SL 5.10 Model displacement, velocity, and acceleration with calculus | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Calculus

Using derivatives and integrals to describe motion along a straight line

A position function describes where an object is at each time. Calculus links its position to how quickly it moves and how its motion changes. In this lesson, motion is along a straight line, with position measured from a chosen origin. Positive and negative signs show direction relative to a chosen positive direction. Always use the stated time interval and include units when interpreting results.

What you will learn

Define displacement, velocity, and acceleration using a position model.
Differentiate position to find velocity and acceleration, and integrate rates of change using initial values.
Interpret signs, units, graphs, and values at particular times.
Use a graphing calculator to check a model while showing the mathematical reasoning.

1. Prior knowledge and the motion model

Write the signed position of an object at time

t

as

s(t)

. For example,

s(t)=8

metres means the object is 8 metres in the positive direction from the origin; a negative position is on the other side. Displacement over an interval is the final position minus the initial position. It is not necessarily the total distance travelled.

If position is measured in metres and time in seconds, velocity is measured in metres per second and acceleration in metres per second squared. Velocity is signed: positive velocity means position is increasing, and negative velocity means position is decreasing. Acceleration describes how velocity changes; its sign alone does not state the direction of motion.

A derivative measures an instantaneous rate of change. For powers of

t

, the power rule is

\frac{d}{dt}(t^n)=nt^{n-1}

. Integration reverses this process for these power functions: for

n\ne -1

, an antiderivative of

t^n

is

\frac{t^{n+1}}{n+1}

. An initial position or velocity determines the constant that appears after integration.

\Delta s=s(t_2)-s(t_1)

State the time interval and units for the model.
Displacement is a change in position; it is not generally the distance travelled.
Signs describe direction relative to the chosen positive direction.

2. Derivatives connect position, velocity, and acceleration

Instantaneous velocity is the rate at which position changes with time. If position is

s(t)

, differentiate it to find velocity. Acceleration is the rate at which velocity changes, so differentiate velocity; equivalently, differentiate position twice.

On a position–time graph, the gradient at a point is the instantaneous velocity. A rising graph has positive velocity, a falling graph has negative velocity, and a horizontal tangent means zero velocity at that instant. On a velocity–time graph, the gradient is acceleration. A rising velocity graph means positive acceleration, whether velocity itself is positive or negative.

Average velocity over an interval is displacement divided by elapsed time. On a position graph, this is the gradient of a line joining the two endpoint positions. Shortening the interval around a particular time gives an average rate closer to the instantaneous velocity at that time.

To find when an object is momentarily at rest, solve

v(t)=0

and keep solutions inside the model's time interval. Check the sign of velocity on either side to decide whether the object changes direction. A zero of velocity does not automatically mean a direction change.

v(t)=\frac{ds}{dt},\qquad a(t)=\frac{dv}{dt}=\frac{d^2s}{dt^2}

Differentiate position once for velocity and twice for acceleration.
Graph gradients represent rates of change: position to velocity, velocity to acceleration.
A change in the sign of velocity indicates a change in direction.

3. Integrals recover changes in motion

When acceleration is given, integrate it with respect to time to obtain velocity. Use a known velocity at a particular time to find the integration constant. Integrating velocity gives position, and a known position fixes the next constant.

The definite integral of velocity over a time interval gives signed displacement. On a velocity–time graph, this is the signed area between the graph and the time axis. Areas below the axis count as negative, so positive and negative displacements may cancel. If asked for total distance, split the interval at any times when velocity changes sign and add the magnitudes of the displacements on the resulting parts.

A graphing calculator can display a model, estimate a zero, or check a calculated value. First identify the function and time interval; then use calculus and algebra to show the method. Use the graph or numerical output as a check, retain adequate intermediate precision, and round only to the requested accuracy.

s(t_2)-s(t_1)=\int_{t_1}^{t_2}v(t)\,dt

Use initial position or velocity to determine an integration constant.
Signed area under a velocity–time graph gives displacement.
For total distance, account for any change in direction before combining distances.

Worked example

Differentiate a position model

An object moves along a straight track with position

s(t)=t^3-6t^2+9t

metres for

0\le t\le4

, where

t

is in seconds. Find its velocity and acceleration functions. Determine both values at

t=1

and interpret them.

Find velocity
Differentiate each term of the position function with respect to time. The resulting unit is metres per second.
$v(t)=3t^2-12t+9$
Find acceleration
Differentiate velocity with respect to time. The resulting unit is metres per second squared.
$a(t)=6t-12$
Evaluate and interpret
Substitute $t=1$ into both functions. Zero velocity means the object is momentarily at rest; negative acceleration means velocity is changing in the negative direction at that instant.
$v(1)=0\text{ m/s},\qquad a(1)=-6\text{ m/s}^2$

Answer: The velocity is

v(t)=3t^2-12t+9

metres per second, and the acceleration is

a(t)=6t-12

metres per second squared. At

t=1

, the object is momentarily at rest and its acceleration is

-6

metres per second squared.

Check: The velocity factors as

v(t)=3(t-1)(t-3)

. It is positive just before

t=1

and negative just after, so the object changes direction at that instant.

Worked example

Integrate acceleration using initial values

An object has acceleration

a(t)=4t-6

metres per second squared. At

t=0

, its velocity is

3

metres per second and its position is

2

metres. Find velocity and position as functions of time, then find displacement from

t=0

to

t=3

.

Integrate acceleration
Integrate each term of the acceleration function. Apply the initial velocity $v(0)=3$ to determine the constant.
$v(t)=2t^2-6t+3$
Integrate velocity
Integrate the velocity function and apply the initial position $s(0)=2$ to determine the new constant.
$s(t)=\frac{2}{3}t^3-3t^2+3t+2$
Find displacement
Evaluate position at the two endpoint times and subtract the initial position from the final position.
$s(3)-s(0)=2-2=0\text{ m}$

Answer: The velocity is

v(t)=2t^2-6t+3

metres per second, the position is

s(t)=\frac{2}{3}t^3-3t^2+3t+2

metres, and the displacement is

0

metres.

Check: The functions give

v(0)=3

metres per second and

s(0)=2

metres, as required. Zero displacement means the endpoints have the same position; it does not mean the object did not move.

Worked example

Use signs and a graphing check

An object has position

s(t)=t^3-3t^2-9t+5

metres for

0\le t\le4

. Find when it is at rest and decide whether it changes direction. Use a graphing calculator to check the result.

Find velocity and its zero
Differentiate position, factor the velocity, and solve for zero velocity. Keep only roots in the given time interval.
$v(t)=3t^2-6t-9=3(t-3)(t+1)$
Check the direction
The root $t=-1$ is outside the domain. At $t=2$ , velocity is negative; at $t=4$ , it is positive. Since the sign changes at $t=3$ , the object changes direction there.
$v(2)=-9\text{ m/s},\qquad v(4)=15\text{ m/s}$
Check with technology
Graph $s(t)$ and $v(t)$ on $0\le t\le4$ . The velocity graph should cross the time axis at $t=3$ , and the position graph should have a horizontal tangent there. Substitution into the position function gives the position at that instant.
$s(3)=-22\text{ m}$

Answer: The object is at rest at

t=3

seconds and changes direction from the negative direction to the positive direction. Its position then is

-22

metres.

Check: The opposite signs of velocity on either side of

t=3

confirm a direction change. The calculator graph provides a visual check of the algebraic result.

Common mistakes and how to avoid them

Treating velocity as speed, or assuming negative velocity is impossible.

Correction: Velocity is signed and indicates direction along the chosen line. Speed is non-negative.

Using the sign of acceleration alone to decide the direction of motion.

Correction: Use velocity to determine direction. Acceleration describes how velocity is changing.

Setting an integration constant to zero without using the given initial information.

Correction: Substitute the initial position or velocity into the integrated function to determine the constant.

Assuming zero displacement means zero distance travelled.

Correction: Displacement compares endpoint positions. Movement in opposite directions can cancel in displacement even though distance travelled is positive.

Lesson summary

Position is modelled by $s(t)$ ; differentiating once gives velocity and twice gives acceleration.
Integrating acceleration gives velocity, and integrating velocity gives position; initial values determine constants.
The gradient of a position graph is velocity, and the gradient of a velocity graph is acceleration.
The signed area under a velocity–time graph gives displacement. Check units, signs, domain, and initial values.

Check your understanding

Question 1

An object's position is

s(t)=2t^2-t

metres. What is its velocity at

t=2

seconds?

$7$ metres per second
$6$ metres per second
$3$ metres per second
$5$ metres per second

Show answer and explanation

7

metres per second

Differentiating gives

v(t)=4t-1

. Substitution gives

v(2)=7

metres per second.

Question 2

An object's velocity is negative throughout a time interval. Which statement must be true, using the chosen positive direction?

Its position is decreasing.
Its acceleration is negative.
Its speed is decreasing.
Its position is negative.

Show answer and explanation

Its position is decreasing.

Negative velocity means position is decreasing. The acceleration, speed change, and sign of position cannot be determined from velocity's sign alone.

Key terms

Position: The signed location of an object relative to a chosen origin.
Displacement: The change in position over a time interval: final position minus initial position.
Velocity: The rate of change of position with time; its sign indicates direction.
Acceleration: The rate of change of velocity with time.
Initial value: A known position or velocity at a particular time, used to determine a constant after integration.

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SL 5.10 · Model displacement, velocity, and acceleration with calculus

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Using derivatives and integrals to describe motion along a straight line

What you will learn

1. Prior knowledge and the motion model

2. Derivatives connect position, velocity, and acceleration

3. Integrals recover changes in motion

Differentiate a position model

Integrate acceleration using initial values

Use signs and a graphing check

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through IB AA SL

About this lesson and its review