2.1 Analyze particle motion in Cartesian components | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Curvilinear and Relative Motion

Position, velocity, and acceleration resolved along fixed axes

Cartesian components describe motion along fixed perpendicular directions, usually horizontal

x

and vertical

y

. This is useful when an object follows a straight or curved path but its motion is measured along those axes. We model the object as a particle when its size and rotation are not needed for the motion being studied. Each example identifies the observer, frame, axes, and positive directions before calculating. The central idea is to treat each component as one-dimensional motion while keeping the components connected as parts of the same position, velocity, or acceleration vector.

What you will learn

Describe a particle’s position, displacement, velocity, and acceleration using Cartesian components.
Use derivatives and integrals to move between position, velocity, and acceleration.
Apply constant-acceleration equations separately along perpendicular axes.
Check component signs, units, initial conditions, and limiting cases.

1. Define the motion and its Cartesian description

The system is the particle whose motion is being analyzed. The observer makes measurements relative to a reference frame; here, use a frame fixed to the ground unless stated otherwise. Choose a fixed origin and perpendicular

x

and

y

axes. State which way is positive on each axis. Coordinates are measured from the origin, so a negative coordinate means the particle is on the negative side of that axis; it is not an error.

The position vector locates the particle. In a plane, write

\mathbf r=x\mathbf i+y\mathbf j

, where

x

and

y

are signed coordinates and

\mathbf i

and

\mathbf j

point along the positive axes. Displacement is the change in position between two times, not the distance travelled along the path. Path length is the total distance along the actual route and is a scalar.

Velocity describes how position changes with time, and acceleration describes how velocity changes with time. Their components follow the same axes:

\mathbf v=v_x\mathbf i+v_y\mathbf j

and

\mathbf a=a_x\mathbf i+a_y\mathbf j

. A component can be negative while speed or acceleration magnitude is positive. Speed is the magnitude of velocity.

\mathbf r=x\mathbf i+y\mathbf j,\quad \mathbf v=\dot{x}\mathbf i+\dot{y}\mathbf j,\quad \mathbf a=\ddot{x}\mathbf i+\ddot{y}\mathbf j

Use one fixed frame and consistent positive directions throughout a calculation.
Position and displacement are vectors; path length and speed are scalars.
A negative component describes direction relative to the chosen positive axis.

2. Governing relationships in components

Differentiate position with respect to time to obtain velocity, then differentiate velocity to obtain acceleration. The reverse operation is integration: integrating acceleration gives a change in velocity, and integrating velocity gives a change in position. The constants of integration are fixed by the initial state, such as position and velocity at

t=0

.

Because the axes are perpendicular, the

x

and

y

component equations can be handled separately. This does not mean the particle has two unrelated motions; both components occur at the same time and together define its path. For time-varying acceleration, integrate each component using its own function of time.

If acceleration is constant, the familiar one-dimensional kinematic equations apply independently along each axis. For example,

v_x=v_{x0}+a_xt

and

x=x_0+v_{x0}t+\tfrac12a_xt^2

, with matching equations in

y

. These equations require constant acceleration over the interval. If acceleration varies, use integration instead.

To find a vector’s magnitude from its components, use the Pythagorean relationship. The direction can be found from the component ratio, but use the signs of both components to identify the correct quadrant. Attach units: position in metres, velocity in metres per second, and acceleration in metres per second squared.

\mathbf v=\frac{d\mathbf r}{dt},\quad \mathbf a=\frac{d\mathbf v}{dt}

Differentiate component by component; integrate component by component.
Use initial conditions to determine integration constants.
Use constant-acceleration equations only when acceleration is constant.

3. A reliable problem-solving sequence

First define the particle, observer, reference frame, origin, axes, positive directions, and time interval. Next sketch the stated path if a visual helps show directions or initial and final positions. Record known initial conditions and the requested quantities. A motion sketch is not a free-body diagram: it shows motion information, not forces.

Choose coordinates that make the given data easy to express. Resolve an initial velocity into signed components before using the kinematic equations. Write a separate equation for each axis, solve symbolically where practical, and substitute SI units once the relationships are clear.

Finish by checking that the result matches the initial conditions and the stated direction of motion. A velocity component may change sign at a turning point. Check dimensions: differentiating metres by seconds gives metres per second, and differentiating again gives metres per second squared. Useful limiting checks include setting time to zero and setting acceleration to zero.

A clear sign convention prevents direction errors.
The same time variable applies to both component equations.
Check units, initial values, and whether the calculated direction is physically consistent.

Worked example

1. Constant acceleration along a straight path

A small cart is modeled as a particle moving along a straight, horizontal track. Relative to an observer fixed to the track, it starts at

x_0=2.0\ \mathrm{m}

with velocity

v_{x0}=3.0\ \mathrm{m/s}

. Its constant acceleration is

a_x=-0.50\ \mathrm{m/s^2}

. Find its position and velocity after

4.0\ \mathrm{s}

, and determine whether it is still moving in the positive direction.

Straight-track motion

The positive direction is to the right; the acceleration is opposite the initial motion.

Set the frame and data
The system is the cart, observed from the ground-fixed track frame. Put the origin at a fixed point $2.0\ \mathrm{m}$ to the left of the cart’s initial position, so the stated initial coordinate remains $x_0=2.0\ \mathrm{m}$ . Choose right as positive $x$ . The initial state is $x_0=2.0\ \mathrm{m}$ and $v_{x0}=3.0\ \mathrm{m/s}$ ; the final state is at $t=4.0\ \mathrm{s}$ . The motion is one-dimensional, so no $y$ component is needed.
Apply constant-acceleration relations
Acceleration is stated to be constant, so use the one-dimensional kinematic equations. The negative sign means the acceleration points left, against the initial positive velocity.
$v_x=v_{x0}+a_xt,\quad x=x_0+v_{x0}t+\tfrac12a_xt^2$
Substitute and interpret
At four seconds, the velocity remains positive, so the cart is still moving right. Its position is measured from the chosen origin, not from the distance travelled.
$v_x=3.0+(-0.50)(4.0)=1.0\ \mathrm{m/s},\quad x=2.0+(3.0)(4.0)+\tfrac12(-0.50)(4.0)^2=10.0\ \mathrm{m}$

Answer: The cart is at

x=10.0\ \mathrm{m}

and has velocity

v_x=1.0\ \mathrm{m/s}

, so it is still moving in the positive direction.

Check: At

t=0

, the equations return

x_0=2.0\ \mathrm{m}

and

v_{x0}=3.0\ \mathrm{m/s}

. The acceleration term in position has units

(\mathrm{m/s^2})(\mathrm{s^2})=\mathrm{m}

.

Worked example

2. Projectile motion resolved into components

A ball is launched from a platform at

x_0=0

,

y_0=1.5\ \mathrm{m}

with speed

20\ \mathrm{m/s}

at

30^\circ

above the horizontal. Ignore air resistance and use

g=9.81\ \mathrm{m/s^2}

. From a ground-fixed observer’s frame, find its position and velocity at

t=1.0\ \mathrm{s}

. Take right and up as positive.

Ball in flight

Gravity gives a downward acceleration while the ball moves to the right.

Resolve the initial velocity
The system is the ball, and the observer is fixed to the ground. Put the origin directly below the launch point on the platform level, so the initial coordinates are $(0,1.5\ \mathrm{m})$ . Resolve the launch velocity using the given angle above positive $x$ .
$v_{x0}=20\cos 30^\circ=17.32\ \mathrm{m/s},\quad v_{y0}=20\sin 30^\circ=10.0\ \mathrm{m/s}$
Write component accelerations
With air resistance neglected, horizontal acceleration is zero and vertical acceleration is downward. This is a stated model assumption for this example.
$a_x=0,\quad a_y=-9.81\ \mathrm{m/s^2}$
Calculate position and velocity
Use the constant-acceleration equations separately in each direction at the same elapsed time. The vertical position includes the initial platform height.
$x=17.32(1.0)=17.32\ \mathrm{m},\quad y=1.5+10.0(1.0)-\tfrac12(9.81)(1.0)^2=6.60\ \mathrm{m}$
$v_x=17.32\ \mathrm{m/s},\quad v_y=10.0-9.81(1.0)=0.19\ \mathrm{m/s}$

Answer: At

1.0\ \mathrm{s}

, the ball is at approximately

(17.32,6.60)\ \mathrm{m}

and its velocity components are

(17.32,0.19)\ \mathrm{m/s}

. It is still moving upward, but only slightly.

Check: The vertical velocity is close to zero because the time is close to the time needed for gravity to reduce the initial upward component. Position units are metres and velocity units are metres per second.

Worked example

3. Variable acceleration and integration

A particle moves along a straight guide in a ground-fixed frame. At

t=0

, it is at

x_0=1.0\ \mathrm{m}

and has velocity

v_{x0}=2.0\ \mathrm{m/s}

. Its acceleration is

a_x=4t\ \mathrm{m/s^2}

, with time in seconds. Find its velocity and position at

t=2.0\ \mathrm{s}

.

Straight-guide motion

The acceleration varies with time, so integrate rather than use constant-acceleration formulas.

Identify the frame and method
The system is the particle; the observer is fixed to the guide. Choose right as positive $x$ . Since acceleration varies with time, integrate it to obtain velocity, then integrate velocity to obtain position. Apply the initial conditions to determine the constants.
Integrate acceleration
Integrating $4t\ \mathrm{m/s^2}$ over time gives the change in velocity. The constant is fixed by the initial velocity.
$v_x(t)=\int 4t\,dt=2t^2+C,\quad v_x(0)=2.0\ \mathrm{m/s}\Rightarrow C=2.0\ \mathrm{m/s}$
Integrate velocity
Position is found by integrating the resulting velocity. The initial position fixes the position constant.
$x(t)=\int(2t^2+2.0)\,dt=\tfrac23t^3+2.0t+C_x,\quad x(0)=1.0\ \mathrm{m}\Rightarrow C_x=1.0\ \mathrm{m}$
Evaluate at the requested time
Substitute $t=2.0\ \mathrm{s}$ . Both acceleration and velocity are positive over the interval, so the particle moves farther in the positive direction.
$v_x(2.0)=10.0\ \mathrm{m/s},\quad x(2.0)=1.0+\tfrac23(2.0)^3+2.0(2.0)=10.33\ \mathrm{m}$

Answer: At

t=2.0\ \mathrm{s}

, the particle has velocity

10.0\ \mathrm{m/s}

in the positive direction and position approximately

10.33\ \mathrm{m}

.

Check: Differentiating the position expression gives

v_x=2t^2+2.0

, and differentiating that gives

a_x=4t

. At

t=0

, the expressions recover both given initial conditions.

Common mistakes and how to avoid them

Treating displacement as the total distance travelled.

Correction: Displacement is final position minus initial position and includes direction; path length adds the distance along the route.

Using a negative velocity to mean the particle is slowing down.

Correction: The velocity sign gives direction relative to the chosen axis. Whether the particle speeds up depends on the relative signs of velocity and acceleration.

Applying constant-acceleration equations when acceleration changes with time.

Correction: For variable acceleration, integrate each component and use the initial conditions.

Using different elapsed times for horizontal and vertical motion.

Correction: Both components describe the same particle and share the same time variable.

Lesson summary

Describe position, velocity, and acceleration with signed Cartesian components in a stated frame.
Differentiate or integrate each component while applying the initial conditions.
Use constant-acceleration equations only when acceleration is constant.
Check signs, units, initial values, and the physical direction of motion.

Check your understanding

Question 1

A particle has

v_x=-3\ \mathrm{m/s}

and

a_x=-2\ \mathrm{m/s^2}

. What happens to its speed at that instant?

It increases because velocity and acceleration point in the same direction.
It decreases because velocity is negative.
It stays constant because both components are negative.
Its direction must be positive because acceleration is nonzero.

Show answer and explanation

It increases because velocity and acceleration point in the same direction.

Both quantities point in the negative direction, so acceleration increases the magnitude of the negative velocity and therefore increases speed.

Question 2

If

a_x=0

throughout an interval, which statement must be true?

$v_x$ is constant throughout the interval.
$x$ is constant throughout the interval.
The particle has zero speed.
The particle’s path length is zero.

Show answer and explanation

v_x

is constant throughout the interval.

Zero acceleration means the derivative of

v_x

is zero, so the velocity component is constant. Position can still change.

Question 3

A particle starts with

y_0=0

and

v_{y0}=0

, and has constant

a_y=-g

. What is its vertical position after time

t

?

$y=-\tfrac12gt^2$
$y=gt$
$y=-gt$
$y=\tfrac12gt^2$

Show answer and explanation

y=-\tfrac12gt^2

With zero initial vertical position and velocity, the constant-acceleration position equation leaves the term

\tfrac12a_yt^2=-\tfrac12gt^2

.

Key terms

Particle: A model of an object whose size and rotation are not needed for the motion description.
Reference frame: The observer’s chosen basis for measuring position and motion.
Displacement: The vector difference between final and initial position.
Cartesian components: The signed parts of a vector along perpendicular coordinate axes.
Initial condition: A known position or velocity at a specified starting time, used to determine constants in a motion calculation.

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2.1 · Analyze particle motion in Cartesian components

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Position, velocity, and acceleration resolved along fixed axes

What you will learn

1. Define the motion and its Cartesian description

2. Governing relationships in components

3. A reliable problem-solving sequence

1. Constant acceleration along a straight path

2. Projectile motion resolved into components

3. Variable acceleration and integration

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through EN PH 131

About this lesson and its review