1 lesson 4.2.3 distance, speed and time distance, speed and time
TRANSCRIPT
1
Lesson 4.2.3Lesson 4.2.3
Distance, Speed and Time
Distance, Speed and Time
2
Lesson
4.2.3Distance, Speed and TimeDistance, Speed and Time
California Standard:Algebra and Functions 4.2Solve multistep problems involving rate, average speed, distance, and time or a direct variation.
What it means for you:You’ll learn the formula for speed, and how to use it to solve problems.
Key words:• speed• distance• time• formula
3
Distance, Speed and TimeDistance, Speed and TimeLesson
4.2.3
Speed is a rate — it’s the distance you travel per unit of time.
There’s a formula that links speed, distance, and time — and you’re going to use it in this Lesson.
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.
4
Distance, Speed and TimeDistance, Speed and Time
Speed is a Rate
Lesson
4.2.3
Speed is a rate. It is the distance traveled in a certain amount of time.
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
timespeed =
2 hours10 miles per hour
20 miles
5
Distance, Speed and TimeDistance, Speed and Time
Example 1
Solution follows…
Lesson
4.2.3
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.
Gila’s average speed was 0.75 miles per hour.
distancetime
speed = =6 miles8 hours
= (6 ÷ 8) miles per hour = 0.75 miles per hour
6
Distance, Speed and TimeDistance, Speed and Time
Rearrange the Equation to Find Other Unknowns
Lesson
4.2.3
You can rearrange the speed formula, and use it to find distance or 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.
distancetime
speed =
distance × timetimespeed × time = distance = speed × time
You can find the equation for time in terms of speed and distance in a similar way.
7
Distance, Speed and TimeDistance, Speed and Time
Example 2
Solution follows…
Lesson
4.2.3
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
8
Distance, Speed and TimeDistance, Speed and Time
Example 3
Solution follows…
Lesson
4.2.3
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
Divide both sides by speedspeed × time
speeddistancespeed
=
Solution continues…
9
Distance, Speed and TimeDistance, Speed and Time
Example 3
Lesson
4.2.3
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:
Andy’s walk should take him 5 hours.
distancespeed
time =15 miles3 mph
= = (15 ÷ 3) hours = 5 hours
10
Distance, Speed and TimeDistance, Speed and Time
Guided Practice
Solution follows…
Lesson
4.2.3
1. Juan ran in a marathon that was 26 miles long. If his time was 4 hours, what was his average speed?
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?
3. Monica travels 6 miles to work at a speed of 30 miles per hour. How long does the journey take her each morning?
Speed = distance ÷ time = 26 ÷ 4 = 6.5 miles per hour
Speed = distance ÷ time = 6 ÷ 0.6 = 10 miles per hour
Time = distance ÷ speed = 6 ÷ 30 = 0.2 hours or 12 minutes
11
Distance, Speed and TimeDistance, Speed and Time
Guided Practice
Solution follows…
Lesson
4.2.3
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?
5. How much longer will he take if he continues at the same speed?
Find how far Josh has already walked. Distance = speed × time = 4 × 5 = 20 milesSo Josh has 22 – 20 = 2 miles left to walk
Time = distance ÷ speed = 2 ÷ 4 = 0.5 hours or 30 minutes
12
So her speed for the last hour = x miles per hour.
Distance, Speed and TimeDistance, Speed and Time
Example 4
Solution follows…
Lesson
4.2.3
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?
You need to write an equation using the information given.
Let the cyclist’s speed for the first two hours be (x + 5) mi/h.
Solution
Solution continues…
13
Distance, Speed and TimeDistance, Speed and Time
Example 4
Lesson
4.2.3
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)
Total distance = +distance traveled in first two hours
distance traveled in last hour
distance = speed × time
58 = + (x + 5) × 2 x × 1
Solution continues…
58 = + 2x + 10 x
14
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
Distance, Speed and TimeDistance, Speed and Time
Example 4
Lesson
4.2.3
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)
58 = 2x + 10 + x
58 = 3x + 10
Solve the equation to find x:
48 = 3x x = 16
15
Distance, Speed and TimeDistance, Speed and Time
Guided Practice
Solution follows…
Lesson
4.2.3
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 citiesTrain A speed = d ÷ 3Train B speed = d ÷ 4
Train A speed = 240 ÷ 3 = 80 mi/h, Train B speed = 240 ÷ 4 = 60 mi/h
Train A speed = Train B speed + 20 mi/hd ÷ 3 = (d ÷ 4) + 204d = 3d + 240 d = 240 mi
16
Distance, Speed and TimeDistance, Speed and Time
Independent Practice
Solution follows…
Lesson
4.2.3
1. A mouse ran at a speed of 3 meters per second for 30 seconds. How far did it travel in this time?
2. A slug crawls at 70 inches per hour. How long will it take it to crawl 630 inches?
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?
90 meters
41 miles
9 hours
17
Distance, Speed and TimeDistance, Speed and Time
Independent Practice
Solution follows…
Lesson
4.2.3
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?
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?
Both travel the same distance
50 mi/h
18
Distance, Speed and TimeDistance, Speed and TimeLesson
4.2.3
Round UpRound Up
You need to remember the formula for speed.
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.
distancetime
speed =