Airplane Question1

Solution!

A.I.) Acceleration Take Off

a(t)1 = 2400v(t)1 = 2400t + C

I.) Acceleration:a(t)1 = 2400 km/h2

v(0)1 = 60 km/hs(0)1 = 10 km

< antidifferentiate a(t)1

*Starting with the acceleration section of the airplane's flight, we antidifferentiate the acceleration function using power rule.


a(t)1 = 2400v(t)1 = 2400t + C60 = 2400(0) + C60 = C


v(0)1 = 60 km/hs(0)1 = 10 km

< antidifferentiate a(t)1< input values of v(t)1 function

*Input the given values of the v(t)1 function to solve for the constant "C".


a(t)1 = 2400v(t)1 = 2400t + C60 = 2400(0) + C60 = Cv(t)1 = 2400t + 60


v(0)1 = 60 km/hs(0)1 = 10 km


< substitute value "C" into v(t), to complete v(t)1 function

*Now that we have the v(t)1 function, our next step is to antidifferentiate this new function to find s(t)1.


a(t)1 = 2400v(t)1 = 2400t + C60 = 2400(0) + C60 = Cv(t)1 = 2400t + 60

s(t)1 = 1200t2 + 60t + C


v(0)1 = 60 km/hs(0)1 = 10 km



< antidifferentiate v(t)1


a(t)1 = 2400v(t)1 = 2400t + C60 = 2400(0) + C60 = Cv(t)1 = 2400t + 60

s(t)1 = 1200t2 + 60t + C10 = 1200(0)2 + 60(0) + C10 = C


v(0)1 = 60 km/hs(0)1 = 10 km



< antidifferentiate v(t)1 < input values of s(t)1 function

*Input the given values of the s(t)1 function to solve for the constant "C".


a(t)1 = 2400v(t)1 = 2400t + C60 = 2400(0) + C60 = Cv(t)1 = 2400t + 60

s(t)1 = 1200t2 + 60t + C10 = 1200(0)2 + 60(0) + C10 = Cs(t)1 = 1200t2 + 60t +10


v(0)1 = 60 km/hs(0)1 = 10 km



< antidifferentiate v(t)1 < input values of s(t)1 function

< substitute value "C" into s(t)1, to complete s(t)1 function

A.II.) Constant Flight

v(t)2 = 660s(t)2 = 660t + C < antidifferentiate v(t)2

II.) Constant Flightv(t)2 = 660 km/h

s(0)2=30 km

*Next we work with the constant flight section of the airplane's flight. We antidifferentiate the velocity function using power rule.


v(t)2 = 660s(t)2 = 660t + C30 = 660(0) + C30 = C

< antidifferentiate v(t)2< input values of s(t)2 function


s(0)2=30 km

*Input the given values of the s(t)2 function to solve for the constant "C".


v(t)2 = 660s(t)2 = 660t + C30 = 660(0) + C30 = Cs(t)2 = 660t + 30




s(0)2=30 km

*Now that we have the s(t)2 function for this section of the flight, we can move onto our final section.

A.III.) Deceleration Landing

a(t)3 = 840v(t)3 = 840t + C< antidifferentiate a(t)3

III.) Decceleration:a(t)3 = 840 km/h2

v(0)3 = 100 km/hs(0)3 = 583 km

*With our final section of the piecewise function, we antidifferentiate a(t)3 using power rule.


a(t)3 = 840v(t)3 = 840t + C100 = 840(0) + C100 = C



v(0)3 = 100 km/hs(0)3 = 583 km

*We input the given values of the v(t)3 function to solve for the constant "C".


a(t)3 = 840v(t)3 = 840t + C100 = 840(0) + C100 = Cv(t)3 = 840t 100


< substitute value "C" into v(t)3, to complete v(t)3 function


v(0)3 = 100 km/hs(0)3 = 583 km

*Now that we have the v(t)3 function, we continue by antidifferentiation to find the s(t)3 function.


a(t)3 = 840v(t)3 = 840t + C100 = 840(0) + C100 = Cv(t)3 = 840t 100

s(t)3 = 420t2 100t + C




v(0)3 = 100 km/hs(0)3 = 583 km

< antidifferentiate v(t)3

*Simply antidifferentiate again using power rule


a(t)3 = 840v(t)3 = 840t + C100 = 840(0) + C100 = Cv(t)3 = 840t 100

s(t)3 = 420t2 100t + C583 = 420(0)2 100(0) + C583 = C




v(0)3 = 100 km/hs(0)3 = 583 km


*Input given coordinate of s(t)3 to solve for the constant "C".


a(t)3 = 840v(t)3 = 840t + C100 = 840(0) + C100 = Cv(t)3 = 840t 100

s(t)3 = 420t2 100t + C583 = 420(0)2 100(0) + C583 = Cs(t)3 = 420t2 100t + 583




v(0)3 = 100 km/hs(0)3 = 583 km



s(t)1 = 1200t2 + 60t +10s(t)2 = 660t + 30s(t)3 = 420t2 100t + 583

*These three functions combine in their respective interval to depict the distance travelled over time by the airplane.

With this set of functions, we can move onto the next part and apply them.

A.) Conclusion

B.I.) Acceleration and Deceleration Sections

Take Off:s(t)1 = 1200t2 + 60t +10 < take off distance function

*Since the times required for take off and landing are the same between each citycity flight, we need to find these distances so that we can find the amount of distance flown at constant flight speed.

0.25 hrs (2/3) hrsx hrs


Take Off:s(t)1 = 1200t2 + 60t +10s(0.25) = 1200(0.25)2 + 60(0.25) +10s(0.25) = 100 km

< take off distance function< input 0.25 hrs

< distance required for a complete take off

*Input the time required by the airplane to take off. This distance is the same for each flight.



Landing:s(t)3 = 420t2 100t + 583



< landing distance function



Landing:s(t)3 = 420t2 100t + 583s(2/3) = 420(2/3)2 100(2/3) + 583s(2/3) = 330 km



< landing distance function< input (2/3) hrs

< distance required for a complete landing

*These values are important, as we will use them in every computation.

B.II.) Route 1: Winnipeg New York Atlanta Mexico City WinnipegNewYork: 2064.87 km < distance between cities2064.87 km 430 km = 1634.87 km < distance @

constant flight speed

*Since it takes a total of 430 km (100km + 130km) for take off and landing, we subtract it from the distance between the cities to find the amount of distance we are going to compute time for.


constant flight speeds(t)2 = 660t + 30< constant flight function

*Use the distance function found from part A for constant flight.


constant flight speeds(t)2 = 660t + 301634.87 = 660t + 300 = 660t 1604.87t = 2.4316 hrs

< constant flight function< input distance @ constant flight speed to compute time

< time in hours

*The amount of time is the time flown between the two cities at constant flight (we will add the already known times for take off and landing afterwards)




< time in hours

New YorkAtlanta: 1202.69 km1202.69 km 430 km = 772.69 km

< distance between cities< distance @ constant flight speed




< time in hours

New YorkAtlanta: 1202.69 km1202.69 km 430 km = 772.69 km

772.69 = 660t + 300 = 660t 742.69t = 1.1253 hrs

< distance between cities< distance @ constant flight speed

< input distance @ constant flight speed to compute time

< time in hours

B.II.) Route 1: Winnipeg New York Atlanta Mexico City AtlantaMexico City: 2157.76 km < distance between cities2157.76 km 430 km = 1727.76 km< distance @


B.II.) Route 1: Winnipeg New York Atlanta Mexico City AtlantaMexico City: 2157.76 km < distance between cities2157.76 km 430 km = 1727.76 km< distance @



< time in hours

*Now we have all three values for time upon which the airplane is at constant speed

B.II.) Route 1: Winnipeg New York Atlanta Mexico City

Total Time:= 3(0.25) + 3(2/3) + 2.4316 + 1.1253 + 2.5724 + 2(0.75)= 10.3793 hours*In total we add:three times the take off timethree times the landing timethree citycity timestwo times the flight delay time at each airport

B.II.) Route 1: Winnipeg New York Atlanta Mexico City

Total Time:= 3(0.25) + 3(2/3) + 2.4316 + 1.1253 + 2.5724 + 2(0.75)= 10.3793 hours

Time of Arrival:= 8.5 + 10.3793= 19.3793= 7:23 pm

< departure time + the total time

< number of hours converted to time

B.II.) Route 2: Winnipeg Seattle Miami Mexico City WinnipegSeattle: 1852.13 km < distance between cities1852.13 km 430 km = 1422.13 km < distance @



constant flight speeds(t)2 = 660t + 301422.13 = 660t + 300 = 660t 1392.13t = 2.1093 hours


< time in hours




< time in hours

SeattleMiami: 4392.16 km < distance between cities4392.16 km 430 km = 3962.16 km < distance @





< time in hours

SeattleMiami: 4392.16 km < distance between cities4392.16 km 430 km = 3962.16 km < distance @

constant flight speed3962.16 = 660t + 300 = 660t 3932.16t = 5.9578 hours


< time in hours

B.II.) Route 2: Winnipeg Seattle Miami Mexico City MiamiMexico City: 2064.40 km < distance between cities2064.40 km 430 km = 1634.40 km < distance @



< time in hours

B.II.) Route 2: Winnipeg Seattle Miami Mexico City

Total Time:= 3(0.25) + 3(2/3) + 2.1093 + 5.9578 + 2.4309 + 2(0.25)= 13.748 hours

*In total we add:three times the take off timethree times the landing timethree citycity timestwo times the flight delay time at each airport

B.II.) Route 2: Winnipeg Seattle Miami Mexico City

Total Time:= 3(0.25) + 3(2/3) + 2.1093 + 5.9578 + 2.4309 + 2(0.25)= 13.748 hours

Time of Arrival:= 6 + 13.748= 19.748= 7:44 pm



B.II.) Route 3: Winnipeg Montreal Dallas Mexico City WinnipegMontreal: 1821.36 km < distance between cities1821.36 km 430 km = 1391.36 km < distance @





< time in hours




< time in hours

MontrealDallas: 2436.25 km < distance between cities2436.25 km 430 km = 2006.25 km < distance @





< time in hours

MontrealDallas: 2436.25 km < distance between cities2436.25 km 430 km = 2006.25 km < distance @

constant flight speed2006.25 = 660t + 300 = 660t 1976.25t = 2.9943 hours


< time in hours

B.II.) Route 3: Winnipeg Montreal Dallas Mexico City DallasMexico City: 1501.38 km < distance between cities1501.38 km 430 km = 1071.38 km < distance @



< time in hours

*Once again, we have all three constant flight times required to solve for the time of arrival of route 3.

B.II.) Route 3: Winnipeg Montreal Dallas Mexico City

Total Time:= 3(0.25) + 3(2/3) + 2.0626 + 2.9943 + 1.5778 + 2(0.5)= 10.3847 hours

*In total we add:three times the take off timethree times the landing timethree citycity timestwo times the flight delay time at each airport

B.II.) Route 3: Winnipeg Montreal Dallas Mexico City

Total Time:= 3(0.25) + 3(2/3) + 2.0626 + 2.9943 + 1.5778 + 2(0.5)= 10.3847 hours

Time of Arrival:= 8.25 + 10.3847= 18.6347= 6:38 pm



*Now we have all the information required to piece the final answer!

B.III.) Final Solution:

6:38 pm > Route 37:44 pm > Route 27:23 pm > Route 1

*The flight path that arrived first of the three was that of route 3. To conclude the solution, a sentence answer must be written for word problems.

Of the three flight paths, route three (from Winnipeg to Montreal to Dallas to Mexico City) arrived the earliest in Mexico City, at 6:38 pm.

Finish!

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Airplane Question1