Answer:
The deceleration must have the engineer to avoid the accident is
a=-5.238 [tex]\frac{m}{s^{2} }[/tex]
Explanation:
[tex]x_{0}=100m\\v_{0}=30 \frac{m}{s} \\t=0.47 s[/tex]
While the engineer reacts the train continue moving so
[tex]x_{f} = v*t= 30\frac{m}{s} *0.47s= 14.1 m[/tex]
[tex]x_{t}= x_{o}+x_{f}\\x_{t}= 100m-14.1m=85.9m[/tex]
Now the final velocity have to be zero so using equation can find deceleration
[tex]V_{f} ^{2} =V_{o} ^{2}+2*a*x_{f}\\ 0= V_{o} ^{2}+2*a*x_{f}\\a=-\frac{V_{o} }{2*x_{f}}\\a=-\frac{(30\frac{m}{s}) ^{2} }{2*85.9m} \\a=-5.238 \frac{m }{s^{2} } } \\[/tex]
