## anonymous 5 years ago help pls.. an elevator car whose floor to ceiling distance is equal to 2.7 m , starts from rest and ascends with a constant acceleration of 1.2 ms^2 .Two seconds later , a bolt drops from the car's ceiling .The time taken by the bolt to hit the floor of the elevator is a. 1 s b. 0.7 s

1. Owlfred

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2. anonymous

b, See it wrt a person standing inside the elevator, and u would get it.

3. anonymous

At the point at which the bolt is dropped, its speed is equal to the speed of the elevator. It begins to accelerate at -9.81 m/s downward, but the elevator is accelerating 1.2 m/s upward. The bolt's acceleration relative to the elevator is 9.81 m/s - 1.2 m/s = 8.61 m/s. It must travel 2.7 m. $\Delta X = (1/2) (a t ^{2})$It follows that $t = \sqrt{\Delta X 2 / (2A)}$I think you can figure it out from there.

4. anonymous

tnx a lot !!

5. anonymous

@LunaEques Well done but please Check ur unit for acceleration it is m/s^2 and not m/s

6. anonymous

At the time of drop, acceleration of bolt w.r.t. car is (-10+ (-1.2)) and ... and the displacement( final position- initial position) is -2.7...since we are taking everthing with respect to the elevator... so the initial velocity of the bolt wrt elevator would be zero..please note that the quantities along the positive X and Y axis are taken to be + and those along the - X and Y axis are taken to be negative... so the Third equation of motion will give you: so the equation is 0 - 2.7= 0.5 X (-11.2) X t X t... which gives you t= 0.69 seconds which is close enuf to the option b..

7. anonymous

Basically you need to decide whether you think the elevator being accelerated up towards the bolt will effectively increase or decrease the bolt's acceleration. The short answer is it will effectively increase, but you need to try and picture this intuitively for the future.

8. anonymous

thank u ol.. :)