3. Using the graph below:
What is the displacement of the object between 0 and 2 seconds? (Show you work and
calculation in detail).
b) What is the acceleration between 2 and 4 seconds? (Show you work and calculation in
detail)

a) a = 1 [m/s²]

b) a = 0.25 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

a)

Between 0 and 2 seconds

where:

Vf = final velocity = 2 [m/s]

Vo = initial velocity = 0

a = acceleration [m/s²]

t = time = 2 [s]

Note: the positive sign of acceleration in the previous equation allows us to appreciate that the velocity is increasing

2 = 0 + a*2

a = 2/2

a = 1 [m/s²]

b)

Between 2 and 4 seconds

where:

Vf = final velocity = 2.5 [m/s]

Vo = initial velocity = 2

a = acceleration [m/s²]

t = time = (4 - 2) = 2 [s]

2.5 = 2 + a*2

2.5 - 2 = 2*a

0.5 = 2*a

a = 0.25 [m/s²]

  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