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Grade 11Mechanics

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.

Profile image of reshma lr
13 Years agoGrade 11
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2 Answers

Profile image of Prajwal kr
13 Years ago

Maximum permissible acceleration = 0.9144 m/s.

If initial velocity=0;

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

=4.67 sec.

Profile image of SOURAV MISHRA
13 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.