1.5 Solve rectilinear motion with variable acceleration | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Particle Kinematics Foundations

Using calculus to connect position, velocity, and acceleration along a straight line

Rectilinear motion is motion along one straight line. The particle may speed up, slow down, or reverse direction; acceleration need not be constant. The key is to describe motion relative to a stated observer and coordinate, then use calculus to connect the quantities that are known to those being sought. In this lesson, the system is a single particle, the observer is fixed to the ground, and the reference frame is treated as inertial. Choose a straight coordinate axis and keep its positive direction fixed throughout. Position gives location on that axis, displacement is a change in position, path length is the total distance travelled, velocity gives the signed rate of position change, and acceleration gives the signed rate of velocity change.

What you will learn

Define a particle’s system, observer, reference frame, coordinate, and positive direction.
Use derivatives and integrals to relate position, velocity, and variable acceleration.
Choose an integration route that matches the given information and unknown.
Apply initial conditions, units, signs, and physical checks to a rectilinear-motion solution.

1. Set up the motion and its signs

Let

x

measure position along the line, with positive direction chosen before solving. The particle’s position is

x(t)

. Its velocity is

v(t)=dx/dt

, and its acceleration is

a(t)=dv/dt=d^2x/dt^2

. A positive velocity means motion in the positive direction; a negative velocity means motion in the opposite direction. Acceleration’s sign describes how velocity changes, not simply whether the particle is moving forward or backward.

The change in position between times

t_1

and

t_2

is displacement,

x(t_2)-x(t_1)

. It may be negative. Path length is nonnegative and counts all travel, including any reversal. If the velocity changes sign, split the interval at the reversal when calculating path length.

v=\frac{dx}{dt},\quad a=\frac{dv}{dt}=\frac{d^2x}{dt^2}

State the observer, frame, coordinate origin, and positive direction.
Keep signed velocity and acceleration distinct from nonnegative path length.

2. Choose the integration route

Acceleration may be specified as a function of time, position, or velocity. Match the given form to a derivative relationship rather than forcing every problem into the same method. When acceleration is given as a function of time, integrate acceleration over time to obtain velocity, then integrate velocity to obtain position.

If acceleration is given as a function of position, use the chain rule:

a=(dv/dt)=(dv/dx)(dx/dt)=v(dv/dx)

. This creates an equation that can be integrated between known and unknown speeds and positions. If acceleration is given as a function of velocity, use

a=dv/dt

to find time as a function of velocity, or use

dt=dv/a(v)

when the goal is elapsed time.

Each integration introduces a constant. Initial conditions determine those constants: for example,

x(0)=x_0

and

v(0)=v_0

. Definite integrals are often convenient because they apply the endpoint data directly.

a(t)=\frac{dv}{dt},\quad a(x)=v\frac{dv}{dx},\quad a(v)=\frac{dv}{dt}

Use $a(t)$ with time integration, $a(x)$ with the chain rule, and $a(v)$ with velocity integration.
Use the stated initial position and velocity; do not assume the particle starts at the origin or at rest.

3. Solve, interpret, and check

Before integrating, list the known function, initial state, requested quantity, and interval. Check dimensions: acceleration has units of metres per second squared, velocity metres per second, and position metres. An integration constant must have units compatible with the terms it accompanies.

A negative result is not automatically an error; interpret it using the chosen positive direction. Check that the result satisfies the initial conditions and that differentiating position recovers velocity and acceleration. For a useful limiting check, a constant acceleration function should reduce to the familiar constant-acceleration relations.

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

Check initial conditions, units, signs, and whether a direction reversal occurs.
For path length, account for every interval of motion rather than using displacement alone.

Worked example

Acceleration given as a function of time

A cart moves on a straight horizontal track. At

t=0

, it is at

x=2.0\,\mathrm{m}

and has velocity

v=1.0\,\mathrm{m/s}

. Its acceleration is

a(t)=3.0t\,\mathrm{m/s^2}

, with

t

in seconds. Find its velocity and position at

t=2.0\,\mathrm{s}

.

Cart on a straight track

The ground-fixed observer measures position along the track.

Define the state
Take the cart as the system and a ground-fixed observer as the reference. Set the track coordinate positive to the right. The initial state is $x_0=2.0\,\mathrm{m}$ and $v_0=1.0\,\mathrm{m/s}$ ; the final time is $2.0\,\mathrm{s}$ .
Integrate acceleration
Since acceleration is specified as a function of time, integrate from the initial time and apply the initial velocity.
$v(t)=1.0+1.5t^2\;\mathrm{m/s}$
Integrate velocity
Integrate the velocity and use the initial position to set the position constant.
$x(t)=2.0+1.0t+0.5t^3\;\mathrm{m}$
Evaluate the final state
At $t=2.0\,\mathrm{s}$ , the cart has positive velocity and remains in the positive direction.
$v=7.0\,\mathrm{m/s},\quad x=8.0\,\mathrm{m}$

Answer: At

2.0\,\mathrm{s}

, the velocity is

7.0\,\mathrm{m/s}

in the positive direction and the position is

8.0\,\mathrm{m}

.

Check: At

t=0

, the expressions give the stated initial position and velocity. Differentiating position gives

1.0+1.5t^2

, and differentiating again gives

3.0t

. The units of each term in position and velocity are consistent.

Worked example

Acceleration given as a function of position

A slider moves along a straight guide, with positive

x

to the right. At

x=0

, its velocity is

6.0\,\mathrm{m/s}

. Over the region considered, its acceleration is

a(x)=-2.0\,\mathrm{s^{-2}}x

. Find its speed at

x=4.0\,\mathrm{m}

and the time taken to reach that position.

Slider along a straight guide

The slider travels to increasing position over the stated interval.

Define knowns and unknowns
The slider is the system, observed from a ground-fixed frame. Position increases to the right. The known initial state is $x_0=0$ and $v_0=6.0\,\mathrm{m/s}$ . We seek the velocity at $x=4.0\,\mathrm{m}$ and elapsed time.
Find velocity from position
Because acceleration is given as a function of position, use $a=v\,dv/dx$ . Integrate from the initial position and speed.
$v^2=36-2x^2\;\mathrm{m^2/s^2}$
Evaluate the speed
At $x=4.0\,\mathrm{m}$ , the expression gives a positive squared speed. Since the slider is moving toward increasing $x$ on this interval, choose the positive root.
$v(4.0\,\mathrm{m})=2.0\,\mathrm{m/s}$
Find elapsed time
Use $v=dx/dt$ and the positive velocity on the interval. Integrating $dt=dx/v$ from $x=0$ to $x=4.0\,\mathrm{m}$ gives the travel time.
$t=\int_0^4\frac{dx}{\sqrt{36-2x^2}}=\frac{1}{\sqrt{2}}\sin^{-1}\!\left(\frac{4\sqrt{2}}{6}\right)\,\mathrm{s}\approx0.870\,\mathrm{s}$

Answer: The slider’s speed at

x=4.0\,\mathrm{m}

is

2.0\,\mathrm{m/s}

, and the elapsed time is approximately

0.870\,\mathrm{s}

.

Check: The units of

v^2

are

\mathrm{m^2/s^2}

because the coefficient multiplying

x^2

has units

\mathrm{s^{-2}}

. The speed remains positive over the interval, and substituting

x=0

recovers

v_0=6.0\,\mathrm{m/s}

. The time integral has units of seconds; its inverse-sine argument is dimensionless.

Worked example

Acceleration given as a function of velocity

A small test trolley moves along a straight line, positive to the right. At

t=0

, its velocity is

2.0\,\mathrm{m/s}

. Its acceleration while it moves is

a(v)=-0.50v\,\mathrm{s^{-1}}

, where

v

is in metres per second. Find its velocity and displacement after

2.0\,\mathrm{s}

.

Trolley on a straight line

Positive velocity indicates motion to the right.

Set the frame and initial state
The trolley is the system; use a ground-fixed observer and a straight coordinate positive to the right. The initial velocity is $v_0=2.0\,\mathrm{m/s}$ . Set the initial position as the displacement reference, so $x(0)=0$ .
Solve for velocity
The acceleration is proportional to velocity, so separate variables in $dv/dt=-0.50v$ and apply the initial velocity.
$v(t)=2.0e^{-0.50t}\;\mathrm{m/s}$
Integrate for displacement
Displacement is the time integral of velocity. Integrate from zero to the requested time; the velocity stays positive, so displacement also increases positively.
$x(t)=4.0\left(1-e^{-0.50t}\right)\,\mathrm{m}$
Evaluate at two seconds
Substitute $t=2.0\,\mathrm{s}$ into both expressions.
$v=0.736\,\mathrm{m/s},\quad x=2.53\,\mathrm{m}$

Answer: After

2.0\,\mathrm{s}

, the trolley’s velocity is approximately

0.736\,\mathrm{m/s}

to the right, and its displacement is approximately

2.53\,\mathrm{m}

to the right.

Check: The exponential argument is dimensionless because

0.50

has units

\mathrm{s^{-1}}

. At

t=0

, the formulas recover the initial velocity and zero displacement. Differentiating the velocity gives

-0.50v

, as specified.

Common mistakes and how to avoid them

Treating acceleration as constant just because a familiar constant-acceleration formula is available.

Correction: Inspect how acceleration depends on time, position, or velocity and integrate the matching relationship.

Choosing a negative velocity root without checking the direction of motion.

Correction: Use initial direction and whether velocity can reach zero on the interval to select the physically consistent sign.

Calling displacement the distance travelled.

Correction: Displacement is signed change in position. Path length adds the distance on each part of the motion, especially if direction reverses.

Dropping an integration constant or assuming the particle starts at the origin or at rest.

Correction: Apply every stated initial condition after integrating.

Lesson summary

Rectilinear motion is described by one coordinate along a straight line, with a fixed positive direction.
The core links are $v=dx/dt$ and $a=dv/dt$ .
Use $a=v\,dv/dx$ when acceleration depends on position; use the given dependence and initial data to choose an efficient integration route.
Check units, signs, initial conditions, and direction before reporting the result.

Check your understanding

Question 1

A particle has

a(t)=4t\,\mathrm{m/s^2}

and initial velocity

v(0)=3\,\mathrm{m/s}

. What is its velocity at

t=2\,\mathrm{s}

?

$7\,\mathrm{m/s}$
$11\,\mathrm{m/s}$
$8\,\mathrm{m/s}$
$19\,\mathrm{m/s}$

Show answer and explanation

11\,\mathrm{m/s}

Integrating acceleration gives

v(t)=3+2t^2\,\mathrm{m/s}

. At

2\,\mathrm{s}

this is

11\,\mathrm{m/s}

.

Question 2

When acceleration is specified as

a(x)

, which relation directly connects it to velocity and position?

$a=v\,dv/dx$
$a=x\,dv/dt$
$a=d^2v/dx^2$
$a=dx/dv$

Show answer and explanation

a=v\,dv/dx

The chain rule gives

(dv/dt)=(dv/dx)(dx/dt)=v(dv/dx)

.

Question 3

A particle moves in the positive direction, slows, stops, then moves in the negative direction. How should its path length be found?

Use the final position minus the initial position.
Take the absolute value of its final displacement.
Add the magnitudes of the displacements before and after the stop.
Use only the displacement before it stops.

Show answer and explanation

Add the magnitudes of the displacements before and after the stop.

Path length counts travel in both parts of the motion, while displacement is the signed net change in position.

Key terms

Rectilinear motion: Motion along a straight line, described by one coordinate.
Displacement: Final position minus initial position; it includes direction through its sign.
Path length: The total distance travelled along the motion, without a direction sign.
Initial condition: A known position or velocity at a specified time or position, used to determine integration constants.

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1.5 · Solve rectilinear motion with variable acceleration

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Using calculus to connect position, velocity, and acceleration along a straight line

What you will learn

1. Set up the motion and its signs

2. Choose the integration route

3. Solve, interpret, and check

Acceleration given as a function of time

Acceleration given as a function of position

Acceleration given as a function of velocity

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