Question

A boy, in an stop accelerator ( by walking) takes time T1 to reach the first floor and takes time T2 in an working accelerator. How much time will he take if he is walking in an working accelerator( both ) .

Answer 1

_____________________________________________________________v-u = at 
here stop acceleration means constant acceleration . 
than the boy have a uniform velocity 
T1= s / v 

now working acceleration
now here the velocity is non-uniform 
so 
T2 = v / a (working )
____________________________________________________________

 

Answer 2

Let ;
stoping or walking acceleration = a
working acceleration = a'
Let he distance of first floor from ground = S
then,
S = 1/2aT1²
a = 2S/T1² -----(2)

similarly ,
S = 1/2a'T2²
a' = 2S/T2² --------(2)

now,
when , he is walking in an accelerator
we use here Pusedo force concept .
S = 1/2arel T²
arel = acceleration of first with respect to second = ( a + a')

S = 1/2(a - a')T²
2S/T² = 2S/T1² - 2S/T2²
1/T² = 1/T1² + 1/T²
1/T² = (T1.T2)²/(T1² + T2²)
T = √(T1² + T2²)/T1.T2