a train has to move 100 M from station a to station B in minimum possible time in product is maximum possible acceleration of the train is 2 metre per second square and maximum possible retardation of train is 4 m per second square then find the minimum possible time
Answers
The train travels from zero speed accelerates for t1 seconds and starts retarding for t2 seconds.
At the end of t2, the train stops in station B.
Initial speed for first t1 seconds = u = 0
Final speed at the end of t1 seconds is v = u + at.
Here acceleration = a = 2 m/sec²
v = 0 + 2t1 = 2t1 m/sec
For the second part, initial velocity = 2t1, final velocity = 0, retardation = a = -4 m/sec²
0 = 2t1 -4t2.
Hence t2 = t1/2 ------------------------------------E1.
Distance traveled by first part of the journey is given by
v² – u² = 2as
for the first part of the journey,
v = 2t1
u = 0
a = 2
s = s1
Hence s1 = (2t1)²/(2*2) = t1²
For the second part of the journey
v = 0
u = 2t1
a = -4
s = s2.
0 – (2t1)² = -2*4*s2
Hence s2 = t1²/2
It is given that distance between two stations is 100m.
S1 + s2 = 100
t1² + t1²/2 = 100
t1² = 200/3
t1 = 10√(2/3)
From E1, t2 = t1 / 2 = 10√(2/3)/2 = 5√(2/3)
Total time traveled = t1 + t2 = 10√(2/3) + 5√(2/3) = 15√(2/3)