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 270 m from the crossing
and its speed is 12 m/s.
if the engineer’s reaction time is 0.79 s,
what should be the magnitude of the minimum deceleration to avoid an accident?
answer in units of m/s
2
.

a=0.27637 m/s²

Explanation:

Hello,

You should remember that ;

Stopping distance=reaction distance + braking distance

First find the reaction distance which is that distance traveled from point of detecting danger to start of breaking process.

Formula: d=(s*r) where s is speed in m/s and r is reaction time in seconds

Given: s=12m/s  and r=0.79 s

d= 12*0.79=9.48 m

Remaining distance to cover and stop

Where distance from locomotive to railroad crossing is 260, then remaining distance will be;

270m-9.48m=260.52 m

To get magnitude of minimum deceleration to avoid an accident you apply the formula:

v²=u²-2as

where

v= final speed = 0 m/s

u=speed of locomotive = 12 m/s

a=acceleration in m/s²

s=remaining distance to cover

v²=u²-2as

0²=12²-2*a*260.52

0=144-521.04 a

144/521.04= 521.04/521.04a

a=0.27637 m/s²

Hope this Helps!

prolly a beacuse the momentuem the car has increases while going down hill