1. What distance is required for a train to stop if its initial velocity is 23 m/s and its deceleration is 0.25 m/s^2? (Assume the train decelerates at a constant rate.) *
Your answer

x=?

dt=?

vi=23m/s

vf=0m/s (it stops)

d=0.25m/s^2

time =

vf=vi+d: 0=23m/s+(0.25m/s^2)t

t=92s

displacement=

vf^2=vi^2+2a(dx)

23^2=0^2+2(0.25m/s^2)x =-1058m

Explanation:

you can find time from vf = vi + a(Dt): 0 = 23 m/s + (0.25 m/s/s)t so t = 92 s and you can find the displacement from vf2 = vi2 + 2a(Dx) and find the answer in one step: 232 = 02 + 2(0.25 m/s/s)x so x = -1058 m

1058

Explanation:

This is the equation that is used : vf^2 = vi ^2 + 2ad

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