An engineer in a locomotive sees a car stuck
on the track at a railroad crossing in front of
the train. When the engineer first sees the
car, the locomotive is 200 m from the crossing
and its speed is 28 m/s.
If the engineer’s reaction time is 0.21 s,
what should be the magnitude of the minimum deceleration to avoid an accident?
Answer in units of m/s^2
Please help :)
d = 200 - Vo*T = 200 - 28*0.21 = 194 m. = Required stopping distance.
V^2 = Vo^2 + 2a*d = 0,
28^2 + 2a*194 = 0,
388a = -784,
a = -
To calculate the minimum deceleration required to avoid an accident, we need to determine the stopping distance and the time available for deceleration.
First, let's calculate the stopping distance:
The stopping distance can be obtained by using the equation of motion:
s = ut + (1/2)at^2
Where:
s = stopping distance
u = initial speed (28 m/s)
t = reaction time (0.21 s)
a = deceleration (unknown)
Since we are looking for the stopping distance, we set the final velocity to zero:
0 = 28 + at
Rearranging the equation, we get:
at = -28
Now, substitute the value of "at" into the equation of motion:
s = ut + (1/2)at^2
s = 28 * 0.21 + (1/2)(-28)(0.21)^2
Calculating the above expression gives us the stopping distance s.
Next, we need to calculate the distance between the locomotive and the crossing when the engineer applies the brakes:
distance = initial distance - stopping distance
distance = 200 m - s
Now, we can calculate the remaining time available for deceleration:
remaining time = reaction time - (distance / initial speed)
remaining time = 0.21 s - (distance / 28 m/s)
Finally, we can calculate the minimum deceleration needed to avoid an accident:
deceleration = (2 * distance) / (remaining time)^2
Plug in the values of distance and remaining time into the equation above to obtain the magnitude of the minimum deceleration required to avoid an accident.
Solving this step-by-step calculation will give you the answer in units of m/s^2.