Question

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 13 m/s.
If the engineer’s reaction time is 0.46 s,

What should be the magnitude of the minimum deceleration to avoid an accident?
Answer in units of m/s²

Answer 1

`-0.44 m/s^2`

Step-by-step explanation:

First of all, we need to calculate the distance covered by the locomotive during the reaction time, which is

t = 0.46 s

During this time, the locomotive travels at

v = 13 m/s

And the motion is uniform, so the distance covered is

`d_1 = vt = (13)(0.46)=6.0 m`

The locomotive was initially 200 m from the crossing, so the distance left to stop is now

`d=200 - 6.0 = 194.0 m`

And now the locomotive has to slow down to a final velocity of
`v=0` in this distance. We can find the minimum deceleration needed by using the suvat equation:

`v^2 - u^2 = 2ad`

where

v = 0 is the final velocity

u = 13 m/s is the initial velocity

a is the deceleration

d = 194.0 m is the distance to stop

Solving for a,

`a=(v^2-u^2)/(2d)=(0^2-13^2)/(2(194))=-0.44 m/s^2`