A train starts from rest and accelerates uniformly for 30 seconds to a...
Solution:
Given:
Initial velocity, u = 0 (train starts from rest)
Time taken to accelerate, t1 = 30 seconds
Final velocity, v = 108 km/h
Time taken to travel with this velocity, t2 = 20 minutes = 1200 seconds
Time taken to decelerate, t3 = 20 seconds
We need to find the total distance covered by the train.
Calculating acceleration:
The formula to calculate acceleration is:
a = (v - u) / t
where,
a = acceleration
v = final velocity
u = initial velocity
t = time taken
Substituting the given values, we get:
a = (108 * 1000 / 3600 - 0) / 30
a = 4 m/s²
Calculating distance covered during acceleration:
The formula to calculate distance covered during acceleration is:
s = ut + 1/2at²
where,
s = distance
u = initial velocity
t = time taken
a = acceleration
Substituting the given values, we get:
s1 = 0 * 30 + 1/2 * 4 * 30²
s1 = 1800 m
Calculating distance covered during constant velocity:
The formula to calculate distance covered during constant velocity is:
s = vt
where,
s = distance
v = velocity
t = time taken
Substituting the given values, we get:
s2 = 108 * 1000 / 3600 * 1200
s2 = 36000 m
Calculating deceleration:
The deceleration will be the same as acceleration, as the train is slowing down uniformly. Therefore, a = 4 m/s².
Calculating distance covered during deceleration:
The formula to calculate distance covered during deceleration is:
s = vt - 1/2at²
where,
s = distance
v = initial velocity
t = time taken
a = acceleration
Substituting the given values, we get:
s3 = 108 * 1000 / 3600 * 20 - 1/2 * 4 * 20²
s3 = 300 m
Calculating total distance:
The total distance covered by the train is the sum of the distances covered during acceleration, constant velocity, and deceleration.
Total distance = s1 + s2 + s3
Total distance = 1800 + 36000 + 300
Total distance = 54300 m
Therefore, the total distance covered by the train is 54300 m.
A train starts from rest and accelerates uniformly for 30 seconds to a...
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