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 170 m from the crossing
and its speed is 25 m/s.
if the engineer’s reaction time is 0.68 s,
what should be the magnitude of the minimum deceleration to avoid an accident?
answer in units of m/s^2

Explanation:

So we want the speed to go from 25 m/s to 0 m/s in 170 m, but the time  needs to incorporate the reaction time, so the slowing down will not start until .68 s pass.  Or, in other words, the train will travel an extra 25 m/s * .68 s = 17 m.  This means, instead of 170 m to slow down it has 153.  Hopefully that makes sense.  With this information we can use the equation vf^2-vi^2=2ad.  If that equation is unfamiliar you need to get a better handle on your physics equations.

Anyway, let's plug in.  

vf = 0 m/s

vi = 25 m/s

a is what we're trying to find

d = 153 m

vf^2-vi^2=2ad

a = (vf^2-vi^2)/(2d)

Can you handle figuring it out from there?  or if there is something you don't understand let me know.  

Explanation:

As we know that during reaction time the train will move with uniform speed

so here the distance moved by the train is given as

now the distance remaining from the position of the car

now we know that train must stop with in the range of above distance

answer: 0.20

explanation: this can only be the answew because all the other measuremnets are not close