REST AND MOTION SOLUTIONS Q31

Q 31. An elevator is descending with uniform acceleration. To measure the
acceleration, a person in the elevator drops a coin at the moment the elevator
starts. The coin is 6 ft above the floor of the elevator at the time it is
dropped. The person observes that the coin strikes the floor in 1second.
Calculate from these data the acceleration of the elevator.




Let acceleration of lift be a.

acceleration of ball with respect to the lift = acceleration of ball with respesct to ground - acceleration of the lift

Or, aball(lift) = g - a = 9.8 m/s^2 - a = 32 ft./s^2 - a [ 9.8m = 32 ft. ]

displacement s = 6 ft, time taken t = 1 s, initial velocity u = 0


Applying relative concept for equation of motions = ut + 1/2 at^2 = 0*1 + 1/2 (32-a)*1*1 = 1/2 (32-a)


Therefore, 6 = 1/2 (32-a), Hence a = 20 ft./s^2



The coin is dropped at the instant when evrything is at standstill . 
the coin drops after one sec that means distance covered by coin 
S=ut+1/2gt^2 
S=0x1+1/2x9.8x1=4.9 m 
but the distance to floor was only 1.82m(6ft) that means the elevator moved 4.9-1.82=3.07m in that one second. 
So substituting this S again we have 
S=ut+1/2at^2 (where a is the acceleration of elevator) 
3.07=0+1/2*a*1 
a=6.14m/sec^2 
Thsi should be the answer..



Height of fall in 1s = 1/2 (t^2 x g), = 4.9m. 
6ft. is 1.829m., so distance elevator has drolled in the 1 sec = (4.9 - 1.829) = 3.071m. 
Acceleration = (2d/t^2), = 6.142m/sec^2.