A cowboy in an old western movie is rescuing a damsel in distress tied to a trai

Question

A cowboy in an old western movie is rescuing a damsel in distress tied to a train track, but a train is on the way! His partner has his ear to the iron train tracks, listening for the train. (Iron has a Young's modulus of 21×1010Pa and a density of 7.874g/cm3 .)  

His partner is first able to hear the sound from the train when it is   11 km    away. How long does it take for this sound from the train to reach his partner?

Part A

Part B

If the train is moving at constant speed   21.1 m/s   , how much time does the cowboy have from when his partner warns him to rescue the damsel and get safely out of the way?

Part C

Unfortunately, the actor playing the parter went on strike. The director decides to have the cowboy listen for the train as he is freeing the damsel. Since the sound dissipates more in air than in the solid tracks, the cowboy can't hear the train until it whistles when it is only   2.2 km    away. If the air temperature is   20.5 C    , how long does it take for the sound to reach the cowboy, propagating through the air? Use =1.40 , and the molar mass of air is 28.8103kg/mol .

Part D

How long does the cowboy have to rescue the damsel between when he hears the train and when it reaches him?

Explanation / Answer

Part A

The expression for the speed of sound in a medium matter is

. V2 = B/ d

B      is the volume module      B = 21 1010 Pa

D      is the density        d= 7.874 g/cm3   1 Kg/1000 g   (100 cm/1 m)3 = 7874 Kg/m3

I believe that exercise there is an error because the speed of sound is related to the module of volume

. V2 = 21 1010/ 7874          V= 5164 m/s

Waves travel at constant speed, therefore with kinematics we can find the time

V=d/t       t= d/V         t= 11 103 /5164       t= 2.13 s

.   t= 3.5 10-2 min

Part b

We calculate the time the train takes to go 11 103 m

VT = d/ tT          tT = d/VT

. tT = 11 103/21.1                    tT= 521.33 s

As the sound slow ts = 2.13 s to get to them, the time to save the maid is the difference between the time the train takes in less time that it takes the sound

Dt = tT –ts     Dt = 521.33 -2.13

Dt = 519.2 s                Dt = 8.65 min

Part C

Y= 1.4

. d= 28.8 10-3 Kg/m3

. T= 20.5°C   

. d2= 2.2 103 m

We use the expression of the speed of air as a function of temperature

V = 331 sqrt( 1+ T/273)                             sqrt = Ö

Vs= 331 sqrt (1+ 20/273)            Vs= 331   1.036

Vs= 342.9 m/s

Vs= d2/t2          t2= d2/Vs              t2= 2.2 103 / 342.9     

. t2= 6.42 s         t2= 1.07 10-2 min

Part D

We calculate the time that it takes the train to travel this distance d2

VT = d2/t2T       t2T = d2/VT              t2T = 2.2 103 /21.1        t2T = 104.27 s

The time remaining for the rescue is

Dt2 = t2T –t2s     Dt2 = 104.27 – 6.42         Dt2 = 97.85 s

Dt2 = 1.63 min

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