Grade 12Mechanics

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

9 Years agoGrade 12

1 Answer

Vikas TU

9 Years ago

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. 4^th year
Thapar University)

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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

1 Answer

Vikas TU

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