Tramcars and a pedestrian
While walking along a tram line I noticed that a tramcar caught up with me every 12 minutes, and every 4 minutes a tramcar coming in the opposite direction passed me. We were both (tramcar and I) moving at a uniform rate.
What was the time interval between two tram cars leaving their terminals?
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3 solutions
Alternately:- Let d = distance between two consecutive tram.
Vt = vel. of tram,...V = vel. of man.
In 12 minutes....... d = distance traveled by tram - distance traveled by man...=...12Vt - 12V.
In 4 minutes...... d = distance traveled by tram + distance traveled by man...=... 4Vt - 4V.
Elemenating V,.... 4d = 24Vt <.........> d/Vt = 6 sec the required time.
Something doesn't compute! If 6 is correct then you will meet 3 trams in 6 minutes, thus 2 minutes between trams, not 4 as the problem states
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Can you explain how you concluded that "you will meet 3 trams in 6 minutes"?
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On further thought you will meet 2 trams in 6 minutes, thus making 3 minutes between trams. I am hampered in my explanation because I'm on my iPad.
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@Guiseppi Butel – My fuzzy thinking! You did meet 3 trams not in 6 minutes but in 12 minutes therefore I agree with the answer.
W e l l d o n e . . . ! ! ! B u t I t h i n k t h e r e i s a m i s t a k e i n y o u r s o l u t i o n . . . T h e t i m e t a k e n b y t r a m t o c o v e r t h e s a m e d i s t a n c e a s u h a v e c o v e r e d i s 4 x − 4 & u h a v e j u s t m a d e a s m a l l m i s t a k e b y w r i t i n g ′ x ′ i n s t e a d o f 4 i n t h e d e n o m i n a t o r , i . e . , u h a v e w r i t t e n x x − 4 . S o , j u s t c o r r e c t i t . R e s t o f i t i s a w e s o m e . . .
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Thanks for correcting my mistake.
how to write this ?
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http://s1.daumcdn.net/editor/fp/service nc/pencil/Pencil chromestore.html
Why is the train you are meeting covering the distance you walked. It is additive not subtractive.
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Let's say I meet an incoming tram at point A. The next incoming tram is at B. I meet the next incoming tram at C.
It takes
x
minutes for an incoming tram to travel from B to A.
It takes 4 minutes for an incoming tram to travel from B to C.
It takes
x
−
4
minutes for an incoming tram to travel from C to A.
It takes me 4 minutes to travel from A to C.
Hence, the relative speeds is 4 x − 4 . Note that as pointed out by @Vighnesh Raut , the fraction in the second paragraph should be 4 x − 4 , and I've updated it accordingly.
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Thanks for replying Calvin. I appreciate it. The question then is why is x now the period the tram rests at the station? This is really saying that it takes 6 minutes for the tram to drive the distance between me and it (when another tram just met me). I guess my logic functions work differently :).
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@Hatim Zaghloul – As defined in the solution, "tramcars leave their terminals every x minutes". There are multiple tramcars that run along the line (think of a bus route). It is not that the tramcar will complete one route in x minutes.
Hence, if an incoming tram is at point A, and the next incoming tram is at point B, then it will take that tram x minutes to travel from B to A.
However, if I were walking, then it will take me 4 minutes to walk from A to C, and it will also take the tramcar 4 minutes to travel from B to C, and hence we will meet in 4 minutes at C.
Something isn't right with this problem, if the time between the first tram which pass beside me and the second is 12min, the same can't be said about the time between the 2 and 3, as the v is constant and the distance change(because i'm walking, i'm not just waiting for the tram) so the time changed, there is no doubt about that unless trams don't start at regular intervals, pls correct me if i'm wrong
let the distance(120 m) let ( v = 1 m/s ) you will meet along the path (10+30 = 40 car) divided by 2 (for going only= 20 car) divided by (120 m) you have (120/20 = 6) minutes for each car.
My calculations are very simple: With 12mts and 4 mts time crossing the speeds of trams are 5 Km and 15 Kms/hr in same and opposite directions .While crossing each other this will be 15-5 10Km/hr that is 6 mts.Ans
K.K.GARG,India
but y 5 km ? doesnt make sense :(
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Wew need to convert interval in to speed in Km/Hr that is how it will be converted in to relative speeds.Thanks.
K.K.GARG.India
What is the logic behind relative speeds??
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Trams are in opposite directions,as pedestrian my speed is similar to tram travelling in same direction I am walking,so basically considering speeds of trams cars in Km/hr ,that is 15 -5 =10 Km/Hr relative speed is calculated and then converted in minutes that is 6 mts interval.Hope this satisfies your querry.Thanks.
K.K.GARG,India
If the tramcars leave their terminals every x minutes, that means that x minutes after I have met a tramcar the following car arrives at that spot. If it is catching up with me then during the remaining 12-x minutes,it has to cover the same distance that I do in 12 minutes. Thus the distance that I cover in one minute is covered by the train in 1 2 1 2 − x minutes.
Now if the tram is coming towards me then it will meet me 4 minutes after the preceding one and during the remaining 4-x minutes, it will cover the same distance that I do in 4 minutes.This means that the distance I cover in one minute is covered by the tram in 4 x − 4 minutes.
We get the equation,
1 2 1 2 − x = 4 x − 4
which yields x=6.Hence the tramcars start out every 6 minutes.