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anonymous

  • 5 years ago

the greatest acceleration or deceleration that a train may have is a.The minimum time in which the train can start from one station to stop at the next is a. 2sqrt{s/a} b. sqrt{s/a}

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  1. anonymous
    • 5 years ago
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    As in the figure, for the train to take minimum time, it should accelerate with it's max acceleration for half the duration and decelerate for the next half. $$s = \frac12 t v_{max} = \frac12 t(\frac{at}{2}) $$ It gives $$ t= 2\sqrt{\frac sa}$$

  2. anonymous
    • 5 years ago
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    tnx a lot.. :)

  3. anonymous
    • 5 years ago
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    also try the v-t (velocity - time) graph method...in complex cases it is really useful...and more easy to understand.just have to keep 2 things in mind: 1)equate the slope of graph to acceleration. 2)equate the area under graph to total displacement....

  4. anonymous
    • 5 years ago
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    thank u.. :)

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