SOLUTION: Ride the peaks. Smith bicycled 45 miles going east from
Durango, and Jones bicycled 70 miles. Jones averaged
5 miles per hour more than Smith, and his trip took one-half
hour lo
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Question 203472This question is from textbook
: Ride the peaks. Smith bicycled 45 miles going east from
Durango, and Jones bicycled 70 miles. Jones averaged
5 miles per hour more than Smith, and his trip took one-half
hour longer than Smith’s. How fast was each one traveling?
I am not sure if i am suppose to use the D=r*t.
I am confused on how to set it up.
I know that smith's distance is 45 miles
and Jones distance is 70 miles
I am confused on how to write the part where Johes goes 5 miles per hour more then Smith and took one half hour longer then smith.
This question is from textbook
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles.
Jones averaged 5 miles per hour more than Smith,
and his trip took one-half hour longer than Smith’s.
How fast was each one traveling?
:
Let s = D's speed
then
(s+5) = J's speed
:
Write a time equation: Time =
:
Smith's time = Jones' time - .5 hrs
= - .5
Multiply equation by s(s+5)
s(s+5)* = s(s+5)* - .5s(s+5)
Cancel out the denominators and you have:
45(s+5) = 70s - .5s^2 - 2.5s
:
45s + 225 = 70s - .5s^2 - 2.5s
arrange as a quadratic equation on the left
.5s^2 + 2.5s + 45s - 70s + 225 = 0
:
.5s^2 - 22.5s + 225 = 0
Multiply equation by 2, get rid of the decimals
s^2 - 45s + 450 = 0
Factors to:
(s - 15)(s - 30) = 0
Two good solutions
s = 15 mph is Smith's speed, then 20 mph is Jones' speed
and
s = 30 mph is Smith's speed, then 35 mph is Jones' speed
:
:
Check solutions, find the time for each trip
Smith: 45/15 = 3 hr
Jones: 70/20 = 3.5 hr
:
You can check the s=30 solution
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