Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership
Page 270 Problem 1 Answer
We have formula for breaking distance that is s2/20
Wheres represents the speed of the car.
Breaking distance = s2/20
First factor that affects the breaking distance is “speed of the car”. if speed of the is high then breaking distance will be more.
Second factor that affects the breaking distance is “break intensity” if the break intensity is more then breaking distance will be less.
Third factor that affects the breaking distance is ” average reaction time” if average reaction time is more then breaking distance will be more.
The factors that affects breaking distance are “speed of the car” , “break intensity”, ” average reaction time”.
Page 270 Problem 2 Answer
Given : Speed of the car = 65 miles/h
The reaction distance of a car traveling at s miles per hour is approximated by using a distance of s feet, the formula can be represented by either of the following.
s+(0.1×s)2×5 or s+s2/20 Where s is the speed of the car.
By using this formula we will find the total stopping distance.
Speed of the car = 65 miles/h
Thus total stopping distance is s+s2/20
Put s = 65 miles/h
We get total stopping distance 65+652/20
=276.25 feet
The total stopping distance is 276.25 feet.
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Page 271 Problem 3 Answer
Given: Speed of the car is 78 Km/h
Here we will use the direct formula that is written above and find the total stopping distance in meters.
Total stopping distance in meters =s2/170+s/5
Put s = 78Km/h
We get Total stopping distance in meters =782/170+78/5
Hence Total stopping distance in meters =51.39m
Hence total stopping distance in meters is 51.39 m.
Cengage Financial Algebra Chapter 5 Exercise 5.8 Solution
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership Page 271 Problem 4 Answer
Given: Speed of Toni’s car = 75 Km/h
Speed of Randy’s Car = 72 Km/h
And Toni notices a family of ducks crossing the road 50 meters ahead of her.
Here first we will find the total stopping distance of Toni’s car if it is less than 50m then Toni will be able to stop before she reaches the ducks.
Total stopping distance in meters =s2/170+s/5
Put s = 75 km/h
We get Total stopping distance of Toni′s car in meters =752/170+75/5
=48.09m
Thus total stopping distance of Toni’s car is less than 50m .
Hence Toni will stop the car before she reaches the ducks.
Now total stopping distance of Randy’s car is
Total stopping distance in meters =s2/170+s/5
Put s = 72 Km/h
We get Total stopping distance of Randy’s car =722/170+72/5
=44.89m
Thus the least distance that Randy’s car can be from Toni’s car to avoid hitting her car is 48.09−44.89=3.2m
Yes, Toni will be able to stop before she reaches the ducks.
the least distance that Randy’s car can be from Toni’s car to avoid hitting her car is 3.2m
Page 272 Problem 5 Answer
We have given 5,280 feet in a mile and speed of the car is 65miles/h
Here we have to find the distance traveled by the car in miles.
Speed of the car is 65miles/h
Thus distance traveled by the car in 1 hour is 65 miles.
The car will go 65 miles in one hour because its speed is 65mi/h.
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership Page 272 Problem 6 Answer
We have given 5,280 feet in a mile And speed of the car is 65 miles/h
Here we will find the distance in feet in 1 hour by using the above data.
Speed of the car is 65 miles/h
Thus distance traveled by the car in one hour is = 65 miles.
Because 1 mile = 5280 feet
Thus Distance in feet is 65 miles =5280×65=343200feet
The car will go 343,200 feet in one hour.
Page 272 Problem 7 Answer
We have given that 5,280 feet in a mile
And speed of the car = 65 miles/h
We also know that 1 hour = 60 minutes
By using these results we will find the distance in one minute.
Speed of car = 65 miles/h
Thus distance in 1 hour = 65 miles
From previous result we have
The car will go 343,200 feet in one hour.
Because 1 hour = 60 minutes
Thus distance traveled in 1 minute = 343200÷60=5720
The car will go 5,720 feet in one minute.
Answers To Exercise 5.8 Automobile Ownership Cengage
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership Page 272 Problem 8 Answer
We have 5,280 feet in a mile
1 hour = 60 minutes
1 minute = 60 seconds
And speed of the car is = 65 miles/h
Here we will calculate the distance covered in one second by using the above results.
Speed of the car = 65 miles/h
From previous part we have
The car will go 5,720 feet in one minute.
Thus 1 minute =60 seconds
5720÷60=95.33
The car will go 95.33 feet in one second.
Page 272 Problem 9 Answer
We have given 5,280 feet in a mile and speed of the car is 42 miles/h
Here we have to find the distance traveled by the car in miles.
Speed of the car is 42 miles/h
Thus distance traveled by the car in 1 hour is 42 miles.
The car will go 42 miles in one hour because its speed is 42mi/h.
Page 272 Problem 10 Answer
We have given 5,280 feet in a mile And speed of the car is 42 miles/h
Here we will find the distance in feet in 1 hour by using the above data.
Speed of the car is 42 miles/h
Thus distance traveled by the car in one hour is = 42 miles.
Because 1 mile = 5280 feet
Thus 42 miles =5280×42=221760 feet
The car will go 221,760 feet in one hour.
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership Page 272 Problem 11 Answer
We have given that 5,280 feet in a mile
And speed of the car = 42 miles/h
We also know that 1 hour = 60 minutes
By using these results we will find the distance in one minute.
Speed of car = 42 miles/h
Thus distance in 1 hour = 42 miles
From previous result we have
The car will go 221,760 feet in one hour.
Because 1 hour = 60 minutes
Thus distance traveled in 1 minute =221760÷60=3696
The car will go 3,696 feet in one minute.
Page 272 Problem 12 Answer
We have 5,280 feet in a mile
1 hour = 60 minutes
1 minute = 60 seconds
And speed of the car is = 42 miles/h
Here we will calculate the distance covered in one second by using the above results.
Speed of the car = 42 miles/h
From previous part we have
The car will go 3,696 feet in one minute.
Thus 1 minute =60 seconds
3696÷60=61.6
The car will go 61.6 feet in one second.
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership Page 272 Problem 13 Answer
We have given 5,280 feet in a mile and speed of the car is x miles/h
Here we have to find the distance traveled by the car in miles.
Speed of the car is x miles/h
Thus distance traveled by the car in 1 hour is x miles.
The car will go x miles in one hour because its speed is xmi/h.
Page 272 Problem 14 Answer
We have given 5,280 feet in a mile And speed of the car is x miles/h
Here we will find the distance in feet in 1 hour by using the above data.
Speed of the car is x miles/h
Thus distance traveled by the car in one hour is = x miles.
Because 1 mile = 5280 feet
Thus 1 mile =5,280 feet
42 miles =5280×x=5280x
The car will go 5,280x feet in one hour.
Chapter 5 Automobile Ownership Worksheet Answers Cengage
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership Page 272 Problem 15 Answer
We have given that 5,280 feet in a mile
And speed of the car = x miles/h
We also know that 1 hour = 60 minutes
By using these results we will find the distance in one minute.
Speed of car = x miles/h
Thus distance in 1 hour = x miles
From previous result we have
The car will go 5,280x feet in one hour.
Because 1 hour =60 minutes
Thus distance in one minute is 5,280x/60
=88x feet
The car will go 88x feet in one minute.
Page 272 Problem 16 Answer
We have 5,280 feet in a mile
1 hour = 60 minutes
1 minute = 60 seconds
And speed of the car is = x miles/h
Here we will calculate the distance covered in one second by using the above results.
Speed of the car = 42 miles/h
From previous part we have
The car will go 88x feet in one minute.
Thus 1 minute =60 seconds
Hence The car will go 5,280x
60×60=88x/60 feet in one second.
The car will go 88x/60 feet in one second.
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership Page 272 Problem 17 Answer
Given a car traveling 55 km/h.
Need to find the distance covered by a car traveling 55km/h in kilometers in one hour.
As the distanced is the product of speeds and time t. Thus,d=s×t. Using this, find the required distance covered by the car.
We have,s=55km/h,t=1h
Then, the distance travelled in one hour in km is,
55×1
=55
The distance covered by a car traveling 55 km/h in kilometers in one hour is 55km.
Page 272 Problem 18 Answer
Given a car traveling 55 km/h.
Need to find the distance covered by a car traveling 55 km/h in meters in one hour.
As the distanced is the product of speeds and time t. Thus,d=s×t. Using this, find the required distance covered by the car.
We have,s=55 km/h,t=1h,
There are1000 meters in1km.
Then, the distance travelled in one hour in km is,
55×1
=55
The distance travelled in one hour in meters is,
55×1000
=55,000
The distance covered by a car traveling 55 km/h in meters in one hour is 55,000 meters.
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership Page 272 Problem 19 Answer
Given a car traveling 55 km/h.
Need to find the distance covered by a car traveling 55 km/h in meters in one minute.
As the distanced is the product of speeds and time t. Thus,d=s×t. Using this, find the required distance covered by the car.
We have, s=55 km/h,t=1minute,
There are1000 meters in 1km.
There are 60 minutes in one hour.
Then, the distance travelled in one hour in km is,
55×1
=55
The distance travelled in one hour in meters is,
55×1000
=55,000
The distance travelled in one minute in meters is,
55,000÷60
=916.666
≈916.7
The distance covered by a car traveling 55 km/h in meters in one minute is approximately916.7
meters.
Page 272 Problem 20 Answer
Given a car traveling 55 km/h. Need to find the distance covered by a car traveling 55 km/h in meters in one second.
As the distanced is the product of speeds and time t. Thus,d=s×t. Using this, find the required distance covered by the car.
We have, s=55 km/h,t=1second,There are1000 meters in 1km,
There are 60minutes in one hour and 60 seconds in one minute.
Then, the distance travelled in one hour in km is,
55×1
=55
The distance travelled in one hour in meters is,
55×1000
=55,000
The distance travelled in one minute in meters is,
55,000÷60
=916.666
≈916.7
The distance travelled in one second in meters is,
916.7÷60
=15.278
≈15.3
The distance covered by a car traveling 55 km/h in meters in one second is approximately15.3
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership Page 272 Exercise 1 Answer
Given a car traveling x km/h.
Need to find the distance covered by a car traveling x km/h in kilometers in one hour.
As the distanced is the product of speeds and time t. Thus,d=s×t. Using this, find the required distance covered by the car.
We have, s=x km/h,t=1h,
Then, the distance travelled in one hour in km is,
x×1
=x
The distance covered by a car traveling x km/h in kilometers in one hour is x km.
Page 272 Exercise 2 Answer
Given a car traveling x km/h.
Need to find the distance covered by a car traveling x km/h in meters in one hour.
As the distanced is the product of speeds and time t. Thus,d=s×t. Using this, find the required distance covered by the car.
We have, s=x km/h,t=1h,There are1000 meters in 1km.
Then, the distance travelled in one hour in km is,x×1=x
The distance travelled in one hour in meters is,x×1000=1000x
The distance covered by a car traveling x km/h in meters in one hour is1000x meters.
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership Page 272 Exercise 3 Answer
Given a car traveling x km/h.
Need to find the distance covered by a car traveling x km/h in meters in one minute.
As the distanced is the product of speeds and time t.
Thus,d=s×t. Using this, find the required distance covered by the car.
We have, s=x km/h,t=1minute,
There are1000 meters in 1km,
There are 60 minutes in one hour.
Then, the distance travelled in one hour in km is,
x×1
=x
The distance travelled in one hour in meters is,
x×1000
=1000x
The distance travelled in one minute in meters is,
1000x÷60
=100x/6
=16.6x
The distance covered by a car traveling x km/h in meters in one minute is approximately 16.6x meters.
Page 272 Exercise 4 Answer
Given a car traveling x km/h.
Need to find the distance covered by a car traveling x km/h in meters in one second.
As the distanced is the product of speeds and time t. Thus,d=s×t. Using this, find the required distance covered by the car.
We have, s=x km/h,t=1second,
There are 1000 meters in1km.
There are 60 minutes in one hour and 60 seconds in one minute.
Then, the distance travelled in one hour in km is,
x×1
=x
The distance travelled in one hour in meters is,
x×1000
=1000x
The distance travelled in one minute in meters is,
1000x÷60
=100x/6
=16.6x
The distance travelled in one second in meters is,
17x÷60
=0.276x
≈0.28x
The distance covered by a car traveling km/h in meters in one second is approximately meters.
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership Page 272 Exercise 5 Answer
Given Mindy is driving 32 mi/h as she nears an elementary school.
A first grade student runs into the street after a soccer ball, and Mindy reacts in about three-quarters of a second.
Need to tell her approximate reaction distance.
As the reaction distanced is the product of speeds and reaction time t.
Thus,d=s×t. Using these, find the reaction distance covered by the car.
We have,s=32 mi/h,the reaction time,t=3/4th of a second
There are5,280 feet in one mile.
There are 60 minutes in one hour and 60 seconds in one minute.
Then, the distance travelled in one hour in miles is,
32×1
=32
The distance travelled in one hour in feet is,
32×5,280
=168,960
The distance travelled in one minute in feet is,
168,960÷60
=2,816
The distance travelled in one second in feet is,
2,816÷60
=46.933
≈46.93
The distance travelled in 3/4th of a second in feet is, 46.93×3/4
=46.93×0.75
=35.197
≈35.2
Given Mindy is driving 32 mi/h as she nears an elementary school.
A firstgrade student runs into the street after a soccer ball, and Mindy reacts in about three-quarters of a second.
Her approximate reaction distance is 35.2 feet.
Page 273 Exercise 6 Answer
Given:
| Speed | Reaction Distance | Braking Distance |
| 40mi/h | a. | f. |
| 30mi/h | b. | g. |
| 20mi/h | c. | h. |
| 15mi/h | d. | i. |
| 5mi/h | e. | j. |
Need to complete the chart for entries a–j.
As the general formula for the braking distance in feet is s2/20, where the speed of the car is s mi/h and the reaction distance of a car travelling at s miles per hour is approximated by using a distance of s feet.
The reaction distance in feet, for speed 40 mi/h is 40 feet.
The reaction distance in feet, for speed 30 mi/h is 30 feet.
The reaction distance in feet, for speed 20 mi/h is 20 feet.
The reaction distance in feet, for speed 15 mi/h is15 feet.
The reaction distance in feet, for speed 5 mi/h is 5 feet.
For given:
| Speed | Reaction Distance | Braking Distance |
| 40mi/h | 40feet | 80feet |
| 30mi/h | 30feet | 45feet |
| 20mi/h | 20feet | 20feet |
| 15mi/h | 15feet | 11.25feet |
| 5mi/h | 5feet | 1.25feet |
The complete chart for entries a–j is:
| Speed | Reaction Distance | Braking Distance |
| 40mi/h | a. | f. |
| 30mi/h | b. | g. |
| 20mi/h | c. | h. |
| 15mi/h | d. | i. |
| 5mi/h | e. | j. |
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Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership Page 273 Exercise 7 Answer
Given: the total stopping distance y=x2/170+x/5, where x represents the speed in km/hour and y represents the total stopping distance in meters.
We need to graph this equation for ≤100km/h.
We get:
Hence, the graph for the equation y=x2/170+x/5 is:
Page 273 Exercise 8 Answer
Given: the total stopping distance y=x2/170+λ/5 where x represents the speed in km/hour and y represents the total stopping distance in meters.
Using the graph we need to approximate the stopping distance for a car traveling at 60km/h.
We draw a vertical line at 60 km/h find its intersection with the graph, draw a horizontal line through this intersection point,
we get y =33.18.
Hence, using the graph
we get the stopping distance for the car as 33.18 m.
Page 273 Exercise 9 Answer
Given: the total stopping distancey=x2/170+x/5 where x represents the speed in km/hour and y represents the total stopping distance in meters.
Using the graph we need to approximate the speed for a car that stops after 60 meters.
We draw a horizontal line at 60m, and find its intersection with the graph, draw a vertical tine through this intersection point,
we get x = 85.42 km/h.
Hence, using the graph
we get the speed of the car as 85.42 km/h.
Automobile Ownership Exercise In Cengage Chapter 5 Guide
Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership Page 273 Exercise 10 Answer
Given: A spreadsheet user inputs a speed in miles per hour into cell A1.
We need to write a formula that would enter the approximate equivalent of
that speed in km/h in cell A2.
We have:
Speed in km/h=1.60934× Speed in miles per hour
⇒A2=1.60934×A1.
Hence, A2=1.60934×A1 the formula that would enter the approximate equivalent of that speed in km/h in cell A2.
Page 273 Exercise 11 Answer
Given: The cells in a spreadsheet where A1= Speed in miles per hour, A3= Total stopping distance.
We need to find a spreadsheet formula that would enter the approximate total stopping distance in feet in cell A3.
We have:
Reaction distance = A1 feet.
Braking distance =A12/20.
Total stopping distance = Reaction distance + Braking distance
A3=A1+A12/20.
Hence, A3=A1+A12/20 is the formula that would enter the approximate total stopping distance in feet in cell A3.
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Cengage Financial Algebra 1st Edition Chapter 5 Exercise 5.8 Automobile Ownership Page 273 Exercise 12 Answer
Given: The cells in a spreadsheet where A3= stopping distance in feet, A4= Total stopping distance in km.
We need to find a spreadsheet formula that would enter the approximate total stopping distance in kilometers in cell A4.
We have:
Total stopping distance in feet = Total stopping distance in km
3281⇒A4=A3/3281.
Hence, A4=A3/3281 is the formula that would enter the approximate total stopping distance in kilometers in cell A4.