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:
[tex]a = -2.04 m/s^2[/tex]
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
[tex]d = vt[/tex]
[tex]d = 25(0.68)[/tex]
[tex]d = 17 m[/tex]
now the distance remaining from the position of the car
[tex]x = 170 - 17[/tex]
[tex]x = 153 m[/tex]
now we know that train must stop with in the range of above distance
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 25^2 = 2a(153)[/tex]
[tex]a = -2.04 m/s^2[/tex]
