A cheetah chases a gazelle, reaching a speed of 28 m/s . A graph of the cheetahs velocity over time is shown below where rightward is the positive velocity direction

The graph of the cheetahs velocity over time is missing, so i have attached it.

Also, we are asked to find out which graph shows the cheetahs acceleration over time.

Option A

Explanation:

From the attached graph of the velocity over time, we can see the velocity line from an initial velocity of 0 to a final velocity of 28 m/s.

We also observe the following from the graph;

after 1 second, the velocity is 4 m/s

After 2 seconds, the velocity is 8 m/s

After 3 seconds, the velocity is 12 m/s

After 7 seconds, the velocity is 28 m/s

Now, from Newton's equation of motion, final velocity is;

v = u + at

u is initial velocity and in this case is zero.

Thus; v = at

a = v/t

At first point above;

a = 4/1 = 4 m/s²

At second point above;

a = 8/2 = 4 m/s²

At third point above;

a = 12/3 = 4 m/s²

At fourth point above;

a = 28/7 = 4 m/²

We can see that the acceleration at each point is 4 m/s². This means it is a constant acceleration and the graph that best represents this is Graph in Option A as it shows the acceleration is 4 m/s² all through.

  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

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