Question

An engineer in a locomotive sees a car stuck

Answer 1

Answer:

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

Answer:

$a = -2.04 m/s^2$

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

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

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

$a = -2.04 m/s^2$