Aspeedboat starts from rest and accelerates at - 2.01 m/s2 for 7.00 s. at the end of this time, the boat continues for an additional 6.00 s with an acceleration of +0.518 following this, the boat accelerates at 1.49 m/s2 for 8.00 s (a) what is the velocity of the boat at t = 21.0 s? (b) find the total displacement of the boat

velocity after 21 sec =  0.96 m/s

total displacement =  164.341 m

Explanation:

given data

accelerates = - 2.01 m/s²

time = 7 s

additional time = 6 s

acceleration = +0.518 m/s²

following boat accelerates = 1.49 m/s²

time = 8 s

to find out

velocity of the boat at time 21 s and the total displacement of the boat

solution

we find velocity after 7 second

velocity = accelerate × time

velocity after 7 sec = - 2.01 × 7 = -14.07 m/s

displacement = ut + 0.5 × at²

here u is initial speed = 0 and a is acceleration and t is time

displacement =   0.5 × (-2.01) (7)² = - 49.245 m

and

velocity after (7 + 6) = 13 second is

velocity after 13 sec = -14.07 +  ( 6 × 0.518)  = -10.96 m/s

displacement = ut + 0.5 × at²

here u is initial speed = -14.07  and a is acceleration and t is time

displacement =  -14.07 (6) +  0.5 × (0.518) (6)² = - 75.096 m

and

velocity after (7 + 6 + 8) = 21 second is

velocity after 21 sec = -10.96 + 1.49 × 8 = 0.96 m/s

displacement =  -10.96 (8) +  0.5 × (1.49) (8)² = - 40 m

so

total displacement = sum of all displacement

total displacement = (- 49.245) + (- 75.096) +  (- 40) = -164.341

total displacement =  164.341 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

the total distance traveled by the car is

30km + 49km = 79 km .