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Tutors Answer Your Questions about Travel Word Problems (FREE)
Question 155510: In a walkathon to raise money for charity, Elisa walked a certain distance at 5 mi/h and then jogged twice that distance at 8 mi/h. Her total time walking and jogging was 2hr and 15min. How many miles long was the walkathon? The answer is 15mi but I cannot formulate the equation.: In a walkathon to raise money for charity, Elisa walked a certain distance at 5 mi/h and then jogged twice that distance at 8 mi/h. Her total time walking and jogging was 2hr and 15min. How many miles long was the walkathon? The answer is 15mi but I cannot formulate the equation.
Answer by Adam(59)  (Show Source):
You can put this solution on YOUR website!There is an important equation of motion  v-velocity, s-distance, t-time. We will use this to solve our problem. First we know the time 2 hours and 15 minutes that is 2.25 hours. We isolate t in the equation and get  , we know two distances a and 2a (twice longer) and two velocities 5 and 8 mi/h and we can formulate 
Cheers,
Adam
p.s. don't forget that a is not a complete distance ;)
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Question 155510: In a walkathon to raise money for charity, Elisa walked a certain distance at 5 mi/h and then jogged twice that distance at 8 mi/h. Her total time walking and jogging was 2hr and 15min. How many miles long was the walkathon? The answer is 15mi but I cannot formulate the equation.: In a walkathon to raise money for charity, Elisa walked a certain distance at 5 mi/h and then jogged twice that distance at 8 mi/h. Her total time walking and jogging was 2hr and 15min. How many miles long was the walkathon? The answer is 15mi but I cannot formulate the equation.
Answer by scott8148(2486) (Show Source):
You can put this solution on YOUR website!r*t=d __ so t=d/r __ 2hr15min is 9/4 hr
"Elisa walked a certain distance at 5 mi/h" __ Tw=d/5
"jogged twice that distance at 8 mi/h" __ Tj=2d/8=d/4
the total time is Tw+Tj __ the total distance is 3d
d/5+d/4=9/4 __ multiplying by 20 __ 4d+5d=45 __ 9d=45 __ d=5 __ 3d=15
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Question 155437This question is from textbook Algebra and Trigonometry
: On the first part of a 363-mile trip, Emily averaged 58 miles per hour. Due to increased traffic volume, she averaged only 52 miles per hour on the last part of the trip. Find the amount of time she traveled at each of the speeds if her total time for the trip was 6 hours and 45 minutes?This question is from textbook Algebra and Trigonometry
: On the first part of a 363-mile trip, Emily averaged 58 miles per hour. Due to increased traffic volume, she averaged only 52 miles per hour on the last part of the trip. Find the amount of time she traveled at each of the speeds if her total time for the trip was 6 hours and 45 minutes?
Answer by nerdybill(466) (Show Source):
You can put this solution on YOUR website!Info from problem:
total miles: 363 miles
two speeds: 58 mph and 52 mph
total time: 6 hrs 45 mins OR 6.75 hrs
.
You must apply the "distance formula":
d = rt
where
d is distance
r is rate or speed
t is time
.
"distance traveled at 58 mph" + "distance traveled at 52 mph" = "total distance"
.
If we:
Let t = time traveling at 58 mph
then
6.75-t = time traveling at 52 mph
.
"distance traveled at 58 mph" = 58t
"distance traveled at 52 mph" = 52(6.75-t)
"total distance" = 363 miles
.
So, now we can rewrite:
"distance traveled at 58 mph" + "distance traveled at 52 mph" = "total distance"
as
58t + 52(6.75-t) = 363
58t + 351 - 52t = 363
6t + 351 = 363
6t = 12
t = 2 hours (time traveling at 58 mph)
.
6.75-t = 6.75-2 = 4.75 hours (time traveling at 52 mph)
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Question 155432: Katy leaves City A for City B at 7 am. driving a 51 mph. at 10am Jensen leaves City B for City A driving at 60 mph. If the cities are 264 miles apart at what time will they meet?: Katy leaves City A for City B at 7 am. driving a 51 mph. at 10am Jensen leaves City B for City A driving at 60 mph. If the cities are 264 miles apart at what time will they meet?
Answer by stanbon(18060) (Show Source):
You can put this solution on YOUR website!Katy leaves City A for City B at 7 am. driving a 51 mph. at 10am Jensen leaves City B for City A driving at 60 mph. If the cities are 264 miles apart at what time will they meet?
------------------------------
Katy DATA:
rate = 51 mph ; time = x hrs. ; distance = 51x miles
--------------------
Jensen DATA:
rate = 60 mph ; time = x-3 hrs ; distance = 60(x-3) miles
-----------------------------
EQUATION:
51x + 60x-180 = 264
111x = 444
x = 4 hours
They will meet at 7 am + 4 hrs = 11 am
==========================================
Cheers,
Stan H.
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Question 155368This question is from textbook College Algebra
: The distance between Majorie and the dude ranch was (4times the square root of 34) miles straight across a rattlesnake-infested canyon. Instead of crossing the canyon, she hiked due north for a long time and then hiked due east for the shorter leg of the journey to the ranch. If she averaged 4mph and it took her 8 hr to get to the ranch, how far did she hike in a northernly direction?
( they show a right triangle and i have the answer but i dont know how to set up the formula.the answer is 20 miles)( they show the hypontenuse as 4xthe sq rt of 34) and the loner leg is going north and the shorter leg is going east)This question is from textbook College Algebra
: The distance between Majorie and the dude ranch was (4times the square root of 34) miles straight across a rattlesnake-infested canyon. Instead of crossing the canyon, she hiked due north for a long time and then hiked due east for the shorter leg of the journey to the ranch. If she averaged 4mph and it took her 8 hr to get to the ranch, how far did she hike in a northernly direction?
( they show a right triangle and i have the answer but i dont know how to set up the formula.the answer is 20 miles)( they show the hypontenuse as 4xthe sq rt of 34) and the loner leg is going north and the shorter leg is going east)
Answer by ankor@dixie-net.com(3953) (Show Source):
You can put this solution on YOUR website!The distance between Marjorie and the dude ranch was (4times the square root of 34) miles straight across a rattlesnake-infested canyon. Instead of crossing the canyon, she hiked due north for a long time and then hiked due east for the shorter leg of the journey to the ranch. If she averaged 4mph and it took her 8 hr to get to the ranch, how far did she hike in a northernly direction?
( they show a right triangle and i have the answer but i don't know how to set up the formula.the answer is 20 miles)( they show the hypotenuse as 4xthe sq rt of 34) and the loner leg is going north and the shorter leg is going east)
:
They give the travel time and the speed, find the total distance
8 * 4 = 32 mi total
:
Let x = the northern leg
then
(32-x) = the shorter eastbound leg
:
The famous right triangle formula:
x^2 + (32-x)^2 = (  )^2
;
x^2 + (1024 - 64x + x^2) = 16(34)
:
x^2 + x^2 - 64x + 1024 = 544
:
2x^2 - 64x + 1024 = 544
:
2x^2 - 64x + 1024 - 544 = 0
:
2x^2 - 64x + 480 = 0
Simplify, divide by 2
x^2 - 32x + 240
Factors to:
(x-12)(x-20) = 0
:
x = 20 mi is the longer leg distance
:
:
Check solution on calc:
Enter  = 23.3238 which is 
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Question 155369This question is from textbook college algebra
: Lisa, an architect from Huntsville, has arranged a business luncheon with her client, Taro from Norwood. They plan to meet at a restaurant off the 300 mile highway connecting their two cities. If they leave their offices simultaneously and arrive at the restaurnat simultaneoulsy, and Lisa averages 50mph while Taro averages 60mph, then how far from Huntsville is the restaurant?This question is from textbook college algebra
: Lisa, an architect from Huntsville, has arranged a business luncheon with her client, Taro from Norwood. They plan to meet at a restaurant off the 300 mile highway connecting their two cities. If they leave their offices simultaneously and arrive at the restaurnat simultaneoulsy, and Lisa averages 50mph while Taro averages 60mph, then how far from Huntsville is the restaurant?
Answer by ankor@dixie-net.com(3953) (Show Source):
You can put this solution on YOUR website! They plan to meet at a restaurant off the 300 mile highway connecting their two
cities. If they leave their offices simultaneously and arrive at the restaurant
simultaneously, and Lisa averages 50 mph while Taro averages 60 mph, then how far from Huntsville is the restaurant?
:
Find the time required for them to meet, then find the distance from that;
t = travel time till they meet
:
Write a distance equation: Dist = speed * time, total distance given as 300 mi
:
60t + 50t = 300
110t = 300
t = 
t = 2.73 hrs
;
Find the distance from Huntsville (Lisa at 50 mph)
:
2.73 * 50 = 136.36 mi
:
:
Check solution, find dist from Norwood
2.73 * 60 = 163.64 mi
:
Total: 136.36 + 163.64 = 300 mi
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Question 155355: Two cars leave from the same point at the same time, traveling in opposite directions. One traveling 15 mph slower than the other. After 6 hours, they are 630 miles apart. Find the rate of each car.: Two cars leave from the same point at the same time, traveling in opposite directions. One traveling 15 mph slower than the other. After 6 hours, they are 630 miles apart. Find the rate of each car.
Answer by jojo14344(379) (Show Source): |
Question 155362: Mike travels ti his friends house at 40miles/hour. He returns at 45 miles/hour.The total trip took 4 hours. How far away is his friend?: Mike travels ti his friends house at 40miles/hour. He returns at 45 miles/hour.The total trip took 4 hours. How far away is his friend?
Answer by jojo14344(379) (Show Source): |
Question 155283: Mr. Newton drives to and from work over the same route each day. On Friday, he drives at an average rate of 52 km/hr and is one minute late to work. ON Monday, he drives at an average rate of 60 km/hr and is one minute early to work. If he left home at the same time each day, what is the distance Mr. Newton travels from home (one way) to work.
Not from a text book, it was on a sheet of paper. thankyou.: Mr. Newton drives to and from work over the same route each day. On Friday, he drives at an average rate of 52 km/hr and is one minute late to work. ON Monday, he drives at an average rate of 60 km/hr and is one minute early to work. If he left home at the same time each day, what is the distance Mr. Newton travels from home (one way) to work.
Not from a text book, it was on a sheet of paper. thankyou.
Answer by checkley77(1780) (Show Source):
You can put this solution on YOUR website!D=RT
D=52(T+1/60)
D=60(T-1/60)
52(T+1/60=60(T-1/60)
52T+52/60=60T-60/60
52T-60T=(-52-60)/60
-8T=-112/60
T=-112/60*-1/8
T=112/480
T=.23333 IS HIS TRAVEL TIME
D=60(.2333-1/60)
D=60(60*.2333-1)/60
D=60(14-1)/60
D=60*13/60
D=13 KM. IS HIS DRIVING DISTANCE.
PROOF:
13=52(.2333+1)/60
13=52(60*.2333+1)/60
13=52(14+1)/60
13=52*15/60
13=780/60
13=13
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Question 155187This question is from textbook Saxon Math Course 3
: On a stretch of road 75 miles long, two trucks approach each other. Truck A is driving 55 mph and Truck B is going 80 mph. What is the distance between the two trucks one minute before their head-on collision?
This question isn't in the textbook - it's the Problem of the Week.This question is from textbook Saxon Math Course 3
: On a stretch of road 75 miles long, two trucks approach each other. Truck A is driving 55 mph and Truck B is going 80 mph. What is the distance between the two trucks one minute before their head-on collision?
This question isn't in the textbook - it's the Problem of the Week.
Answer by stanbon(18060) (Show Source):
You can put this solution on YOUR website!On a stretch of road 75 miles long, two trucks approach each other. Truck A is driving 55 mph and Truck B is going 80 mph. What is the distance between the two trucks one minute before their head-on collision?
----------------------------------------------
The closing speed is 55+80 = 135 mph
In one minute the trucks will cover 135/60 = 2.25 miles
--------------------------------
That is how far they are away from one-another one-minute before
they would have a head-on collision.
===================================
Cheers,
Stan H.
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Question 155108: Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?
: Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?
Answer by ptaylor(1185) (Show Source):
You can put this solution on YOUR website!Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let t=amount of time that passes before the second cyclist catches up with the first from the time the second cyclist starts
distance first cyclist travels=6*3+6*t(3 hour head start)
distance second cyclist travels=10*t
Now when the above two distances are equal, the second cyclist will have caught up with the first cyclist, so our equation to solve is:
10t=18+6t subtract 6t from each side
10t-6t=18-6t
4t=18
t=4.5 hours
CK
3*6+6*4.5=10*4.5
18+27=45
45=45
Hope this helps---ptaylor
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Question 155080: Two cars are traveling toward each other and are 455 miles apart. One car is going 10mph faster than the other. They meet after 3.5 hours. How fast is each car going?: Two cars are traveling toward each other and are 455 miles apart. One car is going 10mph faster than the other. They meet after 3.5 hours. How fast is each car going?
Answer by checkley77(1780) (Show Source):
You can put this solution on YOUR website!3.5(x+x+10)=455
3.5(2x+10)=455
7x+35=455
7x=455-35
7x=420
x=420/7
x=60 mph for the slower car.
60+10=70 mph for the faster car.
Proof:
3.5(60+70)=455
3.5*130=455
455=455
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Question 154987: Two missiles speed directly toward each other, one at 9,000 mph and the other at 21,000 mph. If they start 1,317 miles apart, how far apart are they on minute before they collide?: Two missiles speed directly toward each other, one at 9,000 mph and the other at 21,000 mph. If they start 1,317 miles apart, how far apart are they on minute before they collide?
Answer by stanbon(18060) (Show Source):
You can put this solution on YOUR website!Two missiles speed directly toward each other, one at 9,000 mph and the other at 21,000 mph. If they start 1,317 miles apart, how far apart are they on minute before they collide?
--------------------------------------
Find the closing rate: 9000+21000 = 30,000 mph
Change to miles per minute: 30,000/60 = 500 miles per minute
--------------------------------------
How far apart are they one minute before they collide?
Ans: 500 miles
==================
Cheers,
Stan H.
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Question 154741This question is from textbook college algebra ninth edition
: One morning, John drove 5 hours before stoping to eat. After lunch, he increased his speed by 10 mph. If he completed a 430-mile trip in 8 hours of driving time, how fast did he drive in the morning?This question is from textbook college algebra ninth edition
: One morning, John drove 5 hours before stoping to eat. After lunch, he increased his speed by 10 mph. If he completed a 430-mile trip in 8 hours of driving time, how fast did he drive in the morning?
Answer by ankor@dixie-net.com(3953) (Show Source):
You can put this solution on YOUR website!One morning, John drove 5 hours before stopping to eat. After lunch, he
increased his speed by 10 mph. If he completed a 430-mile trip in 8 hours of
driving time, how fast did he drive in the morning?
:
Let s = his speed in the morning
then
(s+10) = his speed in the afternoon
;
A total of 8 hrs driving time, if he drove 5 hrs in the morning,
He drove 3 hrs in the afternoon
:
Write a distance equation: Dist = time * speed
:
Morning miles + afternoon miles = 430 miles
5s + 3(s+10) = 430
;
5s + 3s + 30 = 430
:
8s = 430 - 30
s = 
s = 50 mph in the morning
:
:
Check solution by finding the distance at the two speeds:
5(50) + 3(60) =
250 + 180 = 430
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Question 154845: A cyclist bikes x distance at 10 miles per hour and returns over the same path at 8 miles per hour. Whast is the cyclist's average rate for the round trip in miles per hour?
: A cyclist bikes x distance at 10 miles per hour and returns over the same path at 8 miles per hour. Whast is the cyclist's average rate for the round trip in miles per hour?
Answer by stanbon(18060) (Show Source):
You can put this solution on YOUR website!A cyclist bikes x distance at 10 miles per hour and returns over the same path at 8 miles per hour. Whast is the cyclist's average rate for the round trip in miles per hour?
--------------
1st leg DATA:
distance = x miles ; rate = 10 mph ; time = d/r = x/10 hrs
----------------
Return leg DATA:
distance = x miles ; rate = 8 mph ; time = d/r = x/8 hrs
----------------
Average rate = (total distance)/(total time)
= 2x/[(x/10) + x/8]
= 2x/[18x/80]
= 2x/[9x/40]
= 80/9
= 8.89 mph
=================
Cheers,
Stan H.
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Question 154757This question is from textbook college algebra ninth edition
: Two cars leave a gas station travling in opposite directions. One car travels at 60 mph and the other at 64 mph. In how many hours will they be 310 miles apart?This question is from textbook college algebra ninth edition
: Two cars leave a gas station travling in opposite directions. One car travels at 60 mph and the other at 64 mph. In how many hours will they be 310 miles apart?
Answer by stanbon(18060) (Show Source):
You can put this solution on YOUR website!Two cars leave a gas station traveling in opposite directions. One car travels at 60 mph and the other at 64 mph. In how many hours will they be 310 miles apart?
--------------
1st car DATA:
distance = x miles ; rate = 60 mph ; time = x/60 hrs.
-----------------
2nd car DATA:
distance = 310-x miles ; rate = 64 mph ; time = (310-x)/64 hrs.
-----------------
EQUATION:
time = time
x/60 = (310-x)/64
64x = 60(310-x)
124x = 60*310
x = 150
1st car time: 150*60 = 15/6 = 5/2 = 2 1/2 hrs
2nd car time is the same.
---------------------------
Cheers,
Stan H.
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Question 154757This question is from textbook college algebra ninth edition
: Two cars leave a gas station travling in opposite directions. One car travels at 60 mph and the other at 64 mph. In how many hours will they be 310 miles apart?This question is from textbook college algebra ninth edition
: Two cars leave a gas station travling in opposite directions. One car travels at 60 mph and the other at 64 mph. In how many hours will they be 310 miles apart?
Answer by nerdybill(466) (Show Source):
You can put this solution on YOUR website!For this problem, you'll need to apply the "distance formula":
d = rt
where
d is distance
r is rate or speed
t is time
.
Let x = time traveled
.
"distance traveled by 60 mph car" + "distance traveled by 64 mph car" = 310
.
60x + 64x = 310
124x = 310
x = 310/124
x = 2.5 hours
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Question 154736This question is from textbook
: The Hudson River flows at a rate of 3 miles per hour. A patrol boat travels 60 miles upriver, and returns in a total time of 9 hours. What is the speed of the boat in still water?This question is from textbook
: The Hudson River flows at a rate of 3 miles per hour. A patrol boat travels 60 miles upriver, and returns in a total time of 9 hours. What is the speed of the boat in still water?
Answer by stanbon(18060) (Show Source):
You can put this solution on YOUR website!The Hudson River flows at a rate of 3 miles per hour.
A patrol boat travels 60 miles upriver, and returns in a total time of 9 hours. What is the speed of the boat in still water?
----------------------------
Let the boat speed be "b".
----------
Upstream DATA:
distance = 60 miles ; rate = b-3 ; time = 60/(b-3) hrs.
------------------
Downstream DATA:
distance = 60 miles ; rate = b+3 ; time = 60/(b+3) hrs
---------------------
EQUATION:
time up + time down = 9 hrs
60/(b-3) + 60/(b+3) = 9
60b + 180 + 60b - 180 = 9(b^2-9)
120b = 9b^2 - 81
9b^2 -120b - 81 = 0
b = -0.643904; b = 13.977237...
Positive solution:
boat speed = 13.977237..
========================
Cheers,
Stan H.
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Question 154700: Two Trains are 2 miles apart and are traveling toward each other on the same track. Each train is going 30 MPH. A fly going 60 MPH starts at the nose of one train then flies toward the other train and upon reaching the second train immediately turns around and flies back towards the first train. The fly buzzes back and forth until all three collide. How far did the fly fly?: Two Trains are 2 miles apart and are traveling toward each other on the same track. Each train is going 30 MPH. A fly going 60 MPH starts at the nose of one train then flies toward the other train and upon reaching the second train immediately turns around and flies back towards the first train. The fly buzzes back and forth until all three collide. How far did the fly fly?
Answer by stanbon(18060) (Show Source):
You can put this solution on YOUR website!Two Trains are 2 miles apart and are traveling toward each other on the same track. Each train is going 30 MPH. A fly going 60 MPH starts at the nose of one train then flies toward the other train and upon reaching the second train immediately turns around and flies back towards the first train. The fly buzzes back and forth until all three collide. How far did the fly fly?
---------------------------------
Train #1 DATA:
distance = x miles ; rate = 30mph; time = d/r = x/30 hrs
-------------------
Train #2 DATA:
distance = 2-x miles ; rate = 30 mph; time = d/r = (2-x)/30 hrs
-------------------
EQUATION:
time of #1 = time of #2
x/30 = (2-x)/30
x = 2-x
2x = 2
x = 1 mile
------------------------------
Time for each train: d/r = 1 mile/30 mph = 1/30 hr = 2 minutes
------------------------------
How far does the fly fly?
distance = time*rate
(2/60)60 mph = 2 miles
==============================
Cheers,
Stan H.
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Question 154650: Jane skates 9 miles each morning. Today, she averaged 10 miles per hour for the first 6 miles but then slowed to 5 miles per hour for the rest of the trip. How long did the entire trip take?: Jane skates 9 miles each morning. Today, she averaged 10 miles per hour for the first 6 miles but then slowed to 5 miles per hour for the rest of the trip. How long did the entire trip take?
Answer by ptaylor(1185) (Show Source):
You can put this solution on YOUR website!Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Time required to travel the first 6 miles =6/10=3/5 hr
Time required for the remaining 3 miles=3/5 hr
Total time=3/5 hr +3/5 hr=6/5 hr=1hr 12min
CK
(3/5)*10+(3/5)*5=9
6+3=9
9=9
Hope this helps---ptaaylor
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Question 154650: Jane skates 9 miles each morning. Today, she averaged 10 miles per hour for the first 6 miles but then slowed to 5 miles per hour for the rest of the trip. How long did the entire trip take?: Jane skates 9 miles each morning. Today, she averaged 10 miles per hour for the first 6 miles but then slowed to 5 miles per hour for the rest of the trip. How long did the entire trip take?
Answer by stanbon(18060) (Show Source):
You can put this solution on YOUR website!Jane skates 9 miles each morning. Today, she averaged 10 miles per hour for the first 6 miles but then slowed to 5 miles per hour for the rest of the trip. How long did the entire trip take?
-------------------------
First 6 miles DATA:
distance = 6 mi ; rate = 10 mph; time = d/r = 6/10 = 3/5 hrs
----------------------
Last 3 miles DATA:
distance = 3 mi ; tate = 4 mph; time = d/r = 3/4 hrs
-----------------------
Total time:
(3/5) + (3/4) = 12/20 + 15/20 = 27/20 = 1 7/20 hrs.
==================
Cheers,
Stan H.
|
Question 154640: Dear Algebra.com tutors,
I'm stumped! Can you help with this bonus problem?
How long would a yellow light have to remain lit to allow a car to safely stop before reaching a 100 ft intersection if the car is traveling at a rate of 50 mph when it begins to decelerate at a steady rate of -20 ft/second squared, after a reaction time of 0.6 seconds?
Use the following two physics formulas to aid in solving this problem:
x = v-subzero * t + 1/2 * a * t-squared
and
v-squared = v-subzero squared + 2 * a * x
(x = distance, a = acceleration, v = velocity, t = time, v-subzero = initial velocity)
Thanks for trying! I appreciated it!
Sam: Dear Algebra.com tutors,
I'm stumped! Can you help with this bonus problem?
How long would a yellow light have to remain lit to allow a car to safely stop before reaching a 100 ft intersection if the car is traveling at a rate of 50 mph when it begins to decelerate at a steady rate of -20 ft/second squared, after a reaction time of 0.6 seconds?
Use the following two physics formulas to aid in solving this problem:
x = v-subzero * t + 1/2 * a * t-squared
and
v-squared = v-subzero squared + 2 * a * x
(x = distance, a = acceleration, v = velocity, t = time, v-subzero = initial velocity)
Thanks for trying! I appreciated it!
Sam
Answer by scott8148(2486) (Show Source):
You can put this solution on YOUR website!50 mph multiplied by 5280 ft per mi and divided by 3600 sec per hr gives 220/3 fps
with a deceleration of 20 fps^2, it takes 11/3 ([220/3]/20) sec to decelerate to a stop
adding the .6 sec reaction time gives a total stopping time of 4.27 sec (approx)
the car travels 44 ft ([220/3]*.6) during the reaction time
__ and [(220/3)/2]*(11/3) or 134.4 ft (approx) during deceleration
for a total stopping distance of 178.4 ft (approx)
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Question 154523: Lita Vance is driving to Lake Tahoe a distance of 662 miles. If she drices at 55 mph. for 5 hours, at what speed must she drive to complete the total trip in 14 hours?: Lita Vance is driving to Lake Tahoe a distance of 662 miles. If she drices at 55 mph. for 5 hours, at what speed must she drive to complete the total trip in 14 hours?
Answer by BrittanyM(80) (Show Source):
You can put this solution on YOUR website!She has already driven for five hours, so the remaining distance of her trip must be completed in nine hours.
To find this, we can subtract the distance that she has already travelled from the total distance of the trip:
55mph * 5hrs = 275miles
And the total distance is 662miles, so:
662mi - 275mi = 387
We know that she wants to make the trip in a total of fourteen hours, but she has already used five hours. So there are nine hours heft for her to complete 387 miles.
Her speed for the remainder of the trip must be 43mph.
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Question 154470: How long will it take me to go 70 mph in 204 mi?: How long will it take me to go 70 mph in 204 mi?
Answer by Alan3354(579) (Show Source): |
Question 154360: I am not sure how to set up the equation or how to solve this word problem.
Elena bicycles 5km/h faster then Carlos. In the same time it takes Carlos to bicycle 39km, Elena can bicycle 54km. How fst does each bicyclist travel?: I am not sure how to set up the equation or how to solve this word problem.
Elena bicycles 5km/h faster then Carlos. In the same time it takes Carlos to bicycle 39km, Elena can bicycle 54km. How fst does each bicyclist travel?
Answer by ankor@dixie-net.com(3953) (Show Source):
You can put this solution on YOUR website!Elena bicycles 5km/h faster then Carlos. In the same time it takes Carlos to bicycle 39km, Elena can bicycle 54km. How fast does each bicyclist travel?
:
Let s = C's speed
then
(s+5) = E's speed
:
it says the times are equal: write a time equation: time = dist/speed
:
 = 
cross multiply
54s = 39(s+5)
:
54s = 39s + 195
:
54s - 39s = 195
:
15s = 195
s = 
s = 13 km/hr is C's speed
then
13 + 5 = 18 km/hr is E's speed
:
:
Check solutions by finding the times
54/18 = 3 hr
39/13 = 3 hr
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Question 154360: I am not sure how to set up the equation or how to solve this word problem.
Elena bicycles 5km/h faster then Carlos. In the same time it takes Carlos to bicycle 39km, Elena can bicycle 54km. How fst does each bicyclist travel?: I am not sure how to set up the equation or how to solve this word problem.
Elena bicycles 5km/h faster then Carlos. In the same time it takes Carlos to bicycle 39km, Elena can bicycle 54km. How fst does each bicyclist travel?
Answer by scott8148(2486) (Show Source):
You can put this solution on YOUR website!"Elena bicycles 5km/h faster then Carlos" __ E=C+5
d=rt __ t=d/r
" In the same time it takes Carlos to bicycle 39km, Elena can bicycle 54km" __ 39/C=54/(C+5)
"cross" multiplying __ 39C+195=54C __ subtracting 39C __ 195=15C __ dividing by 15 __ 13=C
substituting __ E=(13)+5 __ E=18
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Question 154346: Two cars started from the same point and traveled on a straiht course in opposite directions for exactly 2 hours, at which time they were 208 miles apart. If one car traveled, on average, 8 miles per hour faster than the other car, what was the average speed for each car for the 2-hour trip?
: Two cars started from the same point and traveled on a straiht course in opposite directions for exactly 2 hours, at which time they were 208 miles apart. If one car traveled, on average, 8 miles per hour faster than the other car, what was the average speed for each car for the 2-hour trip?
Answer by Alan3354(579) (Show Source):
You can put this solution on YOUR website! Two cars started from the same point and traveled on a straiht course in opposite directions for exactly 2 hours, at which time they were 208 miles apart. If one car traveled, on average, 8 miles per hour faster than the other car, what was the average speed for each car for the 2-hour trip?
-------------
Going in opposite directions, their speeds are added. One car is moving at S mph, the other is S + 8
So S + (S + 8) times 2 = 208
2*(2S + 8) = 208
4S + 16 = 208
4S = 192
S = 48 mph
The faster car moves at 48+8 = 56 mph.
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Question 154325This question is from textbook
: Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.This question is from textbook
: Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
Answer by stanbon(18060) (Show Source):
You can put this solution on YOUR website!Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
---------------------
Original speed data:
distance = 200 miles ; rate = x hours ; rate = 200/x mph
------------------
New speed Data
distance = 200 miles ; rate = x + 10 hrs; rate = 200/(x+10) mph
-------------------
EQUATION
old time - new time = 1
200/x - 200/(x+10) = 1
200x + 2000 - 200x = x^2+10x
x^2 + 10x -2000 = 0
(x-20)(x+10) = 0
Positive answer:
x = 20 mph (original speed)
x+10 = 30 mph (new speed
=================
Cheers,
Stan H.
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Question 154031: I previously asked a question regarding chasing someone with a 10 mile head start and I was running at a rate that is two miles per hour faster than the person I was chasing also with a hornet flying back and forth between us at a rate of 20 miles per hour. How far would the hornet fly by the time I reached the other person? Stan did a great job of answering the question for me, 5 hous to catch up to the person and the hornet flew 5X20=100 miles. My problem is that I still can not see how to write this out using the formulas. Any help is appreciated. Thank you.: I previously asked a question regarding chasing someone with a 10 mile head start and I was running at a rate that is two miles per hour faster than the person I was chasing also with a hornet flying back and forth between us at a rate of 20 miles per hour. How far would the hornet fly by the time I reached the other person? Stan did a great job of answering the question for me, 5 hous to catch up to the person and the hornet flew 5X20=100 miles. My problem is that I still can not see how to write this out using the formulas. Any help is appreciated. Thank you.
Answer by ptaylor(1185) (Show Source):
You can put this solution on YOUR website!Distance(d) equals Rate (r) times Time(t) or d=rt; r=d/t and t=d/r
Let t=time required to catch the front runner
Front runner runs at the rate of r
You run at the rate of r+2
Distance front runner runs =10+r*t (already has a 10 mi head start)
Distance you run=(r+2)*t=rt+2t
Now when the above two distances are equal, you will have caught the front runner, so:
10+rt=rt+2t
subtract rt from each side
10+rt-rt=rt-rt+2t collect like terms
2t=10
t=5 hours
Distance hornet flew (d)=Rate of hornet (20 mph)* amount of time the hornet flew (5 hr)
d=20*5=100 mi
Just as Stan sez
Hope this helps---ptaylor
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Question 154286: If a car is traveling 80 mph how far will it travel in five minutes: If a car is traveling 80 mph how far will it travel in five minutes
Answer by edjones(2169) (Show Source): |
Question 154272: 2 cars start at the same time and start in opposite directions and at what mile marker will they meet if one is going at 50mi/hr and 60mi/hr and the total distance is 100 miles. : 2 cars start at the same time and start in opposite directions and at what mile marker will they meet if one is going at 50mi/hr and 60mi/hr and the total distance is 100 miles.
Answer by stanbon(18060) (Show Source): |
Question 154274: A plane leaves Denver heading due north at 500 mph. Simultaneously, another
plane leaves Denver traveling due east at 1200 mph. After how many minutes
will the planes be 650 miles apart?: A plane leaves Denver heading due north at 500 mph. Simultaneously, another
plane leaves Denver traveling due east at 1200 mph. After how many minutes
will the planes be 650 miles apart?
Answer by stanbon(18060) (Show Source):
You can put this solution on YOUR website!A plane leaves Denver heading due north at 500 mph. Simultaneously, another
plane leaves Denver traveling due east at 1200 mph. After how many minutes
will the planes be 650 miles apart?
-------------------
Draw the picture on a coordinate system; use the y-axis for north
and the x-axis for east.
---------------------------
Let the time in hours be "x" for both planes
North distance = 500x miles
East distance = 1200x miles
-------------------------------
Use Pytagoras with a hypotenuse of 650 miles.
EQUATION:
(500x)^2 + (1200x)^2 = 650^2
250000x^2 + 1440000x^2 = 650^2
1690000x^2 = 650^2
x^2 = 0.25
x = 0.5 hours
------------------
Time = 30 minutes
========================
Cheers,
Stan H.
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Question 153771: erin and susan both took their dogs for a walk. together they walked 15 miles. if erin walked3 milesless than susan, how far did susan walk with her dog?: erin and susan both took their dogs for a walk. together they walked 15 miles. if erin walked3 milesless than susan, how far did susan walk with her dog?
Answer by orca(258) (Show Source):
You can put this solution on YOUR website!Let x be the distance Susan walked, then the distance Erin walked is x - 3.
As the total distance walked is 15, we have:
x + x - 3 = 15
Solving for x, we obtain
2x - 3 = 15
2x = 18
x = 9
So Susan walked 9 miles, and Erin walked x - 3 = 9 - 3 = 6 miles.
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Question 154100: a person on a bike leaves an hour before a person on a motorcycle. the person on the bike is travelling at 10 mph. the person on the motorcycle is travelling at 40 mph. how long before the person on the motorcycle overtakes the person on the bike?: a person on a bike leaves an hour before a person on a motorcycle. the person on the bike is travelling at 10 mph. the person on the motorcycle is travelling at 40 mph. how long before the person on the motorcycle overtakes the person on the bike?
Answer by nerdybill(466) (Show Source):
You can put this solution on YOUR website!a person on a bike leaves an hour before a person on a motorcycle. the person on the bike is travelling at 10 mph. the person on the motorcycle is travelling at 40 mph. how long before the person on the motorcycle overtakes the person on the bike?
.
Let t = time it takes person on motorcycle to overtake
t+1 = time person on bike traveled
.
Applying the "distance formula"
d = rt
where
d is distance
r is rate or speed
t is time
.
For motorcycle person to overtake bicycle person, the distance traveled by both must be the same:
40t = 10(t+1)
40t = 10t + 10
30t = 10
t = 10/30
t = 1/3 hour
Or, in terms of minutes:
t = 1/3 * 60 = 20 minutes
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Question 154100: a person on a bike leaves an hour before a person on a motorcycle. the person on the bike is travelling at 10 mph. the person on the motorcycle is travelling at 40 mph. how long before the person on the motorcycle overtakes the person on the bike?: a person on a bike leaves an hour before a person on a motorcycle. the person on the bike is travelling at 10 mph. the person on the motorcycle is travelling at 40 mph. how long before the person on the motorcycle overtakes the person on the bike?
Answer by checkley77(1780) (Show Source): |
Question 153960: Ryan drives to Pittsburgh at 55 mph and returns by a road 10 miles shorter at 60 mph. The return trip takes 30 minutes less. How long is each road?: Ryan drives to Pittsburgh at 55 mph and returns by a road 10 miles shorter at 60 mph. The return trip takes 30 minutes less. How long is each road?
Answer by ankor@dixie-net.com(3953) (Show Source):
You can put this solution on YOUR website!Ryan drives to Pittsburgh at 55 mph and returns by a road 10 miles shorter at 60 mph. The return trip takes 30 minutes less. How long is each road?
:
let d = distance of the original trip
then
(d-10) = distance of the return trip
:
Convert 30 min to hrs; .5 hrs
:
Write a time equation:
:
to time = return time + one-half hr
 =  + .5
:
Multiply equation by 3300 to get rid of the denominator (55*60)
60d = 55(d-10) + 3300(.5)
:
60d = 55d - 550 + 1650
:
60d - 55d = 1100
d = 
d = 220 mile is the original trip and 210 miles is the return trip
:
:
Check solution by finding the time for each trip
220/55 = 4 hrs
210/60 = 3.5 hrs
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Question 153958: i need help with this equation step by step please
-7(x+2)+4x
and
x-2=-2+x
thanks!
: i need help with this equation step by step please
-7(x+2)+4x
and
x-2=-2+x
thanks!
Answer by stanbon(18060) (Show Source):
You can put this solution on YOUR website!Suppose you are chasing somone with a 10 mile head start, and that you are running at a rate that is two miles per hour faster than the person you are chasing. Also suppose that a hornet is flying back and forth between the two of you at a rate of 20 miles per hour. How far will the hornet fly (if the hornet could fly nonstop all that time) by the time you reached the other person?
----------------
Each hour you will go as far as the lead person plus 2 miles.
So it will take you 5 hours to catch up to that person.
In 5 hours the hornet can fly 5*20 miles = 100 miles.
=================
Cheers,
Stan H.
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Question 153963: The Chunnel connects the UK and Fance by a railway tunnel under the English Channel. The British started tunneling 2.5 months before the French and averaged 0.63 kilometer per month. The French averaged 0.47 kilometer per month. When the two sides met, they had tunneled 37.9 kilometers.
What is the length of tunnel each country built?: The Chunnel connects the UK and Fance by a railway tunnel under the English Channel. The British started tunneling 2.5 months before the French and averaged 0.63 kilometer per month. The French averaged 0.47 kilometer per month. When the two sides met, they had tunneled 37.9 kilometers.
What is the length of tunnel each country built?
Answer by stanbon(18060) (Show Source):
You can put this solution on YOUR website!The Chunnel connects the UK and Fance by a railway tunnel under the English Channel. The British started tunneling 2.5 months before the French and averaged 0.63 kilometer per month. The French averaged 0.47 kilometer per month. When the two sides met, they had tunneled 37.9 kilometers.
-----------------
Distance worked together = 37.9 - 2.5*0.63 = 36.325 km
EQUATION:
0.63x + 0.47x = 36.325 km
1.1x = 36.325 km
x = 33.0227... months
=====================
Cheers,
Stan H.
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Question 153882: An airplane left an airport flying at at 180 mph. A jet that flies at 333 mph left 1 hour later. The jet follows tthe same route as the airplane on parallel altitudes. How many hours will it take the jet to catch up with the airplane?: An airplane left an airport flying at at 180 mph. A jet that flies at 333 mph left 1 hour later. The jet follows tthe same route as the airplane on parallel altitudes. How many hours will it take the jet to catch up with the airplane?
Answer by nerdybill(466) (Show Source):
You can put this solution on YOUR website!An airplane left an airport flying at at 180 mph. A jet that flies at 333 mph left 1 hour later. The jet follows tthe same route as the airplane on parallel altitudes. How many hours will it take the jet to catch up with the airplane?
.
This problem needs you to apply the "distance formula":
d = rt
where
d is distance
r is rate or speed
t is time
.
"distance airplane traveled" = "distnce jet traveled"
.
Let x = hours "airplane" was flying
x-1 = hours "jet" was flying
.
180x = 333(x-1)
180x = 333x - 333
180x + 333 = 333x
333 = 153x
333/153 = x
2.176 hours = x
.
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Question 153884: Two jets leave Dallas and fly in opposite directions. One is flying 50 mph faster than the other. After 2 hours they are 2500 miles apart. Find the speed of each jet.: Two jets leave Dallas and fly in opposite directions. One is flying 50 mph faster than the other. After 2 hours they are 2500 miles apart. Find the speed of each jet.
Answer by orca(258) (Show Source):
You can put this solution on YOUR website!Let x be the speed of the slower jet.
Then the speed of the faster one is x + 50.
The distance that separating them at the end of 2 hours can be expressed in terms of x as:
2x + 2(x + 50)
As the separating distance is 2500, we have
2x + 2(x + 50) = 2500
Solving for x, we have
2x + 2x +100 = 2500
4x = 2400
x = 600
So the speed of the slower jet is 600 miles per hour
the speed of the fast jet is x + 50 = 650 miles per hour.
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Question 153885: On her way to work from home, Jayla averaged only 20 mph. On her drive home, she averaged 40 mph. If the total travel time was 1.5 hours, how long did it take her to drive to work?: On her way to work from home, Jayla averaged only 20 mph. On her drive home, she averaged 40 mph. If the total travel time was 1.5 hours, how long did it take her to drive to work?
Answer by galileo(7) (Show Source):
You can put this solution on YOUR website!On her way to work from home, Jayla averaged only 20 mph. On her drive home, she averaged 40 mph. If the total travel time was 1.5 hours, how long did it take her to drive to work?
Let x = time to drive to work
1.5-x = time to drive home
from your lesson in physics, distance traveled = average speed times time
d = vt
since the distance traveled in driving to work and back home are equal, we have
20(x) = 40(1.5-x)
20x = 60 - 40x
20x + 40x = 60
60x = 60
x = 60/60
x = 1 hr (time to drive to work)
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Question 153779: Mr. Newton drive to and from work over the same route each day. On Friday, he drives at an average rate of 52 mph/km and is one minute late to work. On Monday, he drives at an average rate of 60 mph/km and is one minute early to work. If he left home at the same time each day, what is the distance Mr. Newton travels one way to work? : Mr. Newton drive to and from work over the same route each day. On Friday, he drives at an average rate of 52 mph/km and is one minute late to work. On Monday, he drives at an average rate of 60 mph/km and is one minute early to work. If he left home at the same time each day, what is the distance Mr. Newton travels one way to work?
Answer by ankor@dixie-net.com(3953) (Show Source):
You can put this solution on YOUR website!Mr. Newton drive to and from work over the same route each day. On Friday, he drives at an average rate of 52 mph/km and is one minute late to work. On Monday, he drives at an average rate of 60 mph/km and is one minute early to work. If he left home at the same time each day, what is the distance Mr. Newton travels one way to work?
:
First find the times it takes him to get to work, then find the distance from that
:
Let t = time required if he arrives on time (in hrs)
then
(t+  ) = time if he 1 min late
and
(t- ) = time if he 1 min early
;
Write a distance equation: Dist = speed * time
:
Monday dist = Friday dist
60(t- ) = 52(t+ )
:
60t-  = 52t+ 
:
60t - 52t = +
:
8t =  hrs
we can get rid of that denominator (60); convert it to minutes, (mult by 60)
8t = 112
t = 
t = 14 minutes for him to arrive on time
:
then we can say
It takes 13 min when he is 1 min early
and
15 min when he is 1 min late
:
:
Find the distances
52 *  = 13 miles
60 *  = 13 miles
;
Did this make sense?
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Question 153794: how fast did a car go to overtake a bus in 6 hours if the bus average 45mph and left 3 hours befor the car?: how fast did a car go to overtake a bus in 6 hours if the bus average 45mph and left 3 hours befor the car?
Answer by checkley77(1780) (Show Source): |
Question 153748: If a computer program draws aspiral at a reate of 45% every 20 seconds then how long will it take to complete 25 full turns?: If a computer program draws aspiral at a reate of 45% every 20 seconds then how long will it take to complete 25 full turns?
Answer by checkley77(1780) (Show Source): |
Question 153687: Please help me solve the following word problem (note: this problem came straight from a worksheet):
An object is dropped from an initial height of s feet. The object's height at any time t, in seconds, is given by h= -16t^2 + s. How long does it take for an object dropped from 300 feet to hit the ground? Round your result to two decimal places.: Please help me solve the following word problem (note: this problem came straight from a worksheet):
An object is dropped from an initial height of s feet. The object's height at any time t, in seconds, is given by h= -16t^2 + s. How long does it take for an object dropped from 300 feet to hit the ground? Round your result to two decimal places.
Answer by josmiceli(1851) (Show Source): |
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