Non-constant Acceleration

Dynamics · FE Reference Handbook section

Dynamics

4 formulas

10 exam-style examples

~53 min

All Dynamics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Non-constant Acceleration within Dynamics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what non-constant acceleration describes physically and when it applies.
State every one of the 4 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight.

Lecture

Why this section exists. Non-constant Acceleration is the part of Dynamics that lets you connect a particle or rigid body moving under known forces to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as kinematics first, then a work-energy or impulse-momentum shortcut. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: non-constant acceleration.
Capstone Studio instructional photograph

Dynamics — Non-constant Acceleration: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a particle or rigid body moving under known forces. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 4 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

When non-constant acceleration, a(t), is considered, the equations for the velocity and displacement may be obtained from
For variable angular acceleration
where τ is the variable of integration

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Stopping distance from kinematics

A truck travelling at 60 mph decelerates at 11.2 ft/s². What distance does it need to stop?

Given

v₀ = 60 mph
a = −11.2 ft/s²
v = 0

Find

Stopping distance d

Start with the thinking

Convert mph to ft/s before anything else: 60 mph = 88 ft/s.
Use the velocity-position relation to avoid solving for time.

Step-by-step solution

Convert
$v_{0} = 60 (1.467) = 88.0 f t / s$
Kinematics
$v^{2} = v_{0}^{2} + 2 a d$
Set v = 0
$0 = 88. 0^{2} - 2 (11.2) d$
Rearrange
$d = 7, 744/22.4$
Result
$d = 346 f t$

Answer: d ≈ 346 ft

Why the other options are there

160 ft (mph used directly)
692 ft (factor of 2 dropped)

Reference: FE Reference Handbook — Dynamics — Rectilinear motion

Example 2

Kinetic energy delivered by a drop hammer

A 3,500 lb pile hammer falls freely 4.0 ft. What is its kinetic energy and impact velocity?

Given

W = 3,500 lb
h = 4.0 ft
g = 32.2 ft/s²

Find

KE at impact and velocity

Start with the thinking

Energy conservation: potential energy converts fully to kinetic in free fall.
Mass in slugs equals W/g.

Step-by-step solution

Energy — KE = W h = 3,500(4.0) = 14,000 ft·lb
Mass
$m = W / g = 3, 500/32.2 = 108.7 s l ug$
Velocity relation
$K E = ½ m v^{2}$
Rearrange
$v = (2 K E / m) = (28, 000/108.7)$
Result
$v = 16.0 f t / s$

Answer: KE = 14,000 ft·lb, v = 16.0 ft/s

Why the other options are there

v = 8.0 ft/s (factor of 2 omitted)
KE = 435 ft·lb (weight divided by g twice)

Reference: FE Reference Handbook — Dynamics — Work and energy

Example 3

Braking distance from constant deceleration — Non-constant Acceleration

A vehicle travelling 45 ft/s decelerates at 7.0 ft/s². Find the speed and distance after 4.5 s, and the total stopping distance.

Given

v₀ = 45 ft/s
a = -7.0 ft/s²
t = 4.5 s

Find

v(t), s(t) and total stopping distance

Start with the thinking

Constant acceleration → the three kinematic equations apply.
Deceleration is negative acceleration; keep the sign.

Figure for Braking distance from constant deceleration — Non-constant Acceleration

Step-by-step solution

Velocity
$v = v_{0} + a t$
Substituting
$v = 45 + (- 7.0) (4.5) = 13.50 f t / s$
Displacement
$s = v_{0} t + ½ a t^{2}$
Substituting
$s = 45 (4.5) + 0.5 (- 7.0) (20.25) = 131.6 f t$
Stop time
$t_{s} t o p = - v_{0} / a = 6.43 s$
Stopping distance
$s = v_{0}^{2} / (2∣ a ∣) = 4 5^{2} / (2 \times 7.0) = 144.6 f t$

Answer: v = 13.5 ft/s, s = 131.6 ft; total stop ≈ 144.6 ft

Why the other options are there

202.5 ft (deceleration ignored)
6.4 ft (stop time reported as distance)

Reference: FE Reference Handbook — Dynamics → Non-constant Acceleration

Example 4

Rectilinear motion in Cartesian coordinates — Non-constant Acceleration

A vehicle moving in a straight line has an initial speed of 12 m/s and a constant acceleration of 3.5 m/s². Using the equations of motion in Cartesian coordinates, find the velocity and position after 12 s, and the speed after travelling 200 m.

Given

v₀ = 12 m/s
a = 3.5 m/s²
t = 12 s

Find

v(t), s(t) and v at s = 200 m

Start with the thinking

With constant acceleration the Cartesian equations of motion apply directly — no integration is needed.
The velocity-position relation avoids solving for time.

Step-by-step solution

Formula
$v = v_{0} + a t$
Substituting
$v = 12 + 3.5 (12) = 54.00 m / s$
Formula
$s = v_{0} t + ½ a t^{2}$
Substituting
$s = 12 (12) + ½ (3.5) (12)^{2} = 396.0 m$
Formula
$v^{2} = v_{0}^{2} + 2 a s$
Substituting
$v^{2} = 1 2^{2} + 2 (3.5) (200) = 1, 544$
Evaluate
$v = 39.29 m / s$

Answer: v = 54.00 m/s, s = 396.0 m; at 200 m, v = 39.29 m/s

Why the other options are there

s = 144.0 m (acceleration term dropped)
v = 42.00 m/s (initial speed dropped)

Reference: FE Reference Handbook — Dynamics → Non-constant Acceleration

Example 5

Braking distance from constant deceleration — Non-constant Acceleration (2)

A vehicle travelling 55 ft/s decelerates at 2.5 ft/s². Find the speed and distance after 4.0 s, and the total stopping distance.

Given

v₀ = 55 ft/s
a = -2.5 ft/s²
t = 4.0 s

Find

v(t), s(t) and total stopping distance

Start with the thinking

Constant acceleration → the three kinematic equations apply.
Deceleration is negative acceleration; keep the sign.

Figure for Braking distance from constant deceleration — Non-constant Acceleration (2)

Step-by-step solution

Velocity
$v = v_{0} + a t$
Substituting
$v = 55 + (- 2.5) (4.0) = 45.00 f t / s$
Displacement
$s = v_{0} t + ½ a t^{2}$
Substituting
$s = 55 (4.0) + 0.5 (- 2.5) (16.00) = 200.0 f t$
Stop time
$t_{s} t o p = - v_{0} / a = 22.00 s$
Stopping distance
$s = v_{0}^{2} / (2∣ a ∣) = 5 5^{2} / (2 \times 2.5) = 605.0 f t$

Answer: v = 45.0 ft/s, s = 200.0 ft; total stop ≈ 605.0 ft

Why the other options are there

220.0 ft (deceleration ignored)
22.0 ft (stop time reported as distance)

Reference: FE Reference Handbook — Dynamics → Non-constant Acceleration

Example 6

Rectilinear motion in Cartesian coordinates — Non-constant Acceleration (2)

Given

v₀ = 12 m/s
a = 3.5 m/s²
t = 12 s

Find

v(t), s(t) and v at s = 200 m

Start with the thinking

With constant acceleration the Cartesian equations of motion apply directly — no integration is needed.
The velocity-position relation avoids solving for time.

Step-by-step solution

Formula
$v = v_{0} + a t$
Substituting
$v = 12 + 3.5 (12) = 54.00 m / s$
Formula
$s = v_{0} t + ½ a t^{2}$
Substituting
$s = 12 (12) + ½ (3.5) (12)^{2} = 396.0 m$
Formula
$v^{2} = v_{0}^{2} + 2 a s$
Substituting
$v^{2} = 1 2^{2} + 2 (3.5) (200) = 1, 544$
Evaluate
$v = 39.29 m / s$

Answer: v = 54.00 m/s, s = 396.0 m; at 200 m, v = 39.29 m/s

Why the other options are there

s = 144.0 m (acceleration term dropped)
v = 42.00 m/s (initial speed dropped)

Reference: FE Reference Handbook — Dynamics → Non-constant Acceleration

Example 7

Braking distance from constant deceleration — Non-constant Acceleration (3)

A vehicle travelling 70 ft/s decelerates at 4.5 ft/s². Find the speed and distance after 4.0 s, and the total stopping distance.

Given

v₀ = 70 ft/s
a = -4.5 ft/s²
t = 4.0 s

Find

v(t), s(t) and total stopping distance

Start with the thinking

Constant acceleration → the three kinematic equations apply.
Deceleration is negative acceleration; keep the sign.

Figure for Braking distance from constant deceleration — Non-constant Acceleration (3)

Step-by-step solution

Velocity
$v = v_{0} + a t$
Substituting
$v = 70 + (- 4.5) (4.0) = 52.00 f t / s$
Displacement
$s = v_{0} t + ½ a t^{2}$
Substituting
$s = 70 (4.0) + 0.5 (- 4.5) (16.00) = 244.0 f t$
Stop time
$t_{s} t o p = - v_{0} / a = 15.56 s$
Stopping distance
$s = v_{0}^{2} / (2∣ a ∣) = 7 0^{2} / (2 \times 4.5) = 544.4 f t$

Answer: v = 52.0 ft/s, s = 244.0 ft; total stop ≈ 544.4 ft

Why the other options are there

280.0 ft (deceleration ignored)
15.6 ft (stop time reported as distance)

Reference: FE Reference Handbook — Dynamics → Non-constant Acceleration

Example 8

Rectilinear motion in Cartesian coordinates — Non-constant Acceleration (3)

A vehicle moving in a straight line has an initial speed of 26 m/s and a constant acceleration of 2.5 m/s². Using the equations of motion in Cartesian coordinates, find the velocity and position after 4 s, and the speed after travelling 200 m.

Given

v₀ = 26 m/s
a = 2.5 m/s²
t = 4 s

Find

v(t), s(t) and v at s = 200 m

Start with the thinking

With constant acceleration the Cartesian equations of motion apply directly — no integration is needed.
The velocity-position relation avoids solving for time.

Step-by-step solution

Formula
$v = v_{0} + a t$
Substituting
$v = 26 + 2.5 (4) = 36.00 m / s$
Formula
$s = v_{0} t + ½ a t^{2}$
Substituting
$s = 26 (4) + ½ (2.5) (4)^{2} = 124.0 m$
Formula
$v^{2} = v_{0}^{2} + 2 a s$
Substituting
$v^{2} = 2 6^{2} + 2 (2.5) (200) = 1, 676$
Evaluate
$v = 40.94 m / s$

Answer: v = 36.00 m/s, s = 124.0 m; at 200 m, v = 40.94 m/s

Why the other options are there

s = 104.0 m (acceleration term dropped)
v = 10.00 m/s (initial speed dropped)

Reference: FE Reference Handbook — Dynamics → Non-constant Acceleration

Example 9

Braking distance from constant deceleration — Non-constant Acceleration (4)

A vehicle travelling 65 ft/s decelerates at 7.0 ft/s². Find the speed and distance after 4.0 s, and the total stopping distance.

Given

v₀ = 65 ft/s
a = -7.0 ft/s²
t = 4.0 s

Find

v(t), s(t) and total stopping distance

Start with the thinking

Constant acceleration → the three kinematic equations apply.
Deceleration is negative acceleration; keep the sign.

Figure for Braking distance from constant deceleration — Non-constant Acceleration (4)

Step-by-step solution

Velocity
$v = v_{0} + a t$
Substituting
$v = 65 + (- 7.0) (4.0) = 37.00 f t / s$
Displacement
$s = v_{0} t + ½ a t^{2}$
Substituting
$s = 65 (4.0) + 0.5 (- 7.0) (16.00) = 204.0 f t$
Stop time
$t_{s} t o p = - v_{0} / a = 9.29 s$
Stopping distance
$s = v_{0}^{2} / (2∣ a ∣) = 6 5^{2} / (2 \times 7.0) = 301.8 f t$

Answer: v = 37.0 ft/s, s = 204.0 ft; total stop ≈ 301.8 ft

Why the other options are there

260.0 ft (deceleration ignored)
9.3 ft (stop time reported as distance)

Reference: FE Reference Handbook — Dynamics → Non-constant Acceleration

Example 10

Rectilinear motion in Cartesian coordinates — Non-constant Acceleration (4)

A vehicle moving in a straight line has an initial speed of 11 m/s and a constant acceleration of 1.0 m/s². Using the equations of motion in Cartesian coordinates, find the velocity and position after 8 s, and the speed after travelling 200 m.

Given

v₀ = 11 m/s
a = 1.0 m/s²
t = 8 s

Find

v(t), s(t) and v at s = 200 m

Start with the thinking

With constant acceleration the Cartesian equations of motion apply directly — no integration is needed.
The velocity-position relation avoids solving for time.

Step-by-step solution

Formula
$v = v_{0} + a t$
Substituting
$v = 11 + 1.0 (8) = 19.00 m / s$
Formula
$s = v_{0} t + ½ a t^{2}$
Substituting
$s = 11 (8) + ½ (1.0) (8)^{2} = 120.0 m$
Formula
$v^{2} = v_{0}^{2} + 2 a s$
Substituting
$v^{2} = 1 1^{2} + 2 (1.0) (200) = 521.0$
Evaluate
$v = 22.83 m / s$

Answer: v = 19.00 m/s, s = 120.0 m; at 200 m, v = 22.83 m/s

Why the other options are there

s = 88.0 m (acceleration term dropped)
v = 8.00 m/s (initial speed dropped)

Reference: FE Reference Handbook — Dynamics → Non-constant Acceleration

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a particle or rigid body moving under known forces, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Non-constant Acceleration contains 4 relations; you must be able to find this page in under 15 seconds.
Exam style: kinematics first, then a work-energy or impulse-momentum shortcut.
Unit rule: g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.