Question

A lift performs the first part of its ascent with uniform acceleration a and the remaining with uniform retardation 2a.If t is the time of ascent, find the depth of the shaft. (1.)at^2/4 (2.)at^2/3 (3.)at^2/2 (4.)at^2/8​

Answer 1

Answer:

The depth of the shaft is at^2/3 means option 2

Explanation:

According to the problem the lift has performed first part with acceleration a, and the remaining part is 2a

Therefore retardation time  will be half as it is double

Let t1=acceleration time  

Then t1/2=retardation time  

Let t be the total time

t1+t1/2=t  

∴t1=2t/3

Let d be the depth .  

Now, d=d1+d2  

=(1/2)(a)(t1)^2+(1/2)(2a)(t1/2)^2  

=3/4at1^2  

=(3/4)(a)(2t/3)^2  

=at^2/3

Answer 2

Answer:

Option 2 is the answer that means ^2/3.