SOLUTION: 1)A jogger starts a course at a steady rate of 8 kph.Five minutes later,a second jogger starts the same course at 10 kph.How long will it take the second jogger to catch the first?

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Question 122620: 1)A jogger starts a course at a steady rate of 8 kph.Five minutes later,a second jogger starts the same course at 10 kph.How long will it take the second jogger to catch the first?
2)A man rows downstream at the rate of 5 mph and upstream at the rate of 2 mph.How far downstream should he go if he is to return in 7/4 hourse after leaving?
3)Two planes leave Manila for a southern city,a distance of 900 km.Plane A travels at a ground speed of 90 kph faster than the plane B.Plane A arrives in their destination 2 hours and 15 minutes ahead of Plane B.What is the ground speed of Plane A?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
1)A jogger starts a course at a steady rate of 8 kph.Five minutes later,a second jogger starts the same course at 10 kph.How long will it take the second jogger to catch the first?
:
We are working in km/hr; change 5 min to hrs: 5/60 = 1%2F12 hr
:
Let t = time required by the 2nd jogger to catch up with the 1st
then
(t+1%2F12) = travel time of the 1st jogger
:
When the 2nd jogger catches up with 1st, they will have traveled the same distance.
Write a distance equation: Dist = speed * time
:
10t = 8(t+1%2F12)
10t = 8t + 8%2F12)
10t - 8t = 2%2F3
2t = 2%2F3
t = 1%2F3 hr which is 20 min
:
:
2)A man rows downstream at the rate of 5 mph and upstream at the rate of 2 mph. How far downstream should he go if he is to return in 7/4 hours after leaving?
:
Let d = distance one-way to return in 7/4 hrs
:
Write a time equation: time = Distance%2Fspeed
:
d%2F5 + d%2F2 = 7%2F4
:
Get rid of the denominators, multiply equation by 20:
20*d%2F5 + 20*d%2F2 = 20*7%2F4
Cancel out the denominators and you have:
:
4d + 10d = 5(7)
:
14d = 35
:
d = 35%2F14
d = 2.5 mi he needs to row down stream to return in 7/4 hrs
:
:
3)Two planes leave Manila for a southern city,a distance of 900 km. Plane A travels at a ground speed of 90 kph faster than the plane B. Plane A arrives in their destination 2 hours and 15 minutes ahead of Plane B. What is the ground speed of Plane A?
Change 2 hr 15 min to hrs: 2.25 hrs
:
Let s = speed of plane A
Then
(s-90) = speed of plane B
:
Write a time equation: Time = distance/speed
Plane B's time - Plane A's = 2.25 hr
900%2F%28%28s-90%29%29 - 900%2Fs} = 2.25
:
Get rid of the denominators, multiply equation by s(s-90)
s(s-90)*900%2F%28%28s-90%29%29 - s(s-90)*900%2Fs = s(s-90)*2.25
cancel out the denominator and you have:
900s - 900(s-90) = 2.25(s^2 - 90s
900s - 900s + 81000 = 2.25s^2 - 202.5s
:
Arrange as a quadratic equation:
2.25s^2 - 202.5s - 81000
:
Simplify, divide equation by 2.25
s^2 - 90s - 36000
:
This factors to:
(s-240)(s+150) = 0
:
s = -150
and
s = +240 km/hr is our solution
:
:
I checked all three of these solutions on my calc. The reason I did not include
the calculations here, is because of inexcusable laziness. A