PHYSICS HIGH SCHOOL

When the engineer first sees the car, the locomotive is 100m from the crossing and its speed is 30m/s. If the engineer reaction time is 0.47 s what should the magnitude of the minimum deceleration to avoid an accident?

Answers

Answer 1

Answer:

Answer:

The deceleration must have the engineer to avoid the accident is

a=-5.238 $(m)/(s^(2) )$

Explanation:

$x_(0)=100m\nv_(0)=30 (m)/(s) \nt=0.47 s$

While the engineer reacts the train continue moving so

$x_(f) = v*t= 30(m)/(s) *0.47s= 14.1 m$

$x_(t)= x_(o)+x_(f)\nx_(t)= 100m-14.1m=85.9m$

Now the final velocity have to be zero so using equation can find deceleration

$V_(f) ^(2) =V_(o) ^(2)+2*a*x_(f)\n 0= V_(o) ^(2)+2*a*x_(f)\na=-(V_(o) )/(2*x_(f))\na=-((30(m)/(s)) ^(2) )/(2*85.9m) \na=-5.238 (m )/(s^(2) ) } \n$

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Answers

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Answers

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