a train slows from 108km/h with a uniform retardation of 5m/ssquare how long will it take to reach 18km/h and what is the distances
Vo = 108,000m/3600s = 30 m/s.
V = 18,000m/3600 = 5 m/s.
V = Vo + (-5)t = 5.
30 + (-5)t = 5,
t = 5 s.
d = Vo*t + 0.5a*t^2 = 30*5 + (-2.5)*5^2 =
Itam
To find the time it takes for the train to reach a speed of 18 km/h and the corresponding distance traveled, we can use the equations of uniformly retarded motion.
Let's convert the initial velocity and final velocity into meters per second (m/s) for consistency:
Initial velocity, u = 108 km/h
Final velocity, v = 18 km/h
1 km/h = (1000 m / 3600 s) = 5/18 m/s
So, u = (108 × 5/18) m/s = 30 m/s
v = (18 × 5/18) m/s = 5 m/s
The retardation (deceleration), a, is given as 5 m/s^2.
Now, we can use the equation of uniformly retarded motion to find the time it takes for the train to reach the final velocity:
v = u + at
Rearranging the equation to solve for time, t:
t = (v - u) / a
Substituting the given values:
t = (5 - 30) / -5
Simplifying:
t = -25 / -5
t = 5 seconds
Therefore, it will take 5 seconds for the train to reach a speed of 18 km/h.
To find the distance traveled during this time, we can use the equation of motion:
s = ut + (1/2)at^2
Substituting the known values:
s = 30 × 5 + (1/2) × -5 × 5^2
Simplifying:
s = 150 - 125/2
s = 75 meters
Therefore, the distance traveled by the train during this time is 75 meters.