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An elevator , whose floor to ceiling distance, is 2.7m, starts ascending with a constant acceleration of 1.2m/s^2. 2 seconds later, a bolt begins to fall from ceiling.Displacement and distance covered by bolt during the fall?

Aoishi , 8 Years ago
Grade 12
anser 1 Answers
Vikas TU
Dear Student,
As given, floor to ceiling distance = 2.7m
Acceleration = 1.2 m/s^2
Relative to the acceleration  of the bolt = (g+a) = 9.8+1.2 = 11m/s^2.
Let time t to reach the distance of 2.7 m
U = 0 s = 2.7m , a = 11 m/s^2
From the second equation
S = ut+1/2 at^2
S = ½ *11* t^2
5.4 = 11 t^2
t = 0.7 sec
Displacement :
Upward velocity of the elevator = at =2.4m/sec
Position of  bolt after t = 0.7 sec
S = 2.4*0.7 – ½*-9.8*(0.7)^2
S = 0.7 m
Distance travelled can be calculated from the third eq
  • v^2 – u^2 = 2as
  • v^2 = 2gs
  • (2.4)^2 = 2*9.8 *S
  • S = 5.76/19.6
  • S = 0.3m
    
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
Last Activity: 8 Years ago
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