can u solve these problems for me its urgent?

Question

1. there are 15 poles at equal distance beside the railway track. if a train moves from 1st pole to 10th pole in 10 minutes. how much time it will take to touch 15th pole?

2. a cow was standing on a bridge. 5 feet away from the middle of the bridge. suddenly a lightning express with 90 miles/hr was coming towards the bridge from nearest end of the cow. seeing this the cow ran towards the express and managed to escape when the train is one feet away from the bridge. if it would have ran to opposite direction(ie away from train) it would have been hit the train one ft away from the end of the bridge. calculate the length of bridge.?

can u solve these puzzles tell the answer and the steps of the problems. plz solve it soon its urgent

plz answer with the steps

it is not my homework problem it was asked by my friend

Answer 1

1) 10/9 * 14 = 140/9 = 15.56 minutes = 15 minutes and 33 seconds (first pole to 15th pole)

2)
V1 = velocity of train = 90 ml/hr = 132 ft/s
V2 = velocity of cow (ft/s)
L = length of bridge
S1 = distance from train to bridge when cow sees train initially (ft)
S2 = distance from cows initial position to nearest end of bridge (ft)

L = 2*(S2+5)
S2 = (L/2) - 5

First equation:
(S1 - 1) / V1 = S2 / V2 = ((L/2) - 5) / V2
2*V2*(S1 - 1) = V1*(L - 10)
S1 = (L - 10)*(V1 / 2*V2) + 1

Second equation:
(S1 + L +1) / V1 = (S2 + 10) / V2 = ((L/2) + 5) / V2
2*V2*(S1 + L + 1) = V1*(L + 10)

3 variables V2, S1, L
2 equations

No unique solution

L = (V1 / V2)*10 - 2
L = (1320 / V2) - 2
in feet

That is as far as You can go without defining the problem further, ie cow's velocity or initial train distance.

Making the assumption that the cow is relatively smart (she did have the smarts to run toward the oncoming train afterall), lets say that the cows velocity is high for a cow. Lets say it is 13.2 ft/s even though it is running along a rail road bed.

then V2 = 13.2

therefore L = (1320 / 13.2) - 2 = 98 ft

Pretty smart cow.

Answer 2

1) It took 10 minutes to move 9 gaps and you want to know the time it takes to move 14 gaps. You want gaps times minutes per gap, or 14 * 10 / 9 which is about 15.4.

2) Let the length of the bridge be X. The cow traveled x/2-4 feet in one direction in the same amount of time as the train moved from wherever it started to 1 foot before the start of the bridge. If the cow had run away from the train, then when the train was 1 foot before the start of the bridge, the cow would have still traveled x/2-4 feet. Thus, the cow would have been 9 feet away from the end of the bridge. Thus, the train can cross the bridge in the same amount of time as the cow can travel 8 feet.
From this, you can't tell the length of the bridge. For example, suppose the cow was trotting along at 10 miles per hour. Since the train was moving 9 times faster, the bridge would be 72 feet long. Alternatively, suppose the cow was galloping at 18 miles per hour. Now the train is five times faster than the cow and the bridge is 40 feet long.

Answer 3

Problem 1.
there are 15 poles i.e there are 14 gaps. Now the train reached 10th pole from 1st within 10 seconds, means it passes 9 gaps in 10 seconds. Now the average speed of the train is 10/9 gaps per second. now to reach 15th pole, it has to cross 14 gaps and it will take 14*10/9 seconds or 140/9 seconds or 15.56 seconds.

Problem 2.
There are three unkonws.
How far the train was when the intelligent cow viewed the train.
The speed of cow
and the length of the bridge.
With given conditions, we can form two equations and thus the information you gave is incomplete.

With given condition, the lenght of the bridge is

{1320+2(speed of cow)-2(speed of cow)(the distance of the train at time 0)}/{(2(speed of the cow)-132} feet

or {2(speed of cow)(distance of train at time 0) - 2(speed of cow) +1320}/132 feet

:)

Answer 4

Simple
1. Pretend there is 15 poles that are each 1 meter apart. If the train is at a constant speed, the train would travel at 9 meters per minute. At 9 meters per minute, the train would actually be doing a meter per 6and2/3 of a second. 14 multiplied by 6and2/3 is 1 minute and 33and1/3 seconds.

2.Use trial and error to figure it out.

Answer 5

Sounds like one pole per minute right? 5 ft from the middle going toward the train and 4 ft forward and 5 ft backwards well what does that add up to?

Answer 6

1.The distance between 10 poles is like that of 9 intermediate distances between the poles. If speed's constant, time req to cover each intermediate distance=10/9.To touch 15th pole, train will have to cover 14 intermediate distances. That accounts to 14*10/9min=140*9=15.555min.
2. I am unable to solve the problem .Give its conversions in the SI System.

Answer 7

yes i can solve them

1. assuming the train's speed is constant, it took 10 minutes to go from 1 to 10, so it took 10/9 minute per pole, so it would take 50/9 minutes to go 5 more poles. that is 5.56 minutes.

2. the problem does not make sense, the bridge would have to be negative length

Answer 8

You've got runaway train going.

1. I hope the train doesn't hit the pole.
2. I hope the train doesn't crash into the bridge.

Answer 9

Answers:
1. 15 minutes
2. 10 feet

Answer 10

Yes I can, but I won't..too late you should have knowm the answers before this pitiful plea to a total stranger