Answer:
a)
, b)
, c) 
Explanation:
a) The deceleration experimented by the commuter train in the first 2.5 miles is:
![a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5B%2815%5C%2C%5Cfrac%7Bmi%7D%7Bh%7D%20%29%5Ccdot%20%28%5Cfrac%7B5280%5C%2Cft%7D%7B1%5C%2Cmi%7D%20%29%5Ccdot%20%28%5Cfrac%7B1%5C%2Ch%7D%7B3600%5C%2Cs%7D%20%29%5D%5E%7B2%7D-%5B%2850%5C%2C%5Cfrac%7Bmi%7D%7Bh%7D%20%29%5Ccdot%20%28%5Cfrac%7B5280%5C%2Cft%7D%7B1%5C%2Cmi%7D%20%29%5Ccdot%20%28%5Cfrac%7B1%5C%2Ch%7D%7B3600%5C%2Cs%7D%20%29%5D%5E%7B2%7D%7D%7B2%5Ccdot%20%282.5%5C%2Cmi%29%5Ccdot%20%28%5Cfrac%7B5280%5C%2Cft%7D%7B1%5C%2Cmi%7D%20%29%7D)

The time required to travel is:


b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be
.
c) The final constant deceleration is:


The answer is A. Immediately inform her colleague
Answer:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T*<em> t </em>)
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S = 
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M
Answer:
Explanation:
Fw= y looking symbol sea water (2010.62-168171.2727/h+32.935)ft^3