1.An engineer in a locomotive sees a car stuck on the track at a railroad crossi

Question

1.An engineer in a locomotive sees a car stuck
on the track at a railroad crossing in front of
the train. When the engineer ?rst sees the
car, the locomotive is 300 m from the crossing
and its speed is 23 m/s.
If the engineer’s reaction time is 0.6 s, what
should be the magnitude of the minimum deceleration to avoid an accident?
Answer in units of m/s^2

2.A driver in a car traveling at a speed of
32 mi/h sees a deer 105.9 m away on the
road.
Assuming the deer does not move, what is
the minimum constant acceleration necessary
for the car to stop without hitting the deer?
Answer in units of m/s^2

Explanation / Answer

Ans.1.
Initial velocity of the locomotive, u = 23 m/s

Final velocity of the locomotive, v = 0 m/s

Maximum distance for the locomotive to travel after reaction time passes, s = (300 - 0.6*23)m = 286.2 m

Let the minimum deceration be represented by a m/s2

Now, v2 = u2 + 2*a*s

Hence, a = -232/(2*286.2) m/s2 = -0.92 m/s2

Ans.2.
1 mi = 1.609344 km

32 mi = 51.5 km

51.5 km/h = 143.05 m/s = Initial velocity of the car, u

Final velocity of the car, v = 0 m/s

Displacement, s = 105.9 m

Let the minimum deceration be represented by a m/s2

v2 = u2 + 2*a*s

a = -143.052/(2*105.9) = -96.62 m/s2

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