1.3 Differentiate position to obtain velocity and acceleration | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Particle Kinematics Foundations

EN PH 131 Engineering Mechanics: Dynamics — Study topic 1.3

Kinematics describes motion without asking what causes it. This lesson focuses on one central operation: starting with a particle’s position as a function of time and differentiating it to find velocity and acceleration. The system in each example is the particle being tracked. The observer measures its motion in a fixed Cartesian reference frame, with stated axes and positive directions. Position tells where the particle is relative to the chosen origin; velocity tells how its position changes; acceleration tells how its velocity changes. The signs and directions of the derivatives depend on the coordinate choices, so define those choices before calculating.

What you will learn

Distinguish position, displacement, path length, velocity, and acceleration.
Differentiate a position function with respect to time to find velocity and acceleration.
Use component derivatives to find motion in two dimensions.
Interpret the direction, units, and limiting behaviour of the results.

1. Position, displacement, and path length

In a fixed coordinate frame, the particle’s position is represented by a vector from the origin to the particle. In two dimensions, for example, its position may be written as

\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}

, where

x

and

y

are coordinates and

\mathbf{i}

and

\mathbf{j}

point along the positive axes. The observer and origin remain fixed while the particle moves.

Displacement compares two positions: final position minus initial position. It is a vector and depends only on the endpoints. Path length is the total distance travelled along the route and is a scalar. If a particle moves forward and then back, its path length can be positive even when its net displacement is zero.

Choose axes that make the position components easy to describe. The positive directions are conventions, not claims about which way the particle must move. A negative velocity component means motion toward the negative side of that axis.

\Delta\mathbf{r}=\mathbf{r}(t_2)-\mathbf{r}(t_1)

Position is a vector locating the particle relative to an origin.
Displacement is the change in position; path length is the distance along the route.
State the observer, reference frame, origin, axes, and positive directions.

2. Differentiate to find velocity and acceleration

Average velocity over a time interval is displacement divided by elapsed time. Instantaneous velocity is the limiting value as that interval shrinks to zero; in calculus, it is the derivative of position with respect to time. Velocity points in the direction the position is changing. Its SI unit is metres per second.

Acceleration is the derivative of velocity with respect to time, and therefore the second derivative of position. It describes how velocity changes, including changes in speed, direction, or both. Its SI unit is metres per second squared.

In Cartesian coordinates, differentiate each position component separately. The resulting components combine into vector velocity and vector acceleration. A positive acceleration component does not always mean the particle is speeding up: it means that component of velocity is becoming more positive. For example, negative velocity and positive acceleration can mean the particle is slowing while moving in the negative direction.

For a straight-line coordinate

x(t)

, the same rules apply to signed position: first differentiate to obtain signed velocity, then differentiate again to obtain signed acceleration. Use standard differentiation rules and keep time units consistent.

\mathbf{v}=\frac{d\mathbf{r}}{dt},\qquad \mathbf{a}=\frac{d\mathbf{v}}{dt}=\frac{d^2\mathbf{r}}{dt^2}

Velocity is the time derivative of position.
Acceleration is the time derivative of velocity, or the second time derivative of position.
Differentiate components while preserving their coordinate directions and SI units.

3. Read and check the result

Before differentiating, list what is given and what is requested. Identify whether the position function is one-dimensional or has multiple components, and note the time interval or instant of interest. Use the position function directly rather than trying to infer velocity from a sketch unless the problem provides only graphical information.

After differentiating, substitute the requested time and report components with their directions. In two dimensions, a negative component indicates a direction opposite the corresponding positive axis. The speed is the magnitude of velocity, not a signed component. Acceleration is a vector too, so report its direction or components.

Check dimensions: differentiating a position measured in metres once with respect to seconds gives metres per second; differentiating again gives metres per second squared. Check initial conditions by evaluating the position and velocity at the stated starting time. For a constant position, both derivatives should be zero. These checks can reveal sign, unit, or differentiation errors.

[\mathbf{r}]=\mathrm{m},\qquad [\mathbf{v}]=\mathrm{m/s},\qquad [\mathbf{a}]=\mathrm{m/s^2}

Substitute time only after finding the general derivative.
Use vector components to make directions explicit.
Check units and compare the derivative with initial or limiting behaviour.

Worked example

1. Straight-line motion from a position function

A cart moves along a straight track. Let the fixed observer use an

x

axis positive to the right, with origin at a track marker. Its position is

x(t)=1.5+4.0t-0.50t^2

, where

x

is in metres and

t

is in seconds. Find its position, velocity, and acceleration at

t=3.0\,\mathrm{s}

.

Cart on a straight track

At the stated instant, velocity is rightward and acceleration is leftward.

Set the frame and knowns
The system is the cart, and the observer is fixed to the track. The coordinate origin is the marker and positive $x$ points right. The given position is a function of time; the requested state is at $3.0\,\mathrm{s}$ .
Differentiate position
Velocity is the first time derivative of position. Differentiating the constant term gives zero, and differentiating the quadratic term gives a term linear in time.
$v_x(t)=4.0-1.00t$
Differentiate again
Acceleration is the derivative of velocity. Here it is constant and negative, so it points toward the left under the chosen axis convention.
$a_x(t)=-1.00\,\mathrm{m/s^2}$
Evaluate at the requested time
Substitute $t=3.0\,\mathrm{s}$ into the position and velocity expressions. The position is measured from the origin; the negative velocity indicates motion leftward at that instant.
$x=3.0\,\mathrm{m},\qquad v_x=1.0\,\mathrm{m/s},\qquad a_x=-1.00\,\mathrm{m/s^2}$

Answer: At

3.0\,\mathrm{s}

the cart is

3.0\,\mathrm{m}

to the right of the marker, moving right at

1.0\,\mathrm{m/s}

, with acceleration

1.00\,\mathrm{m/s^2}

to the left.

Check: The units follow from differentiating metres with respect to seconds. At

t=0

, the equations give

x=1.5\,\mathrm{m}

and

v_x=4.0\,\mathrm{m/s}

, consistent with the stated position function and its initial slope. The negative acceleration correctly reduces the positive velocity.

Worked example

2. Two-dimensional component differentiation

A small instrument package is tracked by a fixed observer in a plane. The origin is the initial reference point,

x

is positive east, and

y

is positive north. Its position is

\mathbf{r}(t)=(2.0t^2)\mathbf{i}+(6.0t-1.0t^2)\mathbf{j}

in SI units. Find its velocity and acceleration at

t=2.0\,\mathrm{s}

.

Identify components and directions
The system is the package, viewed from the fixed plane coordinate frame. Its east and north coordinates are given separately. Differentiate each component; no force information is needed to determine these kinematic quantities.
Find component velocities
Apply the position-to-velocity rule independently to the east and north coordinates.
$v_x=4.0t,\qquad v_y=6.0-2.0t$
Find component accelerations
Differentiate each velocity component with respect to time. Both accelerations are constant in this example.
$a_x=4.0\,\mathrm{m/s^2},\qquad a_y=-2.0\,\mathrm{m/s^2}$
Evaluate at two seconds
At the specified instant, the north velocity component is positive but smaller than its initial value. The acceleration points east and south according to the axis definitions.
$\mathbf{v}=(8.0\mathbf{i}+2.0\mathbf{j})\,\mathrm{m/s},\qquad \mathbf{a}=(4.0\mathbf{i}-2.0\mathbf{j})\,\mathrm{m/s^2}$

Answer: The package velocity is

8.0\,\mathrm{m/s}

east and

2.0\,\mathrm{m/s}

north. Its acceleration is

4.0\,\mathrm{m/s^2}

east and

2.0\,\mathrm{m/s^2}

south.

Check: Each velocity component has units of metres per second and each acceleration component metres per second squared. At

t=0

, the northward velocity is

6.0\,\mathrm{m/s}

; the constant southward acceleration makes it decrease to

2.0\,\mathrm{m/s}

after two seconds.

Worked example

3. Position on a circular path

A marker moves around a circle of radius

0.80\,\mathrm{m}

in a fixed plane. A stationary observer places the origin at the circle centre, with

x

positive right and

y

positive up. Its position is

\mathbf{r}(t)=0.80\cos(2.0t)\mathbf{i}+0.80\sin(2.0t)\mathbf{j}

, with SI units. Find velocity and acceleration at

t=0

.

Marker on a circular path

At the rightmost point, position is rightward, velocity upward, and acceleration leftward.

Set the frame and initial position
The marker is the system and the observer is fixed at the circle centre. The given position describes counterclockwise motion. At $t=0$ , cosine is one and sine is zero, so the marker starts at the rightmost point.
$\mathbf{r}(0)=0.80\mathbf{i}\,\mathrm{m}$
Differentiate the position components
Differentiate both trigonometric components with respect to time. The factor $2.0\,\mathrm{rad/s}$ from the angle’s time dependence appears in each derivative.
$\mathbf{v}(t)=(-1.60\sin(2.0t))\mathbf{i}+(1.60\cos(2.0t))\mathbf{j}\;\mathrm{m/s}$
Differentiate to obtain acceleration
Differentiate velocity once more. At the initial point, the resulting acceleration is toward the centre, while velocity is tangent to the circular path.
$\mathbf{a}(t)=(-3.20\cos(2.0t))\mathbf{i}+(-3.20\sin(2.0t))\mathbf{j}\;\mathrm{m/s^2}$
Evaluate at the initial instant
Set $t=0$ . The velocity is upward and the acceleration is leftward, both consistent with the stated counterclockwise path and the selected axes.
$\mathbf{v}(0)=1.60\mathbf{j}\,\mathrm{m/s},\qquad \mathbf{a}(0)=-3.20\mathbf{i}\,\mathrm{m/s^2}$

Answer: At

t=0

, the marker moves upward at

1.60\,\mathrm{m/s}

and accelerates leftward at

3.20\,\mathrm{m/s^2}

.

Check: The velocity is perpendicular to the radius at the initial point, and the acceleration points toward the centre. The units are correct: radius times angular rate gives velocity, and radius times angular rate squared gives acceleration.

Common mistakes and how to avoid them

Treating position, displacement, and path length as interchangeable.

Correction: Position locates the particle, displacement compares two positions, and path length measures the route travelled.

Differentiating only one component of a vector position.

Correction: Differentiate every coordinate component with respect to the same time variable.

Calling positive acceleration a guarantee that the particle is speeding up.

Correction: Acceleration gives the direction of velocity change. Compare its direction with velocity before describing whether speed increases or decreases.

Dropping units or interpreting a negative component as a negative speed.

Correction: Retain SI units and interpret a negative component as direction opposite the chosen positive axis. Speed, the magnitude of velocity, is nonnegative.

Lesson summary

Define the observer, fixed reference frame, origin, coordinates, and positive directions.
Differentiate position with respect to time to obtain velocity; differentiate velocity to obtain acceleration.
For vector position, differentiate each component and interpret signs using the coordinate axes.
Check units, initial conditions, and whether the resulting directions agree with the stated motion.

Check your understanding

Question 1

A particle has position

x(t)=5.0-3.0t+2.0t^2

metres. What are its velocity and acceleration at

t=1.0\,\mathrm{s}

?

$v=1.0\,\mathrm{m/s},\ a=4.0\,\mathrm{m/s^2}$
$v=4.0\,\mathrm{m/s},\ a=2.0\,\mathrm{m/s^2}$
$v=-1.0\,\mathrm{m/s},\ a=4.0\,\mathrm{m/s^2}$
$v=1.0\,\mathrm{m/s},\ a=2.0\,\mathrm{m/s^2}$

Show answer and explanation

v=1.0\,\mathrm{m/s},\ a=4.0\,\mathrm{m/s^2}

Differentiation gives

v=-3.0+4.0t

and

a=4.0\,\mathrm{m/s^2}

. At one second, velocity is

1.0\,\mathrm{m/s}

.

Question 2

If a particle’s position is constant in time, what are its velocity and acceleration?

Velocity is zero and acceleration is zero.
Velocity is constant and nonzero; acceleration is zero.
Velocity is zero and acceleration is nonzero.
Both are nonzero because position is nonzero.

Show answer and explanation

Velocity is zero and acceleration is zero.

The derivative of a constant position is zero, and differentiating that zero velocity also gives zero acceleration.

Key terms

Position: A vector locating a particle relative to a chosen origin in a reference frame.
Displacement: The vector change in position between two times.
Path length: The total distance travelled along a route.
Velocity: The time rate of change of position; a vector with SI unit metres per second.
Acceleration: The time rate of change of velocity; a vector with SI unit metres per second squared.

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1.3 · Differentiate position to obtain velocity and acceleration

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 Engineering Mechanics: Dynamics — Study topic 1.3

What you will learn

1. Position, displacement, and path length

2. Differentiate to find velocity and acceleration

3. Read and check the result

1. Straight-line motion from a position function

2. Two-dimensional component differentiation

3. Position on a circular path

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through EN PH 131

About this lesson and its review