Posted by **chrystabelle** on Sunday, June 19, 2011 at 4:03am.

A train passes a control tower with a velocity u and has a constant acceleration.At 800m beyond the tower,the velocity is 48km/h;and at 1.6km beyond the tower,the velocity is 64km/h,what is the velocity u ?

- physics -
**Damon**, Sunday, June 19, 2011 at 6:49am
V2 = 48000/3600 = 13.33 m/s

V3 = 64000/3600 = 17.78 m/s

average speed between points 2 and 3 =(13.33+17.78)/2

= 15.56 m/s

distance between points 2 and 3 =1600 -800 = 800 m

time between points 2 and 3 = 800/15.56 = 51.4 seconds

acceleration = a = change in speed over time

a = (17.78-13.33)/51.4 = .0866 m/s^2

now between point 0 and point 1

v = u + a t

13.33 = u + .0866 t or u = (13.33-.0866t)

x = 0 + ut +(1/2)at^2

800 = u t +(1/2)(.0866) t^2

so

800 = (13.33-.0866t)t +.0433 t^2

solve quadratic for t, go back and get u

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