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?
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We are working in km/hr; change 5 min to hrs: 5/60 = {{{1/12}}} hr
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Let t = time required by the 2nd jogger to catch up with the 1st
then
(t+{{{1/12}}}) = travel time of the 1st jogger
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When the 2nd jogger catches up with 1st, they will have traveled the same distance.
Write a distance equation: Dist = speed * time
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10t = 8(t+{{{1/12}}})
10t = 8t + {{{8/12}}})
10t - 8t = {{{2/3}}}
2t = {{{2/3}}}
t = {{{1/3}}} hr which is 20 min
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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?
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Let d = distance one-way to return in 7/4 hrs
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Write a time equation: time = {{{Distance/speed}}}
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{{{d/5}}} + {{{d/2}}} = {{{7/4}}}
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Get rid of the denominators, multiply equation by 20:
20*{{{d/5}}} + 20*{{{d/2}}} = 20*{{{7/4}}}
Cancel out the denominators and you have:
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4d + 10d = 5(7)
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14d = 35
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d = {{{35/14}}}
d = 2.5 mi he needs to row down stream to return in 7/4 hrs
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 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
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Let s = speed of plane A
Then
(s-90) = speed of plane B
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Write a time equation: Time = distance/speed

Plane B's time - Plane A's = 2.25 hr
{{{900/((s-90))}}} - {{{900/s}}}} = 2.25
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Get rid of the denominators, multiply equation by s(s-90)
s(s-90)*{{{900/((s-90))}}} - s(s-90)*{{{900/s}}} = 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
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Arrange as a quadratic equation:
2.25s^2 - 202.5s - 81000
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Simplify, divide equation by 2.25
s^2 - 90s - 36000
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This factors to:
(s-240)(s+150) = 0
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s = -150
and
s = +240 km/hr is our solution
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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