Aeronautical Engineering AMT 3102 Module 8: Accelerated Motion

This lecture module, prepared by Engr. Roy Tristan S. Rosales for the Aeronautical Engineering Department at the Philippine State College of Aeronautics, delves into the principles of uniform accelerated motion. The module covers essential topics such as speed, velocity, acceleration, and projectile problems, providing a comprehensive understanding of these concepts. It includes detailed explanations, graphical interpretations, and sample problems to aid in practical application. The module also incorporates learning outcomes aligned with course objectives, encouraging students to apply their knowledge through problem-solving and real-life scenarios. Furthermore, the module references relevant materials, including video and reading resources, and incorporates an honesty clause to promote academic integrity. The content is designed to equip students with the necessary tools to analyze and solve complex engineering mechanics problems related to motion, making it a valuable resource for students studying aeronautical engineering.

PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AERONAUTICAL ENGINEERING DEPARTMENT
Lecture Module 08: Uniform Accelerated Motion
1 | P a g e
LECTURE MODULE
08:
Uniform
Accelerated Motion
AMT 3102 – ENGINEERING
MECHANICS
Prepared by:
ENGR. ROY TRISTAN S. ROSALES

PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AERONAUTICAL ENGINEERING DEPARTMENT
Lecture Module 08: Uniform Accelerated Motion
2 | P a g e
TABLE OF CONTENTS
Title Ref. no Page
1. Uniform Accelerated Motion 1, 2, 3 &
4 5
1.1. Speed 1 & 3 5
1.2. Velocity 1 & 3 6
1.3. Acceleration 1 & 3 6
1.6. Uniformly Accelerated Motion Along a Straight Line 1 & 3 6
1.7. Direction is Important 1 & 3 7
1.8. Instantaneous Velocity 1 & 3 7
1.9. Graphical Interpretations 1 & 3 7
1.10. Acceleration due to Gravity 1 & 3 7
1.11. Velocity Components 1 & 3 8
1.12. Projectile Problems 1 & 3 8
1.13. Sample Problem 8.1. 1 & 5 8
1.14. Sample Problem 8.2. 1 & 5 9
1.15. Sample Problem 8.3. 1 & 5 9
1.16. Sample Problem 8.4. 1 & 5 10
1.17. Enrichment Activity No. 8.1. 1 & 5 11
TABLE OF FIGURES
FIGURE/TABLE REFERENCE
Fig. 8.1. – Fig. 8.2.
Bueche, F.J. & Eugene, H. (1997). Schaums Outlines
College Physic. (9th edition) United States of America:
McGraw-Hill Companies

PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AERONAUTICAL ENGINEERING DEPARTMENT
Lecture Module 08: Uniform Accelerated Motion
3 | P a g e
TABLE OF REFERENCES
References No.
Bueche, F.J. & Eugene, H. (1997). Schaums Outlines College Physic.
(9th edition) United States of America: McGraw-Hill Companies 1
[mcatforme]. (2012, Dec 3). Physic Lecture: Uniform Acceleration
Motion. [Video]. YouTube.
https://www.youtube.com/watch?v=00pRpgfBOjU
2
Meriam, J.L., Kraige L.G., & Bolton J.N. (2015). ENGINEERING
MECHANICS DYNAMICS. (8th Edition). Hoboken, John Wiley & Sons Inc. 3
[The Organic Chemistry Tutor]. (2017, Aug 3). Kinematics In One
Dimension - Distance Velocity and Acceleration – Physics Practice
Problems. [Video]. YouTube.
https://www.youtube.com/watch?v=Po7li9JbEsQ
4
[MATHalino Engineering Mathematics]. (2020). Retrieved From
https://mathalino.com/ 5
TIME COMMITMENT FOR THIS MODULE
Video Materials Time
https://www.youtube.com/watch?v=00pRpgfBOjU 3.5 Minutes
https://www.youtube.com/watch?v=Po7li9JbEsQ 40 Minutes
Reading Materials
Uniform Accelerated Motion 10 Minutes
Speed 5 Minutes
Velocity 5 Minutes
Acceleration 5 Minutes
Uniformly Accelerated Motion Along a Straight Line 5 Minutes
Direction is Important 5 Minutes
Instantaneous Velocity 5 Minutes
Graphical Interpretation 5 Minutes
Acceleration Due to Gravity 5 Minutes
Velocity Components 5 Minutes
Projectile Problems 5 Minutes
Sample Problems 30 Minutes
Activities
Enhancement Activities No. 8.1. 59.2 Minutes
Out of Module Activities
Midterm Plate No. 4 – Uniform Accelerated Motion 100 Minutes
Formative Assessment: Midterm Quiz no. 4 100 Minutes
TOTAL 392.70 Minutes
or 6.545 Hours
HONESTY CLAUSE
As members of the academic community, students are expected to recognize and uphold standards of
intellectual and academic integrity. The state college assumes that as a basic and minimum standard of conduct
in academic matters, the students should be honest and that they submit for credit the products only of their
own efforts.

PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AERONAUTICAL ENGINEERING DEPARTMENT
Lecture Module 08: Uniform Accelerated Motion
4 | P a g e
This module discusses the all about uniform accelerated motion including speed,
velocity, acceleration & projectile. It also discusses all subtopics under it like the
importance of direction, graphical interpretations, acceleration due to gravity and
velocity components. Sample problems are also shown for hands on study.
LEARNING OUTCOME
As an introductory activity, you are encouraged to watch short videos entitled “Physics
lecture: Uniform Acceleration Motion” and “Kinematics In One Dimension - Distance
Velocity and Acceleration – Physics Practice Problems”, using the following YouTube
link: https://www.youtube.com/watch?v=00pRpgfBOjU for the first video and
https://www.youtube.com/watch?v=Po7li9JbEsQ for the latter. The first video is a
three minute long video that seems like a short cut for the formula while the second
video is even longer version but a brief summation of this lesson.
Course Learning Outcomes [CLO]
CLO 01. Apply the knowledge of
mathematics through solving
complex mechanics problems.
CLO 02. Utilize scientific concepts,
laws and theories in solving
problems by relating it to the field of
aircraft maintenance and technology
CLO 08. Analyzing problems
regarding motions of celestial bodies
and motions of objects on earth by
applying Newton’s Law..
CLO 09. Develop sophisticated
mental models through drawings and
diagrams that will serve to describe
the motion of real-world objects.
CLO 10. Demonstrate honesty
through doing individual/group task
required for this course.
Module Learning Outcomes [MLO]
MLO 01. Define & Solve problems
regarding motion in one dimension
through application in practice
problems.
MLO 2. Define & Solve problems
regarding motion in two dimensions
through application in practice
problems.
Topic Learning Outcomes [TLO]
TLO 08. Analyze the application
uniform accelerated motion by
correlating it in real life situations.

PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AERONAUTICAL ENGINEERING DEPARTMENT
Lecture Module 08: Uniform Accelerated Motion
5 | P a g e
UNIFORMLY ACCELERATED MOTION
Since we are going to start tackling the DYNAMICS part of engineering
mechanics at this point onwards, A brief analysis of uniformly accelerating motion may
be in order before studying motion in a resisting medium, since dynamics is the branch
of mechanics which deals with the study of bodies in motion.
In this study, we assume that the motion has a resistance of zero. Any formulas
develop for motion in a resisting medium must first go to the formulas for uniformly
accelerated motion as resistance approaches zero.
Anyone can ask in a situation in which a body starts with speed 𝑣𝑜 and then
accelerates at a rate 𝑎 one of this three questions:
How fast is it moving after time 𝑡?
How far has it moved in time 𝑡?
How fast is it moving after it has covered a distance 𝑥?
Answers to these three questions are well known to any physics students:
𝑣 = 𝑣𝑜 + 𝑎𝑡
𝑣 = 𝑣𝑜𝑡 +1
2𝑎𝑡2
𝑣2 = 𝑣0
2 + 2𝑎𝑥
The following terminologies are a review of all of that equation under uniform
accelerated motion and must be recalled before really tackling dynamics.
SPEED is a scalar quantity that, if an object takes a time interval t to travel a distance
I. then we can get
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑 =
𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑒𝑑
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
or
𝑣𝑎𝑣 = 𝐼
𝑡
As we can see the distance is the total (along-the-path) length traveled. This is how
the car's odometer reads average speeds.

PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AERONAUTICAL ENGINEERING DEPARTMENT
Lecture Module 08: Uniform Accelerated Motion
6 | P a g e
VELOCITY is a vector quantity that if a thing undergoes a vector displacement
denoted as 𝒔⃗ in a time interval t we can get
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝑣𝑒𝑐𝑡𝑜𝑟 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝑣𝑎𝑣 = 𝑠
𝑡
The velocity vector 's direction is the same as that of the displacement vector. The
units of speed (and velocity) are those distance divided by time, such as m/s or km/ h.
ACCELERATION calculates the time rate-of-change of velocity:
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑣𝑒𝑐𝑡𝑜𝑟
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝑎𝑎𝑣 = 𝑣𝑓 − 𝑣𝑖
𝑡
In which 𝑣𝑖 is initial velocity, 𝑣𝑓 is final velocity, and t is the time interval in which change
occurred. Velocity divided by time are the units of acceleration. Particular examples
are (m/s)/s (or m/s 2) and (km/h)/s (or km/h•s). Observe that acceleration is a vector
quantity. It has the direction of the change of velocity𝑣𝑓 − 𝑣𝑖. It is nevertheless
commonplace to speak of the magnitude of the acceleration as just the acceleration,
provided there is no ambiguity.
UNIFORMLY ACCELERATED MOTION ALONG A STRAIGHT LINE is an important
condition. The acceleration vector is constant in this case and lies along the line of the
displacement vector, so that the directions of 𝒗⃗⃗ and 𝒂⃗⃗ may be designated with plus
and minus signs. Representing the displacement by s (plus if in the positive direction,
and minus if in the negative direction), then motion can then be described with this five
equations for uniformly accelerated motion:
𝑆 = 𝑣𝑎𝑣𝑡
𝑣𝑎𝑣 = 𝑣𝑓 − 𝑣𝑖
2
𝑎 = 𝑣𝑓 − 𝑣𝑖
𝑡
𝑣𝑓
2 = 𝑣𝑖
2 + 2𝑎𝑠
𝑠 = 𝑣𝑖𝑡 +1
2𝑎𝑡2

PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AERONAUTICAL ENGINEERING DEPARTMENT
Lecture Module 08: Uniform Accelerated Motion
7 | P a g e
Sometimes 𝑣𝑓 and 𝑣𝑖 are written as 𝑣 and 𝑣𝑜, and often s is replaced by x or y,
respectively.
DIRECTION IS IMPORTANT, and when evaluating motion along a line, a positive
direction must be chosen. Either direction can be chosen as positive. It must be taken
as negative if a displacement, velocity, or acceleration is in the opposite direction.
INSTANTANEOUS VELOCITY is also the average velocity that is evaluated for a time
interval that approaches zero. Therefore, if an object undergoes a displacement
denoted as ∆𝑠in a time denoted as ∆𝑡, then for that object the instantaneous velocity
is as follows
𝑣 = lim
∆𝑡→0
∆𝑠
∆𝑡
In which the notation means that the ratio ∆𝑠/∆𝑡is evaluated for a time interval ∆𝑡that
approaches zero.
GRAPHICAL INTERPRETATIONS for motion along a straight line (the x-axis) are as
given:
• For a certain time, an object's instantaneous velocity is the slope of the
displacement versus time graph at that time. It can be zero, positive, or
negative.
• The slope of the velocity versus time graph at that time is the instantaneous
acceleration of an object at a certain time.
• The x-versus-t graph is a straight line for constant velocity motion while the v-
versus-t graph is a straight line for constant-acceleration motion.
• In general (i.e., one-, two-, or three-dimensional motion) the slope at any
moment of the distance-versus-time graph is the speed.
ACCELERATION DUE TO GRAVITY ( 𝒈): the gravitational (or free-fall) acceleration
(g), which is focused vertically downward, is the acceleration of a body moving only
under the force of gravity. Earth has a gravity of 𝑔 = 9.81 m/s2 (i.e., 32.2 ft/s2); and the
values varies a little from place to place. For added fact, the moon has a free-fall
acceleration of 1.6 m/s2.

PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AERONAUTICAL ENGINEERING DEPARTMENT
Lecture Module 08: Uniform Accelerated Motion
8 | P a g e
VELOCITY COMPONENTS: Suppose that an object travels with a velocity 𝑣 at some
angle 𝜃 up from the x-axis, as we can imagine initially in a case of a ball thrown into
the air. That given velocity then has x and y vector components (see Fig. 8.1) of 𝑣𝑥
and 𝑣𝑦.The scalar components of the corresponding velocity are
𝑣𝑥 = 𝑣𝑐𝑜𝑠𝜃 and 𝑣𝑦 = 𝑣𝑠𝑖𝑛𝜃
and these, depending on 𝜃, can turn out to be positive or negative numbers. As a
regulation, if 𝑣 is on first quadrant, 𝑣𝑥 > 0 and 𝑣𝑦 > 0; if 𝑣 is on second quadrant, 𝑣𝑥 <
0 and 𝑣𝑦 > 0; if 𝑣 is on third quadrant, 𝑣𝑥 < 0 and 𝑣𝑦 < 0; finally, if 𝑣 is on fourth
quadrant, 𝑣𝑥 > 0 and 𝑣𝑦 < 0. It is common to refer to them as velocities because these
quantities have signs, and thus implied directions along known axes. In many texts,
the reader will discover this usage, but it is not without pedagogical drawbacks.
Instead, to anything but a vector quantity (written in boldface with an arrow above)
whose direction is explicitly stated, we shall avoid applying the term velocity.
Therefore, for an object travelling at a velocity 𝑣 = 100 m/s—WEST, the scalar value of
the velocity along the x-axis is 𝑣𝑥 = -100 m/s; and the (always positive) speed is 𝑣
=100 m/s.
PROJECTILE PROBLEMS can easily be solved if air friction can be ignored. We
should just simply consider the motion to consist of two independent parts: horizontal
motion with 𝑎 = 0 and 𝑣𝑓 = 𝑣𝑖 = 𝑣𝑎𝑣 (i.e., constant speed), and vertical motion with 𝑎 =
𝑔 =9.81 m/s2 downward.
Sample Problem No. 1
In a time of 25 s, a runner does one lap around a 200-m track. Determine the
runner's (a) average speed and (b) average velocity?
Solution:
(a) Again from the definition.
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑒𝑑
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 = 200 𝑚
25 𝑠 = 8.0𝑚/𝑠 Ans.
(b) The displacement vector from starting point to end point has zero length since the
run ended at the starting point.. Since 𝑣𝑎𝑣 = 𝑠 /𝑡,
|𝑣𝑎𝑣| = 0 𝑚
25 𝑠= 0 𝑚/𝑠 Ans.

PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AERONAUTICAL ENGINEERING DEPARTMENT
Lecture Module 08: Uniform Accelerated Motion
9 | P a g e
Sample Problem No. 2
An object starts at rest with a constant acceleration along a straight line of 8.00
m/s2. Formulate the (a) speed at the end of 5.00 s, (b) average speed for the 5-s
interval, and (c) distance traveled in the 5.00 s.
Solution:
The motion for the first 5.00 s is of interest to us. Take the direction of motion
to be the +𝑥-𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛(that is 𝑠 = 𝑥). Given that 𝑣𝑖 = 0, 𝑡 = 5.00 s, and 𝑎 = 8.00 m/s2.
the five motion equations apply because the motion is uniformly accelerated.
(a) 𝑣𝑓𝑥 = 𝑣𝑖𝑥 + 𝑎𝑡 = 0 + (8.00 𝑚/𝑠2)(5.00 𝑠) = 40.0 𝑚/𝑠
(b) 𝑣𝑎𝑣 = 𝑣𝑖𝑥+𝑣𝑓𝑥
2 = 0+40.0
2 𝑚/𝑠 = 20.0𝑚/𝑠
(c) 𝑥 = 𝑣𝑖𝑥𝑡 +1
2 (8.00𝑚
𝑠2)(5.00𝑠)2 = 100 𝑚or 𝑥 = 𝑣𝑎𝑣𝑡 + (20.0
𝑚
𝑠2)(5.00𝑠)2 = 100 𝑚
Sample Problem No. 3
A car is moving in a straight line, and in Fig. 8.1. its odometer readings are
plotted against time. Determine the cars instantaneous speed at points A and B, it’s
average speed, and its acceleration.
Figure 8.1.

PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AERONAUTICAL ENGINEERING DEPARTMENT
Lecture Module 08: Uniform Accelerated Motion
10 | P a g e
Solution:
Given the speed is presented by the slope ∆𝑥
∆𝑡 of the tangent line, we take a
tangent to the curve at point A. In this case, the curve itself is the tangent line. A see
at the triangle shown at A, we have
∆𝑥
∆𝑡= 4.0 𝑚
8.0 𝑠= 0.50 𝑚/𝑠
At point B and at every other point on the straight-line graph, this is also the speed so
we can say that 𝑎 = 0 and 𝑣𝑥 = 0.50 m/s = 𝑣𝑎𝑣
Sample Problem No. 4
At a height of 50 m above the ground, a ball is dropped from its resting position.
(a) Compute its speed just before it hits the ground (b) Determine the time it takes to
reach the ground?
Solution:
The ball is uniformly accelerated until it reaches the ground, If we assume that
there are no air friction. Its acceleration is downward and is 9.81 m/s 2 because of
gravity. Assuming down as positive, we can get from the trip:
𝑦 = 50.0 𝑚𝑎 = 9.81 𝑚/𝑠2 𝑣𝑖 = 0
(a) 𝑣𝑓𝑦
2 = 𝑣𝑖𝑦
2 + 2𝑎𝑦 = 0 + 2(9.81 𝑚/𝑠2)(50.0 𝑚) = 981 𝑚2/𝑠2
and so 𝑣 𝑓 = 31.3 𝑚/𝑠.
(b) From 𝑎 = 𝑣𝑓𝑦 − 𝑣𝑖𝑦/𝑡,
𝑡 = 𝑣𝑓𝑦+𝑣𝑖𝑦
𝑎 = (31.3−0)𝑚/𝑠
9.81 𝑚/𝑠2 = 3.19 𝑠 Ans.

PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AERONAUTICAL ENGINEERING DEPARTMENT
Lecture Module 08: Uniform Accelerated Motion
11 | P a g e
Enrichment Activity No. 8.1.
Instructions: Solve the problem in a separate clean sheet of paper. Show all solutions
with free body diagrams (F.B.D.) if needed. Pass a PDF picture entitled “Enrichment
Activity No. 8.1” of your work on your google classroom provided per section.
1. Change the speed 0.200 cm/s to units of kilometers per year. (Show solution)
2. A skier starting from rest, slides 9.0 m down a slope in 3.0 s. Compute the time after
starting will the skier acquire a speed of 24 m/s? Assuming that the acceleration is
constant.
3. A vehicle’s speed increase uniformly from 15 km/h in 20 s. Compute for the (a)
average speed, (b) acceleration, and (c) distance traveled. (all must be in units of
meters and seconds.)
4. A truck travelling at speed of 20 m/s starts to slow at a constant rate of 3.0 m/s each
second. Formulate how far it goes before stopping.
5. An object is thrown straight upward having a speed of 20 m/s. It’s caught on its way
down at a point 5.0 m above on where it was thrown. (a) Determine how fast was it
going when its caught? (b) Determine how long did the trip take?
Figure 8.2.
===== ENF OF MODULE NO. 8 =====