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 ?
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