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. determine and define the distance problem
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: Cooperation, Honesty/Integrity
LEARNING RESOURCES
A. Materials Needed: Power Point Presentation,Chalk,Chalkboard
B. References: Math Plus 7: Nemenzo & Canonigo, (2020). Math Plus 7: Practical
Uses and Solutions, Canonero Publishing, Inc.pp.245-248.
PROCEDURE
Teachers Activity
Students Activity
A. Introductory Activity
- Greetings
Good afternoon, Ma’am
Good afternoon student,
How are you today?
-
-
Prayer
Before we start our discussion this afternoon, please stand for our
Prayer.
Let us bow down our head and put ourselves in the presence of the
Lord.
None Ma’am
Checking of attendance
Who are absent today class?
Okay, Very Good
P
M
76°
62°
Students will Pray
42°
62°
L
S
42°
A
76°
B
Solutions
B. ELICIT
- Reviewing a previous lesson
- Who can recall your previous lesson?
Thank you Very Good
What is Age Problem?
Example we have;
If you are 12 years old today and your father is 50 years older, then what is
Ma’am our lesson last
meeting is about Age
Problems
It is a problem involving
Age.
12 + 56 = 𝑥 + 50
68 − 50 = 𝑥
𝑥 = 18
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
your age when your father is 56 years old?
So, your age would be 18 years old
We used Ratio and Proportion Property.
C. ENGAGE
GAMES: Longest Line
Group the students into three groups and
let them stand and form a line.
After that instruct them not to go back on 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 which 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!
Students will perform
activity
How do you find the activity?
Exciting and enjoyable
What do you call the measure from the start of your line up to 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
Answer:
1. 250 miles
2. 0.4 km
3. 0.5 km
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
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 on our discussion.
E. EXPLAIN
This afternoon I will present to you our new topic 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.
Today let us determine and 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 distanceis 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
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
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.
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.
Example1
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.
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
We can use the distance = rate x time formula to find the distance Lee traveled.
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
1
2
r
60 kph
x kph
t
4
3
d
?
?
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
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?
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
We can picture a car travel 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
55
t
Fill in the values for using the formula d=rt
r
t
bus
50
t
car
55
t
d
d
50t
55t
Since the total distance is 210, we get the equation:
50t + 55t = 210
105t = 210
Isolate variable t
Answer: They will be 210 miles apart in 2 hours.
Thomas leaves his house and drives at a speed of 50kph. His 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 takes Luke to
catch up with Thomas.
Person
Thomas
r
50 kph
t
Luke
60 kph
x hours
𝑥 + 0.5 ℎ𝑜𝑢𝑟𝑠
d
50(x+0.5) km
60x km
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
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.
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,
𝑑 =𝑟∙𝑡
Where d is the distance, r is the rate, and it 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.
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
We’re going to find the time
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
1
60 (𝑥 + ) = 70𝑥
2
60𝑥 + 30 = 70𝑥
60𝑥 − 70𝑥 = −30
−10𝑥 = −30
−10𝑥 −30
=
−10
−10
𝑥 = 3ℎ
Do you understand now class?
Very Good!
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 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 hours
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.
University of Antique
Tario- Lim Memorial Campus
Laboratory High School
Tibiao, Antique
Prepared by: RACHEL ANN S. ARZAGA
Student Teacher
Checked by: MRS. ELLEN JOY D. AMAR
Cooperating Teacher