Acar starts from rest at a stop sign. it accelerates at 2.0 m/s2 for 5.0 seconds, coasts for 3.0 s , and then slows down at a rate of 1.5m/s2 for the next stop sign. part a how far apart are the stop signs?

88.3m

Explanation:

In this problem there are three sections

you must state the equations as appropriate

attached solution

88.33 m

Explanation:

For AB:

u = 0, t = 5 s, a = 2 m/s^2

Let the distance between A and B be s1 and the velocity at B is v.

Use second equation of motion

s1 = 25 m

Use first equation of motion

v = u + a t

v = 0 + 2 x 5 = 10 m/s

For BC:

Let the distance from B to C is s2.

time = 3 s,

speed, v = 10 m/s

s2 = v x t = 10 x 3 = 30 m

For CD:

final velocity, v' = 0

initial velocity, v = 10 m/s

acceleration, a = - 1.5 m/s^2

Let the distance from C to D is s3.

Use third equation of motion

s3 = 33.33 m

The total distance covered is s

s = s1 + s2 + s3 = 25 + 30 + 33.33 = 88.33 m

  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

it would be c. predictability