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A train starts from rest and accelerates uniformly for 30 seconds to acquire a velocity of 108 km/h. It travels with this velocity for 20 minutes. The driver now applies brakes and the train slows down uniformly to stop after 20 seconds. Find the total distance covered by the train?
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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.
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.
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A train starts from rest and accelerates uniformly for 30 seconds to a...
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A train starts from rest and accelerates uniformly for 30 seconds to acquire a velocity of 108 km/h. It travels with this velocity for 20 minutes. The driver now applies brakes and the train slows down uniformly to stop after 20 seconds. Find the total distance covered by the train?
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