The greatest acceleration or deceleration that a train may have is a. The minimum time in which the train can get from one station to the next at a distance s is:
A
√sa
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B
√2sa
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C
2√sa
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D
12√sa
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Solution
The correct option is C2√sa The time will be minimum if it get accelerated to half distance then other hall distance let retreated
So using s=ut+12at2 ; u=0 (initially at rest)
We have s2=12at21
⇒t1=√sa it is time to cover half (i.es2) distance with acceleration a
other half (s2) in all be covered in same time because retarelation is also a so total time t=2t1=2√sa