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A train starts from rest accelerates uniformly for 30s to acquire a velocity of 108 km/h.It travels with this velocity for 20 min.The driver then applies the brake and the train retards uniformly to stop after 20s.Find the total distance covered by the train?
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A train starts from rest accelerates uniformly for 30s to acquire a ve...
Given:
Initial velocity (u) = 0 km/h
Final velocity (v) = 108 km/h
Time taken to accelerate (t1) = 30 s
Time taken to travel at constant velocity (t2) = 20 min = 1200 s
Time taken to decelerate (t3) = 20 s
Calculations:
Step 1: Calculate the acceleration (a) of the train during the first 30 seconds.
Using the formula v = u + at, we can find the acceleration.
v = u + at
108 km/h = 0 km/h + a * 30 s
a = 108 km/h / 30 s = 3.6 km/h/s
Step 2: Calculate the distance (s1) covered during the acceleration period.
Using the formula s = ut + 1/2 at^2, we can find the distance covered during the acceleration period.
s1 = 0 + 1/2 * 3.6 km/h/s * (30 s)^2
s1 = 1620 m
Step 3: Calculate the distance (s2) covered during the constant velocity period.
The distance covered during this period can be calculated using the formula:
s2 = v * t2
s2 = 108 km/h * 1200 s
s2 = 129600 m
Step 4: Calculate the deceleration (b) of the train during the last 20 seconds.
Using the formula v = u + bt, we can find the deceleration.
0 km/h = 108 km/h + b * 20 s
b = -5.4 km/h/s
Note: The negative sign indicates that the train is decelerating.
Step 5: Calculate the distance (s3) covered during the deceleration period.
Using the same formula as in Step 2, we can find the distance covered during the deceleration period.
s3 = 108000 m
Step 6: Calculate the total distance (s) covered by the train.
The total distance can be found by adding up the distances covered during each period.
s = s1 + s2 + s3
s = 1620 m + 129600 m + 108000 m
s = 239220 m
Therefore, the total distance covered by the train is 239220 m.
Initial velocity (u) = 0 km/h
Final velocity (v) = 108 km/h
Time taken to accelerate (t1) = 30 s
Time taken to travel at constant velocity (t2) = 20 min = 1200 s
Time taken to decelerate (t3) = 20 s
Calculations:
Step 1: Calculate the acceleration (a) of the train during the first 30 seconds.
Using the formula v = u + at, we can find the acceleration.
v = u + at
108 km/h = 0 km/h + a * 30 s
a = 108 km/h / 30 s = 3.6 km/h/s
Step 2: Calculate the distance (s1) covered during the acceleration period.
Using the formula s = ut + 1/2 at^2, we can find the distance covered during the acceleration period.
s1 = 0 + 1/2 * 3.6 km/h/s * (30 s)^2
s1 = 1620 m
Step 3: Calculate the distance (s2) covered during the constant velocity period.
The distance covered during this period can be calculated using the formula:
s2 = v * t2
s2 = 108 km/h * 1200 s
s2 = 129600 m
Step 4: Calculate the deceleration (b) of the train during the last 20 seconds.
Using the formula v = u + bt, we can find the deceleration.
0 km/h = 108 km/h + b * 20 s
b = -5.4 km/h/s
Note: The negative sign indicates that the train is decelerating.
Step 5: Calculate the distance (s3) covered during the deceleration period.
Using the same formula as in Step 2, we can find the distance covered during the deceleration period.
s3 = 108000 m
Step 6: Calculate the total distance (s) covered by the train.
The total distance can be found by adding up the distances covered during each period.
s = s1 + s2 + s3
s = 1620 m + 129600 m + 108000 m
s = 239220 m
Therefore, the total distance covered by the train is 239220 m.
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A train starts from rest accelerates uniformly for 30s to acquire a ve...
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A train starts from rest accelerates uniformly for 30s to acquire a velocity of 108 km/h.It travels with this velocity for 20 min.The driver then applies the brake and the train retards uniformly to stop after 20s.Find the total distance covered by the train?
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