Displacement, Velocity, and Acceleration

Introduction

students, imagine watching a bike race 🚴. One rider speeds up, slows down, and sometimes changes direction. To describe what is happening, mathematicians use three key ideas: displacement, velocity, and acceleration. These ideas are central to calculus because calculus helps us describe motion that changes over time.

In this lesson, you will learn how to:

explain the meaning of displacement, velocity, and acceleration,
use derivatives and integrals to connect these quantities,
interpret graphs and signs in motion problems,
apply calculus ideas to real-world motion problems.

By the end, you should see how calculus gives a precise language for motion in a straight line. This is not just about formulas; it is about understanding how position changes, how fast it changes, and how that rate of change itself changes over time.

Displacement: Change in Position

Displacement tells us how far and in what direction an object has moved from its starting point. It is not the same as distance traveled. Distance is the total path length, while displacement is the net change in position.

If an object moves from position $s(0)$ to position $s(t)$, then its displacement over that time interval is

$\Delta s = s(t) - s(0)$

Here, $s(t)$ is the position function, usually measured in units like metres or kilometres. A positive displacement means movement in the positive direction, and a negative displacement means movement in the negative direction.

Example

Suppose a runner starts at $s(0)=2$ m and later is at $s(5)=14$ m. Then the displacement is

$\Delta$ s = 14 - 2 = 12

So the runner’s displacement is $12$ m.

Now suppose the runner goes from $s(0)=2$ m to $s(5)=-4$ m. Then

$\Delta$ s = -4 - 2 = -6

This means the runner has moved $6$ m in the negative direction. The sign matters because it tells us direction.

Displacement is one of the most important ideas in calculus because it is based on a function, and calculus studies how functions change.

Velocity: Rate of Change of Displacement

Velocity tells us how quickly displacement is changing. In straight-line motion, velocity is the derivative of position with respect to time.

If $s(t)$ is the position function, then the velocity is

$v(t) = \frac{ds}{dt}$

This tells us the instantaneous rate of change of position at time $t$. Units of velocity are often metres per second, written $\text{m s}^{-1}$.

There are two important types of velocity:

average velocity,
instantaneous velocity.

Average velocity

Average velocity over a time interval $[a,b]$ is

$\frac{s(b)-s(a)}{b-a}$

This is the slope of the secant line joining two points on the position-time graph.

Instantaneous velocity

Instantaneous velocity at a specific time is the derivative

$v(t)=\frac{ds}{dt}$

This is the slope of the tangent line to the position-time graph at that instant.

Example

Let

$s(t)=t^2+2t$

Then the velocity is

$v(t)=\frac{ds}{dt}=2t+2$

At $t=3$, the velocity is

$v(3)=2(3)+2=8$

So at $3$ seconds, the object is moving at $8\,\text{m s}^{-1}$.

If $v(t)>0$, the object is moving in the positive direction. If $v(t)<0$, it is moving in the negative direction. If $v(t)=0$, the object is momentarily at rest, although it may still be about to change direction.

Acceleration: Rate of Change of Velocity

Acceleration tells us how quickly velocity is changing. It is the derivative of velocity, or the second derivative of position.

If $v(t)=\frac{ds}{dt}$, then acceleration is

$a(t)=\frac{dv}{dt}=\frac{d^2s}{dt^2}$

Its units are usually $\text{m s}^{-2}$.

Acceleration can be positive or negative, but the sign alone does not always tell you whether the object is speeding up or slowing down. That depends on both velocity and acceleration.

Example

For the same position function

$s(t)=t^2+2t$

the velocity is

$v(t)=2t+2$

and the acceleration is

$a(t)=\frac{dv}{dt}=2$

This means the object has constant acceleration of $2\,\text{m s}^{-2}$.

Speeding up or slowing down

If $v(t)$ and $a(t)$ have the same sign, the object speeds up.
If $v(t)$ and $a(t)$ have opposite signs, the object slows down.

For example, if $v(t)=3$ and $a(t)=2$, the object is speeding up. If $v(t)=3$ and $a(t)=-2$, the object is slowing down.

This is a very useful idea in IB Mathematics Analysis and Approaches SL because it connects algebra, graphs, and interpretation.

Connecting Position, Velocity, and Acceleration

The three quantities are linked by derivatives and integrals:

$v(t)=\frac{ds}{dt}$

$a(t)=\frac{dv}{dt}=\frac{d^2s}{dt^2}$

If velocity is known, position can be found by integration:

$s(t)=\int v(t)\,dt + C$

If acceleration is known, velocity can be found by integration:

$v(t)=\int a(t)\,dt + C$

The constant $C$ is important because it represents initial conditions, such as the starting position or starting velocity.

Example with integration

Suppose

$a(t)=6t$

Then velocity is found by integrating:

$v(t)=\int 6t\,dt=3t^2+C$

If the object has initial velocity $v(0)=4$, then

$4=3(0)^2+C$

so $C=4$. Therefore,

$v(t)=3t^2+4$

Now position can be found by integrating again:

$s(t)=\int (3t^2+4)\,dt=t^3+4t+D$

If $s(0)=1$, then

$1=(0)^3+4(0)+D$

so $D=1$, and

$s(t)=t^3+4t+1$

This shows how calculus builds motion step by step.

Interpreting Motion Graphs

Graph interpretation is a major skill in calculus. students, you should be able to connect a graph to motion information.

Position-time graph

slope positive: velocity positive
slope negative: velocity negative
slope zero: object momentarily at rest
steeper slope: faster motion

Velocity-time graph

height above the time axis: positive velocity
height below the time axis: negative velocity
area under the graph: displacement

The displacement over $[a,b]$ is

$\int_a^b v(t)\,dt$

This integral gives the net change in position.

Example

If velocity is

$v(t)=4-t$

then the displacement from $t=0$ to $t=4$ is

$\int_0^4 (4-t)\,dt$

Evaluating gives

$\left[4t-\frac{t^2}{2}\right]_0^4 = 16-8=8$

So the displacement is $8$ units.

This also shows a powerful idea: the net area can be positive, negative, or zero depending on whether the graph lies above or below the time axis.

Real-World Meaning and IB Connections

Displacement, velocity, and acceleration appear in many real situations, not just in textbook problems. A car speeding up on a road, a ball thrown into the air, and a train arriving at a station all involve changing motion. In physics and engineering, these quantities help describe movement precisely.

In IB Mathematics Analysis and Approaches SL, this topic connects to:

derivatives as rates of change,
integrals as accumulated change,
graph interpretation,
solving problems with initial conditions,
understanding increasing and decreasing behavior.

For example, if a particle has velocity function $v(t)$, then finding when it changes direction means solving

$v(t)=0$

and checking the sign of $v(t)$ on either side of those points. This helps identify when the object moves forward or backward.

Another important skill is relating acceleration to curvature. If acceleration is positive, velocity is increasing. If acceleration is negative, velocity is decreasing. When the velocity graph is increasing, the position graph usually becomes steeper over time.

Conclusion

Displacement, velocity, and acceleration are core ideas in calculus because they describe motion through change. Displacement measures net change in position, velocity is the derivative of position, and acceleration is the derivative of velocity. Integration works in the opposite direction by rebuilding position from velocity or velocity from acceleration.

students, understanding these relationships helps you solve motion problems, interpret graphs, and apply calculus to real situations. In IB Mathematics Analysis and Approaches SL, these ideas are essential because they show how derivatives and integrals work together to describe the world.

Study Notes

Displacement is the change in position: $\Delta s=s(t)-s(0)$.
Distance and displacement are different: distance is total path length, displacement is net change in position.
Velocity is the rate of change of position: $v(t)=\frac{ds}{dt}$.
Acceleration is the rate of change of velocity: $a(t)=\frac{dv}{dt}=\frac{d^2s}{dt^2}$.
Average velocity is $\frac{s(b)-s(a)}{b-a}$.
Instantaneous velocity is the derivative of position with respect to time.
Positive velocity means motion in the positive direction; negative velocity means motion in the negative direction.
Speeding up happens when velocity and acceleration have the same sign.
Slowing down happens when velocity and acceleration have opposite signs.
Displacement can be found from velocity using $\int_a^b v(t)\,dt$.
Velocity can be found from acceleration using $v(t)=\int a(t)\,dt + C$.
Initial conditions are needed to determine constants like $C$ and $D$.
On a position-time graph, slope represents velocity.
On a velocity-time graph, area under the curve represents displacement.
These ideas are central to calculus and help model real motion in the physical world 🚀