Question
A train starts from rest at a station and accelerates at the rate of 1.4 m/s2. How long does it take the train to cover a 590 m distance?
Answers
Anonymous
v = Vi + a t
x = Xi + Vi t + (1/2) a t^2
Vi = 0, starts at rest
a = 1.4 m/s^2
so
distance = x - Xi = 590 = 0 t + (1/2)(1.4) t^2
0.7 t^2 = 590
t^2 = 843
t = 29 seconds
x = Xi + Vi t + (1/2) a t^2
Vi = 0, starts at rest
a = 1.4 m/s^2
so
distance = x - Xi = 590 = 0 t + (1/2)(1.4) t^2
0.7 t^2 = 590
t^2 = 843
t = 29 seconds
R_scott
d = 1/2 a t^2
590 = 1/2 * 1.4 * t^2
t = √(2 * 590 / 1.4)
590 = 1/2 * 1.4 * t^2
t = √(2 * 590 / 1.4)