1.4 Solve rectilinear motion with constant 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 1.4

Rectilinear motion is motion along a straight line. If acceleration stays constant, velocity changes at a steady rate and position changes in a predictable way. To use the equations correctly, first define the observer and reference frame, choose an origin and positive direction, and identify the initial and final states. This lesson uses a particle model: the object's size and rotation are unimportant to the motion being studied.

What you will learn

Distinguish position, displacement, path length, velocity, and acceleration in one-dimensional motion.
Choose a coordinate axis and use signs consistently for motion along a straight line.
Apply constant-acceleration equations to find position, velocity, time, or acceleration.
Check solutions using initial conditions, units, direction, and another kinematic relationship.

1. Define the motion and its signs

Choose the system as the particle whose motion you are describing. The observer measures it relative to a reference frame, such as the ground. For straight-line motion, choose an origin and a coordinate axis along the line, then state which direction is positive. Position is the signed coordinate measured from the origin. Displacement is the change in position, while path length is the total distance travelled and is never negative.

Velocity describes how position changes with time, so its sign indicates direction of motion. Acceleration describes how velocity changes with time. Its sign alone does not tell you whether the particle is speeding up: velocity and acceleration with matching signs indicate increasing speed, while opposite signs indicate decreasing speed until velocity reaches zero.

Let the initial state occur at time

t_0

and the final state at time

t

. The elapsed time is

\Delta t=t-t_0

. Use

x_0

and

v_0

for initial position and velocity, and

x

and

v

for final position and velocity. The acceleration is

a

. Position, velocity, acceleration, and displacement are signed according to the chosen axis; elapsed time is nonnegative.

\Delta x=x-x_0

Position depends on the chosen origin; displacement does not.
Velocity and acceleration signs follow the chosen positive axis.
Constant acceleration means the same signed acceleration throughout the interval.

2. Build and choose the constant-acceleration model

Acceleration is the rate of change of velocity. When it is constant, integrating with respect to time gives velocity as a linear function of elapsed time. Integrating velocity gives position as a quadratic function of elapsed time. The equations below apply over an interval with constant acceleration.

Use the velocity-time equation when velocity, acceleration, and elapsed time are relevant. Use the position-time equation when position or displacement is needed and time is known. The velocity-squared equation removes time, which can be useful when elapsed time is not given or required. Before substituting, check that the selected equation contains the known quantities and the unknown you seek.

The average velocity over the interval is the mean of the initial and final velocities when acceleration is constant. Thus displacement is average velocity multiplied by elapsed time. The velocity-squared equation determines signed displacement from the signed quotient

(v^2-v_0^2)/(2a)

when

a\ne0

. It does not by itself determine the direction of velocity: a squared velocity loses its sign, so use the stated motion and other information to choose a velocity direction.

v=v_0+a\Delta t,\quad x=x_0+v_0\Delta t+\tfrac12a(\Delta t)^2,\quad v^2=v_0^2+2a(x-x_0)

Use signed displacement in the position equations, not total path length.
The velocity-squared relation does not specify velocity direction on its own.
Keep the same coordinate direction throughout a solution.

3. Solve and verify

List the knowns and unknowns, including their units. Choose the positive direction and assign signs before inserting values. A known acceleration opposite to the positive direction is negative; a velocity in the positive direction is positive. Solve symbolically first so the sign logic remains visible.

If you take a square root after using the velocity-squared equation, there may be two possible velocity signs. Choose the one consistent with the stated motion and the interval being considered. If a particle reverses direction, distinguish total path length from signed displacement; the simple average-velocity relation gives displacement over the whole constant-acceleration interval.

Check dimensions: acceleration times time has units of velocity, and acceleration times time squared has units of length. Check initial conditions by setting elapsed time to zero. At zero acceleration, the equations must reduce to uniform motion. When possible, verify a result with another applicable relationship and check that its sign agrees with the stated direction.

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

Record signs before substituting numbers.
Check dimensions, direction, and initial conditions.
At zero acceleration, velocity is constant and position changes linearly with time.

Worked example

A vehicle gains speed on a straight track

A vehicle starts from rest at a marker and moves along a straight track with constant acceleration of

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

for

6.0\,\mathrm{s}

. Find its final velocity and displacement from the marker.

Vehicle on straight track

Ground frame; the vehicle moves to the right. The labelled velocity vector shows its direction of motion.

Define frame and knowns
Take the vehicle as the system and the ground as the reference frame. Put the origin at the starting marker and choose forward, to the right, as positive. Rest means the initial velocity is zero; the final state is $6.0\,\mathrm{s}$ later.
$x_0=0,\quad v_0=0,\quad a=2.4\,\mathrm{m/s^2},\quad \Delta t=6.0\,\mathrm{s}$
Find final velocity
Acceleration and elapsed time are known, so use the velocity-time relation. The positive result indicates motion in the chosen positive direction.
$v=v_0+a\Delta t=0+(2.4)(6.0)=14.4\,\mathrm{m/s}$
Find displacement
Use the position-time relation. Since the vehicle moves forward throughout this interval, its displacement is positive.
$\Delta x=v_0\Delta t+\tfrac12a(\Delta t)^2=0+\tfrac12(2.4)(6.0)^2=43.2\,\mathrm{m}$
Check
The mean of the initial and final velocities is $7.2\,\mathrm{m/s}$ . Multiplying by elapsed time gives the same displacement. The units also agree: acceleration times time is velocity, and acceleration times time squared is length.
$\Delta x=\tfrac12(v_0+v)\Delta t=43.2\,\mathrm{m}$

Answer: Final velocity:

14.4\,\mathrm{m/s}

forward. Displacement:

43.2\,\mathrm{m}

forward.

Check: At zero elapsed time, the equations recover zero displacement and zero velocity, as required by the initial state.

Worked example

A cart brakes to a stop

A cart moves along a straight guide at

18.0\,\mathrm{m/s}

and undergoes constant acceleration of magnitude

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

opposite its motion. Find its stopping time and displacement before it stops.

Cart braking on guide

The cart initially moves right; its acceleration points opposite the positive direction.

Set signs and states
Take the cart as the system and the guide as the reference frame. Choose its initial direction of travel as positive. The acceleration points opposite that direction, and the final state is the instant the cart stops.
$v_0=18.0\,\mathrm{m/s},\quad v=0,\quad a=-3.0\,\mathrm{m/s^2}$
Find stopping time
The velocity-time equation relates the known initial and stopping velocities to acceleration and elapsed time. Solving gives a positive duration.
$\Delta t=\frac{v-v_0}{a}=\frac{0-18.0}{-3.0}=6.0\,\mathrm{s}$
Find displacement
The velocity-squared equation gives signed displacement without requiring time. The quotient is positive here, so displacement is in the initial direction. The equation does not, by itself, determine velocity direction; the stated stopping motion does.
$\Delta x=\frac{v^2-v_0^2}{2a}=\frac{0-(18.0)^2}{2(-3.0)}=54.0\,\mathrm{m}$
Check
The average velocity during this interval is $9.0\,\mathrm{m/s}$ . Multiplying by $6.0\,\mathrm{s}$ confirms the displacement. The stopping time is positive, and the cart moves forward until it stops.
$\Delta x=\tfrac12(v_0+v)\Delta t=54.0\,\mathrm{m}$

Answer: Stopping time:

6.0\,\mathrm{s}

. Displacement before stopping:

54.0\,\mathrm{m}

in the initial direction.

Check: Negative acceleration reduces the cart's positive velocity to zero; it does not make the displacement negative.

Worked example

A particle starts at a negative position

A particle is at position

-12.0\,\mathrm{m}

on a straight axis at time zero. Its velocity is

4.0\,\mathrm{m/s}

in the positive direction, and its constant acceleration is

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

. Find its position and velocity after

5.0\,\mathrm{s}

.

Particle on straight axis

The particle starts left of the origin and moves in the positive direction; its velocity and acceleration point right.

Define frame and initial state
Take the particle as the system and use a fixed straight-axis reference frame. The origin is fixed, and right is positive. Position is initially negative, while velocity and acceleration are positive.
$x_0=-12.0\,\mathrm{m},\quad v_0=4.0\,\mathrm{m/s},\quad a=1.5\,\mathrm{m/s^2},\quad \Delta t=5.0\,\mathrm{s}$
Calculate velocity
Apply the velocity-time equation. The positive velocity change adds to the positive initial velocity.
$v=v_0+a\Delta t=4.0+(1.5)(5.0)=11.5\,\mathrm{m/s}$
Calculate position
Position equals initial position plus signed displacement. The negative initial coordinate does not imply a negative displacement.
$x=x_0+v_0\Delta t+\tfrac12a(\Delta t)^2=-12.0+20.0+18.75=26.75\,\mathrm{m}$
Check displacement
Subtract initial position from final position to find displacement. The average-velocity relation gives the same value, and the positive sign matches the direction of travel.
$x-x_0=26.75-(-12.0)=38.75\,\mathrm{m}=\tfrac12(4.0+11.5)(5.0)\,\mathrm{s}$

Answer: After

5.0\,\mathrm{s}

, the particle is at

26.75\,\mathrm{m}

and has velocity

11.5\,\mathrm{m/s}

in the positive direction.

Check: Its positive displacement is greater than the magnitude of its initial negative position, so it has crossed the origin.

Common mistakes and how to avoid them

Treating negative acceleration as proof that a particle is slowing down.

Correction: Compare velocity and acceleration signs. Opposite signs mean decreasing speed; matching signs mean increasing speed while the velocity keeps its sign.

Using path length where an equation requires signed displacement.

Correction: Find displacement from final position minus initial position. If motion reverses, calculate path length separately.

Assuming position must be positive because the particle moves in the positive direction.

Correction: Position depends on the origin; a particle can move positively while its coordinate remains negative.

Taking a square root from the velocity-squared equation without considering direction.

Correction: Use the stated motion and other known information to choose the physically consistent velocity sign. The equation gives signed displacement through its quotient, not velocity direction.

Lesson summary

Define the observer, reference frame, origin, and positive direction before assigning signs.
With constant acceleration, velocity changes linearly with time and position quadratically with time.
Use signed displacement in the equations and distinguish it from path length.
Verify units, direction, initial conditions, and—when possible—a second relationship.

Check your understanding

Question 1

An object has positive velocity and negative constant acceleration. What happens to its speed while its velocity remains positive?

It increases.
It decreases.
It remains constant.
Its direction must reverse immediately.

Show answer and explanation

It decreases.

Velocity and acceleration have opposite signs, so acceleration reduces the positive velocity and therefore decreases speed until velocity reaches zero.

Question 2

A particle starts at

x_0=5\,\mathrm{m}

with velocity

-2\,\mathrm{m/s}

and constant acceleration zero. Where is it after

3\,\mathrm{s}

?

$11\,\mathrm{m}$
$-1\,\mathrm{m}$
$-6\,\mathrm{m}$
$5\,\mathrm{m}$

Show answer and explanation

-1\,\mathrm{m}

With zero acceleration, position is initial position plus velocity times elapsed time:

5+(-2)(3)=-1\,\mathrm{m}

.

Question 3

Which quantity is necessarily nonnegative for motion along a straight line?

Position
Displacement
Velocity
Path length

Show answer and explanation

Path length

Path length is total distance travelled. Position, displacement, and velocity can be positive, negative, or zero depending on the origin and chosen positive direction.

Key terms

Rectilinear motion: Motion along a straight line.
Position: A signed coordinate locating a particle relative to a chosen origin.
Displacement: The change in position between two states.
Path length: The total distance travelled, without a direction sign.
Constant acceleration: Acceleration with the same value and direction throughout the time interval.

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1.4 · Solve rectilinear motion with constant acceleration

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 Engineering Mechanics: Dynamics — Study topic 1.4

What you will learn

1. Define the motion and its signs

2. Build and choose the constant-acceleration model

3. Solve and verify

A vehicle gains speed on a straight track

A cart brakes to a stop

A particle starts at a negative position

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