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 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

Answer 1

Step-by-step 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.

Answer 2

`a = -2.04 m/s^2`

Step-by-step 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

`d = vt`

`d = 25(0.68)`

`d = 17 m`

now the distance remaining from the position of the car

`x = 170 - 17`

`x = 153 m`

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

`v_f^2 - v_i^2 = 2 a d`

`0 - 25^2 = 2a(153)`

`a = -2.04 m/s^2`