An object initially at rest experiences an acceleration of 2.20 m/s2 on a level surface. Under these conditions, it travels 5.00 m . Let’s designate the first 2.50 m as phase 1 with a subscript of 1 for those quantities, and the second 2.50 m as phase 2 with a subscript of 2. The times for traveling each phase should be related by which condition?

t2  = Time for whole trip - time t1  for first phase trip = 0.6 sec

Explanation:

The object starts from rest;

Vi=0 m/s

a = 2.20 m/sec2

S= distance traveled

t= time

Here we can first  calculate time expression in terms of distance S;

We know

Putting values

==>S= 0+ ×2.2×

==> = 2S/2.2

==> t = 0.953462589

Lets suppose it completes the trip without 'phases' then S here is 5 meters as a whole ;

==> t = 0.953462589 ×

==> t = 0.953462589 × 2.236067977= 2.132007163 Sec

Now let's consider phases

Phase 1:

Let S = S1 = 2.5 m

t=t1

==> t1 = 0.953462589 × =  1.507556723 sec

So for the second phase the time t2  = Time for whole trip - time t1 for first phase trip

==> t2 = 2.132007163 - 1.507556723 = 0.62445044 = 0.6 sec

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂