shilpi84
asked on
cow and the bridge
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 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 be three feet. But it moved in opposite direction to the train and saved by one feet. Then What is the length of the bridge?
David_Ward
16ft
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how
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https://www.experts-exchange.com/questions/21082175/Cow-and-Bridge.html
In ath above solution assumed that...
d is the distance of train from the bridge....
x is the near distansr so far distanse in x + 10
Center of bridge is x + 5 and length will be 2x + 10;
Assume cows speed is c feet per sec so time taken is (x + 7)/c seconds
which is equivalent to (2x + 7 + d)/132
(x + 7)/c = (2x + 7 + d)/132
132x + 924 = 2cx + 7c + 7d
(2c - 132)x + 7c + 7d = 924
Another condition
IF cow moves in opposite to train it is safe be 1 feet means mov
(d - x - 1)/132 = x/c
c = 132x/(d - x - 1)
(x + 7)/c = (2x + 7 + d)/132
(x + 7)(d - x - 1)/132x = (2x + 7 + d)/132
(x + 7)(d - x - 1) = x(2x + 7 + d)
dx + 7d -xX - 7x - x - 7 = 2xx + 7x + dx
7d - 7 = 3xx + 15x
3x(x+5) = 7(d - 1)
x > 0 && d > 0 so
3x(x+5) = Multiple of 7
so x of (x+5) should be multiple of
so value so x can be .... 7,14,21,28...
or 2,9,16,23...
length of bridge is 2x + 10 ss it can be 24,38,52,66.....
or 14, 28, 42 .....
d is the distance of train from the bridge....
x is the near distansr so far distanse in x + 10
Center of bridge is x + 5 and length will be 2x + 10;
Assume cows speed is c feet per sec so time taken is (x + 7)/c seconds
which is equivalent to (2x + 7 + d)/132
(x + 7)/c = (2x + 7 + d)/132
132x + 924 = 2cx + 7c + 7d
(2c - 132)x + 7c + 7d = 924
Another condition
IF cow moves in opposite to train it is safe be 1 feet means mov
(d - x - 1)/132 = x/c
c = 132x/(d - x - 1)
(x + 7)/c = (2x + 7 + d)/132
(x + 7)(d - x - 1)/132x = (2x + 7 + d)/132
(x + 7)(d - x - 1) = x(2x + 7 + d)
dx + 7d -xX - 7x - x - 7 = 2xx + 7x + dx
7d - 7 = 3xx + 15x
3x(x+5) = 7(d - 1)
x > 0 && d > 0 so
3x(x+5) = Multiple of 7
so x of (x+5) should be multiple of
so value so x can be .... 7,14,21,28...
or 2,9,16,23...
length of bridge is 2x + 10 ss it can be 24,38,52,66.....
or 14, 28, 42 .....
CAN YOU CLARIFY, WHERE COULD THE TRAIN BEING TWICE THE LENGTH OF THE BRIDGE COME INTO IT?? THERE IS APPARENTLY A MISSING CONSTRAINT
cow is x feet from nearest end of bridge, so sets off to this at velocity v, walk takes x/v seconds, train velocity is 132 feet per second, so it reaches a point R from it's origin
R= 132(x/v) (R is one foot short of the bridge) (1)
now going the other way,
cow walks 5 feet to middle, plus 5 feet to postion symmetrical with start, plus x - 3 to where it got hit, so x + 7 feet walked in time (x+7)/v
And this point is R + 1 + x + 5 + 5 + x-3 = R + 2x +8
132(x + 7)/v = R + 2x + 8 (2)
so substituting 1 into 2
132(x + 7)/v = 132 (x/v) + 2x + 8
so 924/v = 2x + 8
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difference between time of death and time of escape = (2x + 8)/132=(x + 7)/v - x/v = 7/v - leads to same equation
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My conclusion is that there are different ways of approaching the problem which lead to the same relation between length of bridge and cow velocity being found, it's easy to make a mistake in one method and appear to solve for x and v, but in fact there is no full solution.
In fact if you take a cow velocity of 1, then the bridge is 926 feet long, the cow takes 458 seconds to escape and the train has travelled 60456 feet. The other way the cow takes 465 seconds to die, the train has travelled 61380 feet, which is 924 feet further - which is what we wanted
At 6 feet per second for the cow, the bridge is 156 feet long, the cow escapes in 12 1/6 seconds and the train went 1606 feet. The other way the cow dies in 13.33333s and the train went 1760 feet, which again is the one foot short + length - 3 foot short.
The cow velocity, brdge length and train origin are all within the questions parameters
cow is x feet from nearest end of bridge, so sets off to this at velocity v, walk takes x/v seconds, train velocity is 132 feet per second, so it reaches a point R from it's origin
R= 132(x/v) (R is one foot short of the bridge) (1)
now going the other way,
cow walks 5 feet to middle, plus 5 feet to postion symmetrical with start, plus x - 3 to where it got hit, so x + 7 feet walked in time (x+7)/v
And this point is R + 1 + x + 5 + 5 + x-3 = R + 2x +8
132(x + 7)/v = R + 2x + 8 (2)
so substituting 1 into 2
132(x + 7)/v = 132 (x/v) + 2x + 8
so 924/v = 2x + 8
-----------------
difference between time of death and time of escape = (2x + 8)/132=(x + 7)/v - x/v = 7/v - leads to same equation
-----------------
My conclusion is that there are different ways of approaching the problem which lead to the same relation between length of bridge and cow velocity being found, it's easy to make a mistake in one method and appear to solve for x and v, but in fact there is no full solution.
In fact if you take a cow velocity of 1, then the bridge is 926 feet long, the cow takes 458 seconds to escape and the train has travelled 60456 feet. The other way the cow takes 465 seconds to die, the train has travelled 61380 feet, which is 924 feet further - which is what we wanted
At 6 feet per second for the cow, the bridge is 156 feet long, the cow escapes in 12 1/6 seconds and the train went 1606 feet. The other way the cow dies in 13.33333s and the train went 1760 feet, which again is the one foot short + length - 3 foot short.
The cow velocity, brdge length and train origin are all within the questions parameters
Train will hit from the from not from the back....no use of length of train