2.5 Relate positions, velocities, and accelerations between translating frames | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Curvilinear and Relative Motion

EN PH 131 Engineering Mechanics: Dynamics — Study topic 2.5

A moving observer can describe the same particle differently from a stationary observer. A person walking on a moving train, for example, has one velocity relative to the train and another relative to the ground. To relate these descriptions, identify the particle, the observers, and their reference frames. In this lesson, one frame translates relative to another but does not rotate: its axes keep the same directions. The central idea is that the particle’s motion measured from the moving frame combines with the frame’s motion to give the motion measured from the fixed frame.

What you will learn

Describe one particle’s position from either of two reference frames whose axes remain parallel.
Relate the particle’s velocity and acceleration to those of a translating frame.
Choose coordinates and signs consistently, then check the result’s direction, units, and limiting cases.

1. Define the frames and position vectors

Let frame

A

be used by an observer fixed to the ground, and frame

B

be used by an observer whose origin moves relative to

A

. The system of interest is a single particle,

P

. Both observers describe that same particle at the same instant. Assume frame

B

translates without rotating, so its coordinate axes remain parallel to those of frame

A

.

Choose common positive directions, such as

+x

to the right and

+y

upward. Let

O_A

and

O_B

be the frame origins. The vector from

O_A

to

O_B

gives the moving origin’s position relative to

A

. The vector from

O_B

to

P

gives the particle’s position relative to

B

. Adding these vectors head-to-tail gives the particle’s position relative to A:

\boldsymbol r_{P/A}=\boldsymbol r_{B/A}+\boldsymbol r_{P/B}

.

Position is a vector from a specified origin; it is not the distance travelled. Displacement is the change in position, path length is the total distance along the route, and velocity is the time rate of change of position. Keeping origins and observers explicit helps prevent treating a relative quantity as though it were measured from the ground.

\boldsymbol r_{P/A}=\boldsymbol r_{B/A}+\boldsymbol r_{P/B}

Name the fixed frame, translating frame, and particle before writing equations.
Position vectors depend on the chosen origin; the particle itself does not.
The translation-only relations require the frames’ coordinate axes to remain parallel.

2. Relate velocity and acceleration

Differentiate the position relation with respect to time. Since frame

B

does not rotate, its coordinate directions do not change as the particle is described. The particle’s velocity measured in frame

A

is therefore the velocity of the moving origin plus the particle’s velocity relative to frame B:

\boldsymbol v_{P/A}=\boldsymbol v_{B/A}+\boldsymbol v_{P/B}

.

Differentiate once more to obtain the acceleration relation:

\boldsymbol a_{P/A}=\boldsymbol a_{B/A}+\boldsymbol a_{P/B}

. These equations use the same time measurement in both frames and assume that frame

B

translates without rotating.

Use vector components to handle directions. In one dimension, choose right as positive; a leftward velocity or acceleration then has a negative sign. In two dimensions, apply the relations along the shared horizontal and vertical axes. Do not add speeds without signs or directions: velocity is a vector.

A useful check is a particle at rest relative to the moving frame. Its relative velocity and acceleration are zero, so the observer in frame

A

measures the same velocity and acceleration as the moving origin. If the frames are at rest relative to each other, their descriptions agree.

\boldsymbol v_{P/A}=\boldsymbol v_{B/A}+\boldsymbol v_{P/B},\quad \boldsymbol a_{P/A}=\boldsymbol a_{B/A}+\boldsymbol a_{P/B}

Absolute velocity equals translating-frame velocity plus velocity relative to that frame.
Absolute acceleration equals translating-frame acceleration plus relative acceleration.
These simple relations assume no rotation of the moving frame.

3. A reliable solution method

Begin by stating what is observed and from where. For example, the system might be a person, frame

A

the ground, and frame

B

a train moving along a straight track. Set the positive direction and record which positions, velocities, or accelerations are known. A motion diagram can clarify directions when the signs are not obvious.

Write the position relation first, then differentiate only if velocity or acceleration is needed. This keeps the sign convention consistent. For one-dimensional motion, signed scalar equations are often simplest; for planar motion, use vector components along the shared axes.

Check dimensions before reporting a result: position uses metres, velocity metres per second, and acceleration metres per second squared. Interpret the sign using the chosen coordinates. Test a simple limiting case, such as a particle at rest relative to the translating frame or a frame with zero velocity. These are kinematic relations; they do not by themselves describe forces on the particle.

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

Use one sign convention in every term of an equation.
A negative component indicates a direction opposite to the chosen positive axis.
Translation-only relations describe motion, not the forces producing it.

4. Apply the relations in one and two dimensions

For straight-line motion, choose a single axis along the direction of travel. If a vehicle moves right and a passenger walks left relative to it, the passenger’s ground velocity is the vehicle velocity plus a negative relative velocity. The passenger can still move right relative to the ground if the vehicle’s speed is greater.

For planar motion, the translating frame may move in one direction while the particle moves relative to it in another. Add the horizontal and vertical components separately. The same method applies to acceleration: a particle can accelerate relative to a vehicle while the vehicle itself accelerates along its track.

The frame assumption matters. If the moving axes turn, their directions change with time and these translation-only equations are not sufficient. This lesson treats only frames that translate without rotating.

\boldsymbol r_{P/A}=\boldsymbol r_{B/A}+\boldsymbol r_{P/B}

Relative velocity can oppose frame velocity without reversing the particle’s ground motion.
In two dimensions, add vector components rather than magnitudes.
Do not use translation-only relations for rotating axes.

Worked example

A person walking inside a moving train

A train moves right at

12\,\mathrm{m/s}

. A passenger walks left at

1.5\,\mathrm{m/s}

relative to the train. Find the passenger’s velocity relative to the ground.

Train and passenger motion

The passenger’s relative velocity is leftward while the train moves right.

Define frames and signs
The system is the passenger. Frame $A$ is fixed to the ground, and frame $B$ moves with the train. Choose right as positive. The passenger’s velocity relative to the train is negative because it points left.
Use the velocity relation
The passenger’s ground velocity is the train’s velocity plus the passenger’s velocity relative to the train.
$v_{P/A}=v_{B/A}+v_{P/B}$
Substitute and interpret
Substitute the signed values. The positive answer means the passenger moves right relative to the ground.
$v_{P/A}=12-1.5=10.5\,\mathrm{m/s}$

Answer: The passenger moves right relative to the ground at

10.5\,\mathrm{m/s}

.

Check: Both inputs are velocities, so the result has units of metres per second. If the passenger walked left at

12\,\mathrm{m/s}

relative to the train, the ground velocity would be zero, consistent with the relation.

Worked example

Position and velocity from a moving cart

A cart’s origin is at

x=3\,\mathrm{m}

and moves right at a constant

2\,\mathrm{m/s}

. At the instant considered, a small sensor is

4\,\mathrm{m}

to the left of the cart origin and moves right at

0.5\,\mathrm{m/s}

relative to the cart. Find the sensor’s position and velocity relative to the ground.

Sensor relative to cart

The sensor is left of the cart origin, but its relative velocity points right.

Set up the observation
The system is the sensor; frame $A$ is fixed to the ground and frame $B$ translates with the cart. Right is positive. The sensor’s relative position is negative because it is left of the cart origin; its relative velocity is positive.
Find ground position
Add the cart origin’s ground position to the sensor’s position relative to the cart. This position relation applies at the stated instant.
$x_{P/A}=3+(-4)=-1\,\mathrm{m}$
Find ground velocity
Add the cart’s velocity to the sensor’s velocity relative to the cart. The positive result indicates motion to the right.
$v_{P/A}=2+0.5=2.5\,\mathrm{m/s}$

Answer: The sensor is at

x=-1\,\mathrm{m}

relative to the ground and moves right at

2.5\,\mathrm{m/s}

.

Check: The sensor’s position is left of the ground origin because its coordinate is negative. Its velocity is greater than the cart’s because it also moves right relative to the cart.

Worked example

Acceleration observed from an accelerating vehicle

A vehicle accelerates right at

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

. A package slides on its floor and has acceleration

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

to the left relative to the vehicle. Find the package’s acceleration relative to the ground.

Vehicle and package motion

The package’s relative acceleration points left while the vehicle accelerates right.

Define the frames
The system is the package. Frame $A$ is fixed to the ground; frame $B$ translates with the vehicle. Choose right as positive. The vehicle acceleration is positive, while the package’s relative acceleration is negative.
Relate accelerations
For translating, nonrotating frames, the package’s ground acceleration equals the vehicle’s acceleration plus its acceleration relative to the vehicle.
$a_{P/A}=a_{B/A}+a_{P/B}$
Calculate the signed result
Substitute the signed accelerations. A positive result means the package accelerates to the right relative to the ground.
$a_{P/A}=1.2-0.4=0.8\,\mathrm{m/s^2}$

Answer: The package accelerates right relative to the ground at

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

.

Check: The units are acceleration units. If the package’s leftward relative acceleration equalled the vehicle’s rightward acceleration, its ground acceleration would be zero.

Common mistakes and how to avoid them

Adding velocity magnitudes even when their directions oppose.

Correction: Choose a positive direction and use signed components, or add the velocity vectors.

Treating a position relative to a moving origin as the particle’s ground position.

Correction: Add the moving origin’s position to the particle’s position relative to that origin.

Using translation-only equations when the moving frame’s axes rotate.

Correction: These equations require the moving frame to translate without rotating.

Confusing path length with displacement.

Correction: Position and displacement have direction and depend on coordinates; path length is the nonnegative distance travelled along the route.

Lesson summary

Specify the particle, observers, frame origins, shared axes, and positive directions.
For translating frames with parallel axes, add origin motion and relative motion as vectors.
Differentiate the position relation to obtain velocity and acceleration relations.
Use signs or vector components consistently, then check units, direction, and simple limiting cases.

Check your understanding

Question 1

A cart moves left at

3\,\mathrm{m/s}

. A rider moves left at

2\,\mathrm{m/s}

relative to the cart. What is the rider’s velocity relative to the ground? Take right as positive.

$-5\,\mathrm{m/s}$
$-1\,\mathrm{m/s}$
$1\,\mathrm{m/s}$
$5\,\mathrm{m/s}$

Show answer and explanation

-5\,\mathrm{m/s}

The cart velocity is

-3\,\mathrm{m/s}

and the rider’s relative velocity is

-2\,\mathrm{m/s}

. Their signed sum is

-5\,\mathrm{m/s}

, so the rider moves left.

Question 2

A translating frame has acceleration

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

right. A particle has acceleration

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

left relative to that frame. What is its acceleration in the fixed frame?

$1.4\,\mathrm{m/s^2}$ right
$0\,\mathrm{m/s^2}$
$0.7\,\mathrm{m/s^2}$ left
$1.4\,\mathrm{m/s^2}$ left

Show answer and explanation

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

The equal and opposite signed acceleration components add to zero.

Key terms

Reference frame: A chosen origin and set of coordinate directions used by an observer to describe motion.
Translating frame: A frame whose origin moves while its axes keep the same directions.
Relative position: A particle’s position measured from a specified origin, which may be fixed or moving.
Relative velocity: The time rate of change of a particle’s position as measured in a particular frame.

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2.5 · Relate positions, velocities, and accelerations between translating frames

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 Engineering Mechanics: Dynamics — Study topic 2.5

What you will learn

1. Define the frames and position vectors

2. Relate velocity and acceleration

3. A reliable solution method

4. Apply the relations in one and two dimensions

A person walking inside a moving train

Position and velocity from a moving cart

Acceleration observed from an accelerating vehicle

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