Transcript Distance, Speed and Time Lesson 4.2.3 Lesson 4.2.3 Distance, Speed and Time California Standard: What it means for you: Algebra and Functions 4.2 Solve multistep problems involving rate, average.

Distance, Speed
and Time
Lesson 4.2.3
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Lesson
4.2.3
Distance, Speed and Time
California Standard:
What it means for you:
Algebra and Functions 4.2
Solve multistep problems involving
rate, average speed, distance, and
time or a direct variation.
You’ll learn the formula for
speed, and how to use it to
solve problems.
Key words:
•
•
•
•
speed
distance
time
formula
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Lesson
4.2.3
Distance, Speed and Time
Speed is a rate — it’s the distance you travel per unit of time.
55 miles per hour is the speed limit on some roads. If you
drive steadily at this speed, you’ll travel 55 miles every hour.
There’s a formula that links speed, distance, and time
— and you’re going to use it in this Lesson.
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Lesson
4.2.3
Distance, Speed and Time
Speed is a Rate
Speed is a rate. It is the
distance traveled in a
certain amount of time.
2 hours
10 miles per hour
20 miles
Speed can be measured in lots of different units, such as
miles per hour, meters per second, inches per minute...
The formula for speed is:
distance
speed =
time
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Lesson
Distance, Speed and Time
4.2.3
Example
1
Gila walked 6 miles in 8 hours.
What was Gila’s average speed?
Solution
Use the formula, and substitute
in the values from the question.
distance
6 miles
speed =
=
time
8 hours
= (6 ÷ 8) miles per hour = 0.75 miles per hour
Gila’s average speed was 0.75 miles per hour.
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Solution follows…
Lesson
4.2.3
Distance, Speed and Time
Rearrange the Equation to Find Other Unknowns
You can rearrange the speed formula,
and use it to find distance or time.
distance
speed =
time
To change the equation into an equation that gives
distance in terms of speed and time, multiply both
sides of the equation by time.
distance × time
speed × time =
time
distance = speed × time
You can find the equation for time in terms of speed
and distance in a similar way.
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Lesson
4.2.3
Example
Distance, Speed and Time
2
Alyssa runs for 0.5 hours at a speed of 11 kilometers per hour.
How far does she run?
Solution
Use the formula for distance, and substitute
the values for speed and time.
Distance = speed × time
= 11 kilometers per hour × 0.5 hours
= 5.5 kilometers
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Solution follows…
Lesson
4.2.3
Example
Distance, Speed and Time
3
Andy is planning a walk. He walks at an average speed
of 3 miles per hour, and plans to cover 15 miles.
How long should his walk take him?
Solution
You need to rearrange the speed formula.
distance = speed × time
distance
speed × time
=
speed
speed
Divide both sides by speed
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Solution
Solution
continues…
follows…
Lesson
4.2.3
Example
Distance, Speed and Time
3
Andy is planning a walk. He walks at an average speed
of 3 miles per hour, and plans to cover 15 miles.
How long should his walk take him?
Solution (continued)
Now you can use the formula to answer the question:
distance 15 miles
time =
=
= (15 ÷ 3) hours = 5 hours
speed
3 mph
Andy’s walk should take him 5 hours.
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Lesson
4.2.3
Distance, Speed and Time
Guided Practice
1. Juan ran in a marathon that was 26 miles long.
If his time was 4 hours, what was his average speed?
Speed = distance ÷ time = 26 ÷ 4 = 6.5 miles per hour
2. Moesha goes to school every day by bike.
The journey is 6 miles long, and takes her 0.6 hours.
What is her average speed?
Speed = distance ÷ time = 6 ÷ 0.6 = 10 miles per hour
3. Monica travels 6 miles to work at a speed
of 30 miles per hour.
How long does the journey take her each morning?
Time = distance ÷ speed = 6 ÷ 30 = 0.2 hours or 12 minutes
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Solution follows…
Lesson
4.2.3
Distance, Speed and Time
Guided Practice
Josh has been walking for 5 hours
at a speed of 4 miles per hour.
4. His walk is 22 miles long.
How far does he have left to walk?
Find how far Josh has already walked.
Distance = speed × time = 4 × 5 = 20 miles
So Josh has 22 – 20 = 2 miles left to walk
5. How much longer will he take
if he continues at the same speed?
Time = distance ÷ speed = 2 ÷ 4 = 0.5 hours or 30 minutes
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Solution follows…
Lesson
4.2.3
Example
Distance, Speed and Time
4
On a three-hour bike ride, a cyclist rode 58 miles.
The first two hours were downhill, so the cyclist rode
5 miles per hour quicker than she did for the last hour.
a) What was her speed for the first two hours?
b) What was her speed for the last hour?
Solution
Let the cyclist’s speed for the first two hours be (x + 5) mi/h.
So her speed for the last hour = x miles per hour.
You need to write an equation using the information given.
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Solution
Solution
continues…
follows…
Lesson
4.2.3
Example
Distance, Speed and Time
4
On a three-hour bike ride, a cyclist rode 58 miles.
The first two hours were downhill, so the cyclist rode
5 miles per hour quicker than she did for the last hour.
a) What was her speed for the first two hours?
b) What was her speed for the last hour?
Solution (continued)
distance traveled distance traveled
Total distance =
+
in first two hours
in last hour
58 =
(x + 5) × 2
+
x×1
distance = speed × time
58 =
2x + 10
+
x
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Solution continues…
Lesson
4.2.3
Example
Distance, Speed and Time
4
On a three-hour bike ride, a cyclist rode 58 miles.
The first two hours were downhill, so the cyclist rode
5 miles per hour quicker than she did for the last hour.
a) What was her speed for the first two hours?
b) What was her speed for the last hour?
Solution (continued)
Solve the equation to find x: 58 = 2x + 10 + x
58 = 3x + 10  48 = 3x  x = 16
a) The speed for the first two hours was (x + 5)
= 16 + 5 = 21 mi/h
b) So the speed for the last hour was x = 16 mi/h
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Lesson
4.2.3
Distance, Speed and Time
Guided Practice
6. Train A travels 20 mi/h faster than Train B.
Train A takes 3 hours to go between two cities, and
Train B takes 4 hours to travel the same distance.
How fast does each train travel?
Let d = distance between the two cities
Train A speed = d ÷ 3
Train B speed = d ÷ 4
Train A speed = Train B speed + 20 mi/h
d ÷ 3 = (d ÷ 4) + 20
4d = 3d + 240  d = 240 mi
Train A speed = 240 ÷ 3 = 80 mi/h,
Train B speed = 240 ÷ 4 = 60 mi/h
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Solution follows…
Lesson
Distance, Speed and Time
4.2.3
Independent Practice
1. A mouse ran at a speed of 3 meters per second
for 30 seconds. How far did it travel in this time?
90 meters
2. A slug crawls at 70 inches per hour.
How long will it take it to crawl 630 inches?
9 hours
3. A shark swims at 7 miles per hour for 2 hours, and then at
9 miles per hour for 3 hours. How far does it travel altogether?
41 miles
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Solution follows…
Lesson
4.2.3
Distance, Speed and Time
Independent Practice
4. Bike J moves at a rate of x miles per hour for 2 hours.
Bike K travels at 0.5x miles per hour for 4 hours.
Which bike travels the furthest?
Both travel the same distance
5. On a two-day journey, you travel 500 miles in total.
On the first day you travel for 5 hours at an average speed
of 60 mi/h. On the second day you travel for 4 hours.
What’s your average speed for these 4 hours?
50 mi/h
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Solution follows…
Lesson
4.2.3
Distance, Speed and Time
Round Up
You need to remember the formula for speed.
distance
speed =
time
If you know this, you can rearrange it to figure out the
formulas for distance and time when you need them
— so that’s two less things to remember.
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