solve physics question. a train moves 3.6km from rest to rest in 3minutes. the greatest speed is 90km/h, and the acceleration and retardation are uniform. find the distance travelled at full speed.
It sounds like the acceleration and deceleration are equal in value. So, since
90km/hr = 25m/s
3 min = 180 sec
3.6km = 3600m
If acceleration/deceleration take t seconds, leaving u seconds for full-speed travel,
2t+u = 180
at = 25
at^2 + 25u = 3600
a = 25/3
t = 36
u = 108
so, going 25m/s for 108s = 2700m = 2.7 km
To find the distance traveled at full speed, we need to break down the given information and use some physics equations.
Given:
Initial velocity (u) = 0 km/h (since the train starts from rest)
Final velocity (v) = 90 km/h
Acceleration (a) and retardation (b) are uniform
Total distance (s) = 3.6 km
Time taken (t) = 3 minutes = 3/60 = 1/20 hours
To solve for the distance traveled at full speed, we can use the kinematic equation:
s = ut + (1/2)at^2
First, let's calculate the acceleration. Since the acceleration and retardation are uniform, and the train starts from rest, we can use the equation:
v^2 = u^2 + 2as
Rearranging the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
Plugging in the values:
a = (90^2 - 0^2) / (2 * 3.6)
a = 8100 / 7.2
a = 1125 m/h^2 (meters per hour squared)
Now, let's calculate the distance traveled during the acceleration phase. We'll use the equation:
s1 = (1/2)at^2
Plugging in the values:
s1 = (1/2) * 1125 * (1/20)^2
s1 = 1125 * (1/400)
s1 = 2.8125 m
Since the total distance is 3.6 km and the distance traveled during acceleration is 2.8125 m, the remaining distance at full speed is:
s2 = total distance - distance traveled during acceleration
s2 = 3.6 km - 2.8125 m
s2 = 3.6 km - 0.0028125 km
s2 = 3.5971875 km
Therefore, the distance traveled at full speed is approximately 3.597 km.