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:


Answer:
I am in 6th grade why am i in high school things.
Explanation:
Answer:

Just draw a line from point D join to point E
The triangle formed DME will be congruent to AMC
Answer:
41.5° C
Explanation:
Given data :
1025 steel
Temperature = 4°C
allowed joint space = 5.4 mm
length of rails = 11.9 m
<u>Determine the highest possible temperature </u>
coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C
Applying thermal strain ( Δl / l ) = ∝ * ΔT
( 5.4 * 10^-3 / 11.9 ) = 12.1 * 10^-6 * ( T2 - 4 )
∴ ( T2 - 4 ) = ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6
hence : T2 = 41.5°C
Answer:
See attachment for detailed answer.
Explanation: