It takes a 9 wagon−wagon train 12 seconds to pass a person standing outside. how many hours will it take the wagon train to pass a route of 216 kilometers, if the length of each wagon is 16 meters?

Answers

i don't know if this is what you are needing but hoping it helps


Find the length of AC if AB = 16 and BC = 12.
Find the length of AC if AB = 16 and BC = 12.

change interval is 3.93 sec

clearance interval is 1.477 sec

Explanation:

Given data:

upgrade of intersection 4%

street total width at intersection is 60 ft

vehicle length of approaching traffic = 16 ft

speed of approaching traffic =40 mi/hr

85th percentile speed is calculated as

S_{85} = S +5

S_{ 85} = 40 + 5  = 45 mi/h

15th Percentile speed

S_{15} =S-5

          = 40 - 5 = 35 mi/hr

change in interval is calculated as

y = t + \frac{1.47 S_{85}}{2a +(64.4\times 0.01 G}

t is reaction time is 1.0,

deceleration rate is given as 10 ft/s^2

y = 1.0 +\frac{1.47\times 45}{2a +(64.4\times 0.01\times 4}

y = 3.93 s

clearance interval is calculated as

a_r = \frac{W+ L}{1.47\times S_{15}}

a_r = \frac{60+16}{1.47\times 35} = 1.477 s

  5 hours 12 seconds

Step-by-step explanation:

For the entire train to pass the person, the head of the lead wagon must travel (9·16 m) in 12 seconds. It is traveling at a speed of ...

  (9·16 m)/(12 s) = (144/12) m/s = 12 m/s

For the train to travel 216 km + 144 m = 216,144 m will take ...

  (216,144 m)/(12 m/s) = 18,012 s = 5 hours 12 seconds

Comment on the speed

The speed of this "wagon train" is about 26.8 miles per hour, faster than any human and most animals can run for long periods. The Pony Express cross-country mail delivery service maybe set a reasonable standard for cross-country travel. Its average speed was about 10 miles per hour.



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