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Two Similar Work/Speed Problems

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Two Similar Work/Speed Problems

by kakz » Sat Nov 05, 2011 9:07 am
1.While on a straight road, Car X and Car Y are traveling at different constant rates. If Car X is now 1 mile ahead of Car Y, how many minutes from now will Car X be 2 miles ahead of Car Y, if Car X is traveling at 50 miles per hour and Car Y is traveling at 40 miles per hour?

2.While on a straight road, Car X and Car Y are traveling at different constant rates. If Car X is now 1 mile ahead of Car Y, how many minutes from now will Car X be 2 miles ahead of Car Y, if three minutes ago Car X was 1/2 mile ahead of Car Y?

DS problem that I converted into PS...Getting very very confused with the calculations so kindly help[/b]
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by Anurag@Gurome » Sat Nov 05, 2011 9:38 am
kakz wrote:1.While on a straight road, Car X and Car Y are traveling at different constant rates. If Car X is now 1 mile ahead of Car Y, how many minutes from now will Car X be 2 miles ahead of Car Y, if Car X is traveling at 50 miles per hour and Car Y is traveling at 40 miles per hour?
As the cars are moving in the same direction, relative speed of the cars = (50 - 40) = 10 miles per hour.

Total distance between them cover = (2 - 1) = 1 mile

Hence, time required = 1/10 hour = 6 minutes
kakz wrote:2. While on a straight road, Car X and Car Y are traveling at different constant rates. If Car X is now 1 mile ahead of Car Y, how many minutes from now will Car X be 2 miles ahead of Car Y, if three minutes ago Car X was 1/2 mile ahead of Car Y?
In three minutes, car X is (1 - 1/2) = 1/2 miles ahead of car Y.
Hence, relative speed of the cars = 1/2 miles per 3 minute = 1 mile per 6 minute

Total distance between them cover = (2 - 1) = 1 mile

Hence, time required = 6 minutes
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by rijul007 » Sat Nov 05, 2011 9:41 am
1.While on a straight road, Car X and Car Y are traveling at different constant rates. If Car X is now 1 mile ahead of Car Y, how many minutes from now will Car X be 2 miles ahead of Car Y, if Car X is traveling at 50 miles per hour and Car Y is traveling at 40 miles per hour?
Speed of car X with respect to car Y = 10 miles per hour
Initial distance = 1 miles
Final = 2 miles

Time taken = 1/10 hrs = 6 mins

2.While on a straight road, Car X and Car Y are traveling at different constant rates. If Car X is now 1 mile ahead of Car Y, how many minutes from now will Car X be 2 miles ahead of Car Y, if three minutes ago Car X was 1/2 mile ahead of Car Y?
Relative speed of X w.r.t Y = v = (1/2)/(3/60) = 10 miles per hour
Time taken for X to be 2 mile ahead of Y = 1/10 hrs = 6 mins
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by kakz » Sat Nov 05, 2011 9:48 am
Question to Mr. Anurag
Sir if i will change small part of the problem i.e. instead of saying 2 miles I will say, How many minutes from now will Car X be 5 miles ahead of Car Y? Hope you can please show the solution for that for both problems.
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by Anurag@Gurome » Sat Nov 05, 2011 9:58 am
kakz wrote:Question to Mr. Anurag
Sir if i will change small part of the problem i.e. instead of saying 2 miles I will say, How many minutes from now will Car X be 5 miles ahead of Car Y? Hope you can please show the solution for that for both problems.
I hope you've noticed that in both the cases their relative speeds are determinable and equal. 1 mile per 6 minutes is same as 10 miles per hour.

Hence, whatever is the distance between them is to cover, it'll take same time for both the cases as follows.

Total distance between them to cover = (5 - 1) = 4 mile
Hence, time required = 4/10 hour = 24 minutes
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by user123321 » Sat Nov 05, 2011 8:01 pm
1)In one hour x can get ahead of y by 10 miles
so to get ahead by 2-1 = 1 mile, it will take 1/10 hr = 6 minutes

2)it took 3 minutes to get ahead 1/2 mile...
So to get a mile more ahead it will take 6 more minutes.

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