Time, distance formula

BoyetAluan 5,460 views 37 slides Jan 24, 2014
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About This Presentation

discussed speed, time and distance.


Slide Content

Speed, Time and Distance
Formulas
Boyet B. Aluan
San Roque Elementary School

What are Time and Distance formulas?
Time and Distance
Formulas relate time,
distance, and speed.
These relationships
have many practical
applications.

Why do you need to know the Speed,
Time and Distance Formula?
To figure how long a trip will
take
To see how far you can go in
a set amount of time
To see how fast you took a trip
To compare different trips

Steps in solving Speed, Time and Distance
Formulas
Step 1. Translate the question
into mathematical terms. For
example, if you are asked miles
per hour, write the question as
miles divided by hours.
Ex m/h or mph

Steps in solving Speed, Time and Distance
Formulas
Step 2. Put everything in
constant units. For example, if
the question asks miles per
hour, every time must be placed
in terms of hours, and every
distance must be placed in
terms of miles.

Conversion Units
Remember: There are 60 seconds in
a minute, 60 minutes in an hour, and
24 hours in a day.
Remember: There are 12 inches in a
foot, 3 feet in a yard, and 5280 feet in
a mile.

Time Conversions
To convert minutes to hours, divide by 60.
To convert hours to minutes, multiply by 60
To convert seconds to minutes, divide by 60
To convert minutes to seconds, multiply by
60
To convert days to hours, multiply by 24
To convert hours to days, divide by 24

Measurement Conversions
To convert inches to feet, divide by 12
To convert feet to inches, multiply by 12
To convert feet to miles, divide by 5280
To convert miles to feet, multiply by 5280

Step 3. Write the equation you want
to solve.
Use the correct formula to solve it.
Be careful about what you multiply,
and what you divide!

What are the basic formulas?
Distance = Rate * Time (d = r * t)
Rate = Distance / Time (r =d/ t)
Time = Distance / Rate (t = d / r)
Be sure you use the right formula!

To find distance
Distance = Rate * Time (d = r * t)
For example, to find miles, multiply miles per
hour (rate) times the number of hours.
Miles = Miles x Hours
Hours
The hours cancel, you are left with miles.

To find the rate of speed
Rate = Distance /Time (r =d/ t)

rate- constant/ average
speed. To find miles per hour,
divide miles driven by the
number of hours driven.

To find the time it takes to travel.
Time = Distance / Rate (t = d / r)
Hours = Miles_____
Miles/Hour
Remember, to divide fractions, flip and multiply
Hours = Miles x Hours
Miles
Miles cancel, you are left with hours

Quiz 1-finding the distance
A girl cycles for 3h at a
speed of 40 km/h. what is
the distance did she
travel?

Step one
Put in constant units
Rate was given in km/h,

Question 1
What formula should be used?
We are looking for distance so use
Answer
A. d=r*t
B. r=d/t
C. t=d/r

Solve
D=rt
D=?
R=40km/h
T=3hrs
Substitute the value
D=rt
D=40km/h(3h)
Cancel h
D=120km

Quiz2 finding the time
A train travels at a speed of
30mph and travel a distance of
240 miles. How long did it take
the train to complete its
journey?

Question 1
What formula should be used?
We are asked to find how long did it
take the train to complete its journey so
used
Answer
A. d=rt
B. r=d/t
C. t=d/r

Solve
T=d/r
T=?
R= 30mph
D=240m
T= 240M/30mph
Cancel m(miles)
T=8h is the time to complete the
journey

Quiz 3
 A car travels a distance
of 540km in 6 hours.
What is the speed did it
travel at?

Question 3
We are asked to find the rate to travel the
distance of 540km in 6 hours. What
formula should be used?
Answer
A. d=rt
B. r=d/t
C. t=d/r

Solution
R=d/t
R=?
D= 540 m
T=6h
Substitute:
R=d/t
R=(540m)/6h
R=90mph

Quiz4
John is a runner. He runs the
100m sprint in 10.6s. What speed
did he travel at?
The unit is (m/s)
Use r=d/t where r=?, t=10.6s,
d=100meter
Substitute
R=(100m)/10.6s
R= 9.43m/s

Quiz 5
At 11:00 am, a car(1) leaves
city “A” at a constant rate of
60m/h toward city “B”. At the
same time a second car(2)
leaves city “B” toward city “A”
at the constant speed of
50mph. The distance between
city A and B is 220miles and
these cities are connected by a
highway used by the two cars.
At what time will the two cars
cross each other?

Solution -construct a tabular
presentation
Speed (R) TIME (T) Distance (D)
Car 1 60mph
car2 50mph
total 110mph ? 220m
T=d/t
T=220m/110mph
T=2h
So at constants speed, cars cross each other at
1:00pm

Quiz 6
Kali left school and traveled toward
her friend’s house at an average
speed of 40km/h. Matt left one hour
later and traveled in the opposite
direction with an average speed of
50km/h. find the number of hours
Matt needs to travel before they are
400km apart.

Solution-construct tabular presentation
1
st
hour2
nd
hour3
rd
hour4
th
hour5
th
hourtotal
Kali
Matt -left
total 40
From the table, Matt needs 4
hours so they can be 400km
apart
40 4040 40 40
50 505050 200
200
400km

Quiz 7
Chelsea left the White House and
traveled toward the capital at an
average speed of 34km/h. Jasmine left
at the same time traveled in the
opposite direction with an average
speed of 65km/h. Find the number of
hours Jasmine needs to travel before
they are 59.4 km apart.

Solution
R=34km/h+64km/h=99km/h
D=59.4km
T=?
T=d/r
T=59.4km/99kmph
T=0.6h
Speed Time Distance
Chelsea 34km/h ?
Jasmin 64km/h ?
total 99km/h ? 59.4km apart

Quiz 8
A train leaves Deb’s house and travels
at 50mph. Two hours later, another
train leaves from Deb’s house on the
track beside or parallel to the first train
but it travels at 100mph. How far away
from Deb’s house will the faster train
pass the other train?

Solution
Since their distance is equal,
Let n-time takes faster train to take the distance
N+2-time takes slower train to cover the distance
Thus
50(n+2)=100n
50n+100=100n-distibutive property
-100n+50n=-100-transposition
-50n=100
(-50n/-50)=-(100/-50)-cancelation/property of sign
numbers
N=2
Speed Time Distance
Slower train 50kmph N+2 50(n+2)
Faster train 100kmph n 100n
total
So means, by substitution
method
=100n
=100(2)
=200km
The faster train is at 200miles
away from slower train.

Quiz 9
A train left Chicago and traveled towards
Dallas. Five hours later another train left
for Dallas traveling at 40mph with a goal
or catching up with the first train bounded
for Dallas. The second train finally caught
up with the first train after traveling for
three hours. How fast was the train that
left first going?

Solution –Completing the table/graph
Since 2
nd
train speed is
40mph travelled for 3h..
D=rt
D=40mph*3h
D=120m
Speed Time distance
1
st
train 15mph 5+3 120miles
2
nd
train 40mph 3 120miles
Thus 1
st
train d=120m
T=3h+5h
R=d/t
R=120m/8h
R=15mph

Quiz 10.Catching up same direction,
has equal distance!
A jet took off Toronto, heading west at a
speed of 405mph. Another jet left for Toronto
from the same airport sometime after the first
jet took off and it was traveling at a speed of
486mph. Ten hours later, the second jet
caught up with the first jet. How long did the
jet fly before the second jet caught up?

Solution
405(n+10)=4860
405n+4050=4860
405n=4860-4050
405n=810
405n/(405)=810/(405)
N= 2
Speed Time Distance
Jet1 405mph n+10 405(n+10)
Jet2 486mph 10h 4860mph
So n+10
=2+10
=12
Time =12h
 1
st
jet took 12h before it
just get caught.

Thanks!!!
Boyet B. Aluan
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