University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
7 e’s DETAILED LESSON PLAN
I.
II.
III.
IV.
OBJECTIVES
At the end of the lesson the students must be able to:
1. identify distance
2. solve problems involving distance
3. solve work problems that involves distance, time and rate
4. take and pass the test with mastery level of 75%
CONTENT
A. Topic: More Application of linear Equation
B. Subject Integration: Science
C. Value Focus:
LEARNING RESOURCES
A. Materials Needed: Power Point Presentation, Virtual Photos
B. References: Math Plus 7: Nemenzo & Canonigo, (2020).Math Plus 7: Practical Uses
and Solutions, Nemenzo.C&E Publishing, Inc.pp.245-248.
PROCEDURE
Teachers Activity
A. Introductory Activity
- Greetings
Good afternoon student,
How are you today?
-
Prayer
To formally start our discussion this afternoon, please
stand for our Prayer.
Let us bow down our and put ourselves in the presence
of the Lord
-Energizer
Before you will let us have first a short energizer to
awaken your mind and body before we start the class
Students Activity
Good Afternoon Ma’am
Students will Pray
The student will stand
Our Gracious and
heavenly loving Father,
thank you for this
wonderful day you have
given to us. Thank you for
ways in which you provide
for us all. For Your
protection and love we
thank you. Help us to focus
our hearts and minds now
on what we are about to
learn. Inspire us by Your
Holy Spirit as we listen and
write. Guide us by your
eternal light as we discover
more on ourselves in
learning about the subject.
This is all we ask in Jesus’
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
discussion.
name.
Amen.
Sowing A Dance video
-
Checking of attendance
Who is absent today class?
Okay, Very Good
B. ELICIT
- Reviewing a previous lesson
- Who can recall your previous lesson?
None Ma’am
Student will their Hands
Thank you Very Good
As a review, what is the formula in finding the average?
Very good!
Ma’am lesson last meeting
is about Age Problems
The formula for age
problem is ;
Average = sum of terms
number of terms
C. ENGAGE
GAMES: Longest Line
Group the students into three groups and instruct then
let them stand and form a line.
After that instruct them to not go back to their bags or
sets anymore.
(Activity done as instructed by the teacher)
Rules to follow;
1. Do not use or remove clothes in this activity
2. Use the things that is only available in your body (ex.
bracelets, ballpens,pencils, ties, etc.)
3. The group that has unusual noise will be disqualified
4. And the winner will be the group with the longest line
Do you have any question?
If none let us do the activity and this is called the Longest line.
Thank you for your cooperation class. Let us announce the
winner of this game or activity is the Group _____.
Let us give them a Victory clap
(1, 2,3,4,5) Victory!
How do you find the activity?
What do you call the measure from the start of your line up to
Students will perform
activity
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
the end of the line?
Any idea class?
Okay very good idea.
Ma’am it is called a
distance
Base on your activity how do you define distance or in your
own opinion?
Very Good.
Okay when we talk about distance it is a numerical
measurement of how far apart objects or points are. In Physics
or everyday usage, distance may refer to a physical length or an
estimation based on other criteria
D. EXPLORE
ACTIVITY:
Using the given Illustration and data below, answer the
following situations. Write your answers on a sheet of
paper.
1. Distance of Market to Cecile’s house.
2. Distance from Cecile’s House to Park
3. Distance from Peter’s house to Park
Students will share their
opinion
For answering the activity I will give you 5mins.
After that pass your paper forward.
Now let us answer your activity
Ask 1 student to answer
From number 1 answer we have _______.
Thank you, Very Good.
For number 2 and 3 do the same thing.
Now you already know how to identify the distance
Any question?
If none let us proceed
E. EXPLAIN
This afternoon I will present to you our new topic
Answer:
1. 250 miles
2. 0.4 km
3. 0.5 km
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
which is about Distance Problem. Base on your
previous lesson do you already encounter this topic?
In your other subject you tackle distance in Science
right Distance and the displacement.
But in this subject our focus on this topic is to Solve
distance Problems.
Before that let us first define what is a distance
problem?
Distance Word Problems, which are also called
uniform rate problems, involve the distance traveled by
an object at some rate for a certain period of time.
These problems usually ask how fast an object is
moving, how far an object has traveled, or for how
long an object has been moving.
- These are often called train problems because one of the most
famous types of distance problems involves finding out when
two trains heading toward each other cross paths.
Let’s look at some basic principles that apply to any distance
problem.
There are three basic aspects to movement and travel: distance,
rate, and time. To understand the difference among these,
think about the last time you drove somewhere.
The distance is how far you traveled. The rate is how fast you
traveled. The time is how long the trip took.
The relationship among these things can be described by this
formula:
distance = rate x time
d = rt
In other words, the distance you drove is equal to the rate at
which you drove times the amount of time you drove.
For an example of how this would work in real life, just
imagine your last trip was like this:
You drove 25 miles
—
that's the distance.
You drove an average of 50 mph—that's the rate.
The drive took you 30 minutes, or 0.5 hours—that's the time.
According to the formula, if we multiply the rate and time, the
product should be our distance.
And it is! We drove 50 mph for 0.5 hours—and 50 mph x 0.5
equals 25, which is our distance.
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
What if we drove 60 mph instead of 50? How far could we
drive in 30 minutes?
We could use the same formula to figure this out.
60 x 0.5 is 30, so our distance would be 30 miles.
Solving distance problems
When you solve any distance problem, you'll have to do what
we just did—use the formula to find distance, rate, or time.
The following formula is used when solving distance
word problems:
𝑑 = 𝑟. 𝑡
Where d, is the distance r is the rate, and t is the time.
Let's try another simple problem.
Example 1
On his day off, Lee took a trip to the zoo. He drove an average
speed of 65 mph, and it took him two-and-a-half hours to get
from his house to the zoo. How far is the zoo from his house?
First, we should identify the information we know. Remember,
we're looking for any information about distance, rate, or time.
According to the problem:
The rate is 65 mph.
The time is two-and-a-half hours, or 2.5 hours.
The distance is unknown—it's what we're trying to find.
You could picture Lee's trip with a diagram like this:
This diagram is a start to understanding this problem, but we
still have to figure out what to do with the numbers for
distance, rate, and time. To keep track of the information in the
problem, we'll set up a table. (This might seem excessive now,
but it's a good habit for even simple problems and can make
solving complicated problems much easier.) Here's what our
table looks like:
Distance
Rate
Time
d
65
2.5
We can put this information into our formula:
distance = rate x time.
We can use the distance = rate x time formula to find the
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
distance Lee travelled.
The formula d = rt looks like this when we plug in the numbers
from the problem. The unknown distance is represented with
the variable d.
d = rt
d = 65 x 2.5
d = 162.5
To find d, all we have to do is multiply 65 and 2.5. 65 x 2.5
equals 162.5.
d = 162.5
We have an answer to our problem: d = 162.5.
In other words, the distance Lee drove from his house to the
zoo is 162.5 miles.
Example 2
A bus that is traveling at an average rate of 60 kph
takes 4 hours to make a trip, if it needs to make the trip
in 1 hour less, at what speed should it travel
Solution:
There are two given cases in the problem. Set up a
table showing the rate, the time, and the distance for
each case.
Let x be the speed or rate at which the bus should
travel to make the trip in 1hour less.
Case
r
t
d
60 kph
4
?
1
x kph
3
?
2
Use the formula 𝑑 = 𝑟𝑡 to fill in the third column
Case
1
2
r
t
d
60 kph
4
(60)(4)=240km
x kph
3
3x km
In both cases, the distance is the same. Such
information leads you to the following equation:
3𝑥 = 240
Solving for x you will have x=80.
Hence the bus should increase its speed to 80 kph. To
make trip in only 3 hours.
Is the answer in the previous example reasonable?
Note that when you need to cover the same distance in
a shorter period of time, you should move at a faster
rate.
Solving for rate and time
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
In the problem we just solved we calculated for distance, but
you can use the d = rt formula to solve for rate and time too.
Or we can also have this formula;
Average speed= distance traveled
Time taken
𝑑
𝑟=𝑡
For example, take a look at this problem:
If a car travels 100 miles in 2 hours. What was her average
speed in miles per hour?
We can picture a car travels as something like this:
And we can set up the information from the problem we know
like this:
distance
rate
time
100
r
2hours
As always, we start with our formula. Next, we'll fill in the
formula with the information from our table.
d=rt
100=r.2
r=100
2
r=50mph
r = 50, so 50 is the answer to our problem. A car travel 50
miles per hour.
In the problems on this page, we solved for distance and rate of
travel, but you can also use the travel equation to solve for
time. You can even use it to solve certain problems where
you're trying to figure out the distance, rate, or time of two or
more moving objects.
Example:
If the bus is traveling at 50 mph and the car is traveling at
55mph, in how many hours will they be 210 miles apart?
Now let t=time when they are 210 miles apart.
r
t
bus
50
t
car
50
t
Fill in the values for using the formula d=rt
r
t
bus
50
t
car
50
t
d
d
50t
50t
Since the total distance is 210, we get the equation:
Thomas leaves his house and drives at a speed of 50kph. His
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
brother, Luke, leaves the same house 30 minutes later and
drives the same route at a speed of 60 kph. How many minutes
will it take Luke to catch up with Thomas?
Solution:
Organize the given information in a table. Let x be the time it
take Luke to catch up with Thomas.
Person
Thomas
r
50 kph
t
𝑥 + 0.5 ℎ𝑜𝑢𝑟𝑠
d
50(x+0.5) km
Luke
60 kph
x hours
60x km
In the previous table, why is Thoma’s time represented by𝑥 +
0.5 ?
The illustration maybe illustrated as follows:
In order for Luke to catch up with Thomas he should travel the
same distance as what Thomas has traveled .Such information
leads to the following equation: 50(𝑥 + 0.5) = 60𝑥;
Solve for x as follows;
50(𝑥 + 0.5) = 60𝑥
50𝑥 = 25 = 20𝑥
10𝑥 = 25
𝑥 = 2.5
The problem asks for the number of minutes it will take Luke
to catch up with Thomas; so you need to convert the obtained
of x, which is 2.5 hours, to minutes.
60 𝑚𝑖𝑛.
2.5 ℎ𝑟 ·
= 150𝑚𝑖𝑛
1 ℎ𝑟
Hence, Luke will catch up with Thomas 150 minutes.
Working systematically is important in problem solving. A
table may help you think of one idea at a time without getting
confused by other data. To solve the next types of word
problems, tables would be useful.
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
This is not the same as the simple average of the two given
rates.
F. ELABORATE
To further understand our topic this afternoon always
remember the key concept or the formula in solving the
distance word problems is 𝑑 = 𝑟 ∙ 𝑡
Where d is the distance, r is the rate, and t is time.
While for solving the rate and time we can use the
same formula in finding the distance problem.
Try this;
1. Mr. Rivera headed south, driving his Car 60kph.
After 30 minutes, his son followed driving another
car at a rate of 70kph.
How long will it take Mr. Rivera’s son to overtake
him?
In this problem what are we going to find?
Okay, Very Good.
Now I want you try answering this Problem.
Get a piece of Paper.
Solution:
Given: Mr. Rate =60kph+1/2h
Son rate: 70kph
Representation: Let x be the time travelled of the
son and then x +1/2 with the time travelled of
Mr. Rivera.
Equation:
Distance travelled of Mr. Rivera= Distance
travelled of the son
Solve the equation: 𝑑 = 𝑟𝑡
1
60 (𝑥 + ) = 70𝑥
2
60𝑥 + 30 = 70𝑥
60𝑥 − 70𝑥 = −30
−10𝑥 = −30
−10𝑥 −30
=
−10
−10
𝑥 = 3ℎ
The student who got the correct answer will be given a
reward after.
G. EVALUATION
Direction: Analyze and Solve the following distance
Problem using its formula.
1. Karla is driving 50kph. How far will she travel in 4
Time
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
hours?
Given: rate(50kph)
Time (4hours)
Distance=?
2. Carlo travelled 260km.What was his rate if he
made the trip in 10h.?
Given: distance(260km)
Time: 10 hour
Rate=?
3. You are biking for 30 miles at a rate of 8 miles per
hour. How long have you been biking?
Given : distance(30 miles)
rate(8mph)
time=?
4. You have been in a boat for 8 hours and have
traveled 56 miles. How fast is your boat going?
Given : time(8hours)
Distance(56miles)
rate=?
5. A family’s car trip took 5 hours. If they travelled
50miles each hour, how far did they travel?
Given: time(5hours)
rate(50mph)
distance=?
H. EXTEND
Assignment
Direction: Find the answer to each problem.
1. An express train travel 60kph from Pasay to
monumento. A local train travelling at 40 kph
takes 2hours longer for the same trip. How far
apart are Pasay and monument?
Given : rate(60kph) and (40kph)
Time(2hours)
Distance=?
2. At 6 a.m., two buses leave the same station and
travel in opposite directions. Bus A is traveling
at 80 kph. Bus B is traveling at 65 kph. At
what time will they be 435 km apart? Assume
that both of them are traveling in straight
paths.
Prepared by: RACHEL ANN S. ARZAGA
Student Teacher
Checked by: MRS. ELLEN JOY D. AMAR
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
Cooperating Teacher