Question

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?

Answer 1

The deceleration must have the engineer to avoid the accident is

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

Step-by-step explanation:

`x_(0)=100m\\v_(0)=30 (m)/(s) \\t=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)\\x_(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)\\  0= V_(o) ^(2)+2*a*x_(f)\\a=-(V_(o) )/(2*x_(f))\\a=-((30(m)/(s)) ^(2) )/(2*85.9m) \\a=-5.238 (m )/(s^(2) ) } \\`