Physics
posted by Danny on .
A train traveling at a speed of 110 km/h uniformly decelerates for 20 seconds. During that time it travels 450m.
A) What is the acceleration? (m/s^2)
B) What is its speed after deceleration? (km/h)
A ANS = .805m/s^2
B ANS = 52km
s = 450m
t = 20 seconds
u = 110km/h*1000m/km /60s/min*60 min/h
u = 110000/ 3600
u = 30.55
A) s = ut + 1/2 at^2
a = s  ut/ 1/2 * t^2
a = 450m  30.55 * 20 sec/ 1/2 * 20^2
a = 450  611/ 1/2 * 400
a = 161/ 200
a = .805m/s^2
I get stuck trying to figure out how to use the formula to get the answer to B which is 52 km.
I believe I would start with the formula: s = v + u/ 2 *t.
I would really appreciate the help with the second part of the question.
Thank you.

v= v₀+at=30.550.805•20=14.5 m/s =14.5•3600/100=52.04 km/h