Diff. Avg Speeds

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Diff. Avg Speeds

by kanha81 » Wed Apr 01, 2009 4:23 pm
I always get this type of question wrong. It's like when I see this kind of question, I forget how I should solve! What's a good approach?

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?

A. 2 hours
B. 4 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours

Thank you.
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by kaf » Wed Apr 01, 2009 4:40 pm
I think the answer is B this is how i solve it though


D = RT

After 3 hours there would be a distance of 18 miles between the two cyclist
Distance covered by first cyclist

D = 6*3 = 18

The difference in speeds at which the two are cycling is 4 mph

Therefore we assume Second cyclist has to cover a distance of 18 miles at a speed of 4 mph to catch up with the first cyclist.

Hence D = RT
18 = 4T
18 / 4 = T
4/2/4 = T
4 / 1/2 =T

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by vittalgmat » Wed Apr 01, 2009 11:55 pm
At 6mph, in 3 hrs, the first cyclist would have travelled 3*6 = 18 miles.

The second cyclist will take 18/(10-6) = 4.5 hrs to catchup with the first cyclist.

The procedure to solve the "catchup" type problems is below.


“Catch up” Problem: Two objects (A: r1 , B: r2) moving in same direction; A faster than B (r1 > r2 ) but B ahead of A.
Procedure:
• Find the catch up rate: ie difference of the two rates: r1 - r2
• Find catch up distance: ie the distance by which A is behind B.
• Divide the distance by catchup rate : ie d / (r1-r2) to get time (t) taken for A to catch up with B (catchup time)
• If the question asks how long it would take to overtake by x miles. Then the prev cal will be modified to have total distance = (catchup distance + overtake distance).
• If the question asks how far had A travelled when A and B meet and if A started x hrs earlier than B, then A’s distance = (x*r1)+catchup distance.


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Re: Diff. Avg Speeds

by dtweah » Sun Apr 05, 2009 3:41 am
kanha81 wrote:I always get this type of question wrong. It's like when I see this kind of question, I forget how I should solve! What's a good approach?

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?

A. 2 hours
B. 4 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours

Thank you.
One thing clear is they would have travelled the same distance by the time second cyclist catches up. So equate both distance but remember the second guy starts 3 hours behind.

Let t be time the first cyclist starts, then t-3 be time of 2nd cyclist. So

10(t-3)=6t
t=7.5. So from the time the second guy starts 7.5 hrs would have elasped, but remember he starts 3 hours later. So 7.5-3= 4.5hrs.

The second probably better or easier way in which you don't have to worry about subtracting elasped time is to let t be the time of the 2nd cyclist. Then t+3 woud be time of first.
10t=6(t+3)
t=4.5

Hope this helps.