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mixer [17]
2 years ago
7

A train starts from rest at station A and accelerates at 0.6 m/s^2 for 60 s. Afterwards it travels with a constant velocity for

25 min. It then decelerates at 1.2 m/s^2 until it is brought to rest at station B. Determine the distance between the stations.
Engineering
1 answer:
Aleks [24]2 years ago
7 0

Answer:

The distance between the station A and B will be:

x_{A-B}=55.620\: km  

Explanation:

Let's find the distance that the train traveled during 60 seconds.

x_{1}=x_{0}+v_{0}t+0.5at^{2}

We know that starts from rest (v(0)=0) and the acceleration is 0.6 m/s², so the distance will be:

x_{1}=\frac{1}{2}(0.6)(60)^{2}

x_{1}=1080\: m

Now, we need to find the distance after 25 min at a constant speed. To get it, we need to find the speed at the end of the first distance.

v_{1}=v_{0}+at

v_{1}=(0.6)(60)=36\: m/s

Then the second distance will be:

x_{2}=v_{1}*1500

x_{2}=(36)(1500)=54000\: m        

The final distance is calculated whit the decelerate value:

v_{f}^{2}=v_{1}^{2}-2ax_{3}

The final velocity is zero because it rests at station B. The initial velocity will be v(1).

0=36^{2}-2(1.2)x_{3}

x_{3}=\frac{36^{2}}{2(1.2)}  

x_{3}=540\: m

Therefore, the distance between the station A and B will be:

x_{A-B}=x_{1}+x_{2}+x_{3}  

x_{A-B}=1080+54000+540=55.620\: km  

I hope it helps you!

 

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1. For ball bearings, determine: (a) The factor by which the catalog rating (C10) must be increased, if the life of a bearing un
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Answer:

(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.

Equation 11-1: F*L^(1/3) = Constant

Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483

Explanation:

(a)The Catalog rating(C)

   Bearing life:L_1 = L , L_2 = 2L

   Catalog rating: C_1 = C , C_2 = ? ,

From given equation bearing life equation,

F\times\frac{1}{3} (L_1)  = C_1   ...(1)   \\\\    F\times\frac{1}{3} (L_2)  =C_2...(2)

we Dividing eqn (2) with (1)

\frac{C_2}{C_1} =\frac{1}{3}  (\frac{L_2}{L_1})\\\\ C_2 = C*(\frac{2L}{L})\frac{1}{3} \\\\   C_2 = 1.26 C

The Catalog rating increased by factor of 1.26

(b) Reliability Increase from 0.9 to 0.99

R_1 = 0.9 , R_2 = 0.99

Now calculating life adjustment factor for both value of reliability from Weibull parametres

a_1 = x_o + (\theta - x_o){ ln(\frac{1}{R_1} ) }^{\frac{1}{b}}

= 0.02 + 4.439{ ln(\frac{1}{0.9} ) }^{\frac{1}{1.483}} \\\\     = 0.02 + 4.439( 0.1044 )^{0.67}\\\\a_1 = 0.9968

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a_2 = x_o + (\theta - x_o){ ln(\frac{1}{R_2} ) }^{\frac{1}{b} }\\\\   = 0.02 + 4.439{ ln(1/0.99) }^{\frac{1}{1.483} }\\\\     = 0.02 + 4.439( 0.0099 )^{0.67}\\\\a_2 = 0.2215

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Catalog rating increased by factor of 0.61

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