Answer:
a) [tex]t = 277.477\,s\,(4.625\min)[/tex], b) [tex]v_{f} = 0\,\frac{mi}{h}[/tex], c) [tex]a = -0.128\,\frac{ft}{s^{2}}[/tex]
Explanation:
a) The deceleration experimented by the commuter train in the first 2.5 miles is:
[tex]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} )}[/tex]
[tex]a = -0.185\,\frac{ft}{s^{2}}[/tex]
The time required to travel is:
[tex]t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }[/tex]
[tex]t = 277.477\,s\,(4.625\min)[/tex]
b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be [tex]v_{f} = 0\,\frac{mi}{h}[/tex].
c) The final constant deceleration is:
[tex]a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}[/tex]
[tex]a = -0.128\,\frac{ft}{s^{2}}[/tex]
