consider a train which can accelerate with an acceleration of 20cm/sec2 and slow down with a deceleration of 100cm/sec2. Find the minimum time for the train to travel between the stations 2.7 km apart.
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Answers
Answered by
88
Let the train accelerate for t1 seconds when it starts from rest
from 1st railway station and then decelerate for t2 seconds to stop at
the next station. The train does not travel with uniform velocity in between.
Then Its maximum speed is v = u + a t = 0 + 0.20 t1
So v = 0.20 t1 m/sec
Again for the decelerating part: v = u+at => 0 = v - 1.00 t2 => v = t2
so v = 0.20 t1 = t2
t1 = t2/0.2 = 5 t2
so durations of accelerating phase and decelerating phase are inversely proportional to the magnitudes of acceleration.
distance traveled s = ut+1/2 a t²
Sum of the distances traveled in the two phases :
0 t1 + 1/2 0.20 t1² + (t2)* t2 - 1/2 *1*t2² = 2700 meters
substitute the value of t1
0.1 (5 t2)² + (t2)² - 1/2 (t2)² = 2700
3 (t2)² = 2700
t2² = 900 t2 = 30 sec t1 = 150 sec
total time of travel between stations = 180 sec
Then Its maximum speed is v = u + a t = 0 + 0.20 t1
So v = 0.20 t1 m/sec
Again for the decelerating part: v = u+at => 0 = v - 1.00 t2 => v = t2
so v = 0.20 t1 = t2
t1 = t2/0.2 = 5 t2
so durations of accelerating phase and decelerating phase are inversely proportional to the magnitudes of acceleration.
distance traveled s = ut+1/2 a t²
Sum of the distances traveled in the two phases :
0 t1 + 1/2 0.20 t1² + (t2)* t2 - 1/2 *1*t2² = 2700 meters
substitute the value of t1
0.1 (5 t2)² + (t2)² - 1/2 (t2)² = 2700
3 (t2)² = 2700
t2² = 900 t2 = 30 sec t1 = 150 sec
total time of travel between stations = 180 sec
Answered by
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