Problem:
A cow is standing 5 feet from the middle of the bridge.
A train is coming towards the cow at speed of 90 miles per hour from the near end
(Note #1) Distance between the near end of the bridge and the train is twice as long as the bridge.
If the cow had tried to move forward in the same direction as of the train then the cow would have got hit by (Note #2) three inches.
But it moved in opposite direction to the train and saved by one feet.
Then What is the length of the bridge and The velocity of the cow?
This problem is misquoted in many websites and forums.
#1 "Distance between the near end of the bridge and the train is twice as..." is misquoted as "the train is twice as..."
#2. "three inches" is misquoted as "three feet" or "one feet".
My source is Mathematical puzzles of Sam Loyd, Volume 2 By Sam Loyd, Martin Gardner Page 117. I saw this version in Google Book Review of this book.
Step by Step solution:
T...........................1...A........C...M........2...B
T = Position of the train just before cow moved
C = Position of the cow just before cow moved
A = Nearer (to train and cow) end of Bridge
B = Far (to train and cow) end of the Bridge
1 = Position of the train just after cow escaped. [Note: Cow did not go to 1; it escaped at A] (If cow moved towards train)
2 = Position of the train just after cow was hit (If cow moved towards far end)
M = Mid point of Bridge
Cv = Speed (V=Velocity) of the cow
Tv = Speed (V=Velocity) of the train
Let us assume length of the bridge AB is 2x
Assuming 2x rather than x makes it easier to calculate, as we need to deal with midpoint based calculation.
And, let us calculate distance traveled
T1 = TA - A1 = 2 times AB - 1 = (4x - 1)
T2 = TA + A2 = 2 times AB + AB - B2 = 3 times AB - 0.25 = 6x - 0.25
Similarly
CA = AM - CM = (x - 5) = x - 5 [Note: Cow did not go to 1; it escaped at A]
C2 = CB - B2 = CM + MB - B2 = 5 + x - 0.25 = x + 4.75
Time taken by Cow / Train is same in a given option.
CA / Cv = T1 / Tv ...(1)
C2 / Cv = T2 / Tv ....(2)
Adding (1) and (2)
(C2 + CA) / Cv = (T1 + T2) / Tv
i.e.
Cv = Tv * (CA + C2) / (T1 + T2)
Cv = 90 * (2x - 0.25) / (10x - 1.25) = 90 / 5 = 18 mph is the speed of the cow
Subtracting (1) and (2)
(C2 - CA)/Cv = 9.75ft/18mph = (T2 - T1)/Tv = (2x + 0.75) ft/90mph
i.e. 2x + 0.75 = 48.75 ft or 2x = 48 ft is the length of the bridge AB
Variation 1: CM is not given and asked to find out just Cv.
There is no change in Train related calculation. But
CA = AM - CM = (x - d) = x - d
C2 = CB - B2 = CM + MB - B2 = d + x - 0.25 = x + d - 0.25
So, C2 + CA is still the same; 2x - 0.25. So, Cv is still 18
A cow is standing 5 feet from the middle of the bridge.
A train is coming towards the cow at speed of 90 miles per hour from the near end
(Note #1) Distance between the near end of the bridge and the train is twice as long as the bridge.
If the cow had tried to move forward in the same direction as of the train then the cow would have got hit by (Note #2) three inches.
But it moved in opposite direction to the train and saved by one feet.
Then What is the length of the bridge and The velocity of the cow?
This problem is misquoted in many websites and forums.
#1 "Distance between the near end of the bridge and the train is twice as..." is misquoted as "the train is twice as..."
#2. "three inches" is misquoted as "three feet" or "one feet".
My source is Mathematical puzzles of Sam Loyd, Volume 2 By Sam Loyd, Martin Gardner Page 117. I saw this version in Google Book Review of this book.
Step by Step solution:
T...........................1...A........C...M........2...B
T = Position of the train just before cow moved
C = Position of the cow just before cow moved
A = Nearer (to train and cow) end of Bridge
B = Far (to train and cow) end of the Bridge
1 = Position of the train just after cow escaped. [Note: Cow did not go to 1; it escaped at A] (If cow moved towards train)
2 = Position of the train just after cow was hit (If cow moved towards far end)
M = Mid point of Bridge
Cv = Speed (V=Velocity) of the cow
Tv = Speed (V=Velocity) of the train
Let us assume length of the bridge AB is 2x
Assuming 2x rather than x makes it easier to calculate, as we need to deal with midpoint based calculation.
And, let us calculate distance traveled
T1 = TA - A1 = 2 times AB - 1 = (4x - 1)
T2 = TA + A2 = 2 times AB + AB - B2 = 3 times AB - 0.25 = 6x - 0.25
Similarly
CA = AM - CM = (x - 5) = x - 5 [Note: Cow did not go to 1; it escaped at A]
C2 = CB - B2 = CM + MB - B2 = 5 + x - 0.25 = x + 4.75
Time taken by Cow / Train is same in a given option.
CA / Cv = T1 / Tv ...(1)
C2 / Cv = T2 / Tv ....(2)
Adding (1) and (2)
(C2 + CA) / Cv = (T1 + T2) / Tv
i.e.
Cv = Tv * (CA + C2) / (T1 + T2)
Cv = 90 * (2x - 0.25) / (10x - 1.25) = 90 / 5 = 18 mph is the speed of the cow
Subtracting (1) and (2)
(C2 - CA)/Cv = 9.75ft/18mph = (T2 - T1)/Tv = (2x + 0.75) ft/90mph
i.e. 2x + 0.75 = 48.75 ft or 2x = 48 ft is the length of the bridge AB
Variation 1: CM is not given and asked to find out just Cv.
There is no change in Train related calculation. But
CA = AM - CM = (x - d) = x - d
C2 = CB - B2 = CM + MB - B2 = d + x - 0.25 = x + d - 0.25
So, C2 + CA is still the same; 2x - 0.25. So, Cv is still 18
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