A train is moving with an initial velocity of 30m/s the brakesapplied to produce retardation of 1.5 m/s calculate the time when it comes to rest
assuming you meant 1.5 m/s^2, then it will of course take
(30m/s)/(1.5 m/s^2) = 20 s
Think about it. If it loses velocity by 1.5 m/s every second, wit will take 20 s to lose all 30 m/s.
V = Vo + a*t = 0,
30 + (-1.5)t = 0,
t = 20 s.
To calculate the time it takes for the train to come to rest, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s since the train comes to rest)
u = initial velocity (30 m/s)
a = acceleration (retardation in this case, -1.5 m/s^2)
s = distance traveled
Plugging in the values, the equation becomes:
0^2 = 30^2 + 2(-1.5)s
Simplifying:
0 = 900 - 3s
Rearranging the equation to solve for s:
3s = 900
s = 900 / 3
s = 300 m
Now we can use the equation of motion:
v = u + at
Where:
v = final velocity (0 m/s)
u = initial velocity (30 m/s)
a = acceleration (retardation in this case, -1.5 m/s^2)
t = time
Plugging in the values, the equation becomes:
0 = 30 + (-1.5)t
Rearranging the equation to solve for t:
-1.5t = -30
t = -30 / -1.5
t = 20 seconds
So, it takes 20 seconds for the train to come to rest.