Question
A moving train slows down uniformly from 40 m/s to 20 m / s in 5 second. What is the distance it covers during the 4th second?
Answers
Anonymous
if a = constant
v = Vi + a t
20 = 40 + 5 a
-20 = 5 a
a = -4 m/s^2
x(t) = Vi t + (1/2) a t^2 = 40 t - 2 t^2
x(3) = 40*3 - 18 = 102
x(4) = 40*4 - 32 = 128
128 - 102 = 26 meters
v = Vi + a t
20 = 40 + 5 a
-20 = 5 a
a = -4 m/s^2
x(t) = Vi t + (1/2) a t^2 = 40 t - 2 t^2
x(3) = 40*3 - 18 = 102
x(4) = 40*4 - 32 = 128
128 - 102 = 26 meters