam questions without explanation.
Fill the blank: if $x = t^2 - 6t + 5$ then the particle is at rest at $t = $ .
Velocity is $v = \dfrac{dx}{dt} = \dot{x}$, and speed is $|v|$. For $x = t^2 - 6t + 5$, $v = 2t - 6$, so at $t = 1$ the velocity is $-4$ and the speed is $4$.
Pause, copy $v = \dot{x}$, the worked $v = 2t - 6$, and the distinction between velocity and speed, into your book.
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At rest, and finding total distance
core concept
We just saw how to get velocity. That raises a question: when does the particle stop, and how do you total the distance? This card answers it → solve $v = 0$, then work stage by stage.
A particle is at rest when $v = 0$. For $v = 2t - 6$ that is $t = 3$, and the position there is $x = 9 - 18 + 5 = -4$.
Those times matter because the particle can reverse at them. To find total distance from $t = 0$ to $t = 5$: from $5$ to $-4$ is $9$ m, then from $-4$ back to $0$ is $4$ m, giving $13$ m altogether.
The displacement over the same interval is only $0 - 5 = -5$ m. Subtracting endpoints gives displacement; adding stage magnitudes gives distance.
Always find the turning times first. A distance question answered by subtracting the endpoints is wrong whenever the particle reversed inside the interval, and that is the case the question is usually testing.
Which statement is FALSE for a particle with $v = 2t - 6$?
Solve $v = 0$ to find when the particle is at rest, then compute distance stage by stage between those times and add the magnitudes. Subtracting endpoint positions gives displacement, not distance.
Pause, copy the at-rest condition, the staged distance calculation giving $13$ m, and the contrast with a displacement of $-5$ m, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · VELOCITY FROM DISPLACEMENT
A particle has displacement $x = t^3 - 3t^2$ metres at time $t$ seconds. Find its velocity at $t = 4$.
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$v = \dfrac{dx}{dt} = 3t^2 - 6t$
Differentiate with respect to time.
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$v(4) = 3(16) - 6(4) = 48 - 24$
Substitute $t = 4$.
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$v = 24$ metres per second
Positive, so it is moving in the positive direction.
PROBLEM 2 · WHEN IS IT AT REST
For $x = t^2 - 6t + 5$, find when the particle is at rest and where it is at that time.
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$v = 2t - 6$, and at rest means $v = 0$
Set the velocity to zero.
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$2t - 6 = 0$ gives $t = 3$ seconds
Solve for $t$.
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$x(3) = 9 - 18 + 5 = -4$ metres
Substitute into $x$, not into $v$.
PROBLEM 3 · DISTANCE AGAINST DISPLACEMENT
For $x = t^2 - 6t + 5$, find the displacement and the total distance travelled from $t = 0$ to $t = 5$.
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$x(0) = 5$, $x(5) = 0$, so the displacement is $0 - 5 = -5$ m
Endpoints give displacement.
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The particle is at rest at $t = 3$, where $x = -4$
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
If $x = t^2 + 3t$, find $v$.
A particle has $v = -7$. What is its speed?
For $v = 4t - 12$, when is the particle at rest?
A particle moves $3$ m right then $8$ m left. Give its displacement and the distance travelled.
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Revisit the particle that went out and came back
Run the quick drill and copy the summary into your book.
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Revisit the particle that went out and came back
At the start a particle moved $6$ m right then $4$ m left. State its displacement and the distance travelled, and explain in one sentence why a question about total distance cannot be answered by subtracting the endpoints.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
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Short answer
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Short answer
ApplyBand 44 marks
Q1. A particle has displacement $x = t^2 - 8t + 7$ metres at time $t$ seconds. Find its velocity, the time at which it is at rest, and its displacement at that time. (4 marks)
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ApplyBand 54 marks
Q2. For the particle in Q1, find the total distance travelled from $t = 0$ to $t = 6$, and explain why this differs from the change in displacement. (4 marks)
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UnderstandBand 42 marks
Q3. Explain the difference between velocity and speed, and give an example where two particles have different velocities but the same speed. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $v = 2t + 3$. Practice 2: $7$. Practice 3: $4t - 12 = 0$, so $t = 3$. Practice 4: displacement $-5$ m, distance $11$ m.
Q2 (4 marks): $x(0) = 7$ and $x(6) = 36 - 48 + 7 = -5$, so the change in displacement is $-12$ m [1]. The particle is at rest at $t = 4$, where $x = -9$, so it reverses inside the interval [1]. Stage 1: $|{-9} - 7| = 16$ m; stage 2: $|{-5} - (-9)| = 4$ m [1]. Total distance $= 20$ m, which differs from $12$ m because the $4$ m travelled back is added rather than cancelled [1].
Q3 (2 marks): Velocity carries a sign showing the direction of motion, while speed is its magnitude and is never negative [1]. A particle with $v = 5$ and one with $v = -5$ are moving in opposite directions but both have a speed of $5$ [1].
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Review and finish
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Boss battle · Motion Master
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Differentiate displacement, find when a particle is at rest, and separate distance from displacement. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.