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# If the greatest admissible acceleration of a train is 3 feet/sec square, calculate the least time taken from one station to another at a distance of 10m.

Grade:11

## 2 Answers

Prajwal kr
49 Points
7 years ago

Maximum permissible acceleration = 0.9144 m/s.

If initial velocity=0;

t=((2x10)/0.9144)1/2

=4.67 sec.

SOURAV MISHRA
37 Points
7 years ago

i assume that the train starts from the first station and crosses the second station with the same acceleration.

so the initial velocity is zero.

the acceleration is 0.9m/s2.

time taken is (2*10/0.9)(1/2) =  4.71 s.

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