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#1 2008-09-21 14:01:29

DrSimple
Member
Registered: 2008-09-21
Posts: 1

Problems!!

Problem 1
A commercial jet took off from city 1 for city 2, a distance of 2550 km, at a speed of 800 km/h. At the same time a private jet, traveling 900 km/h, left City 2 for City 1. How long after takeoff will the jet pass each other?

Problem 2
Wendy took a trip from Davenport to Omaha, a distance of 300 mi. She traveled part of the way by bus, which arrived at the train station just in time for Wendy to complete her journey by train. The bus averaged 40 mi/h and the train 60 mi/h. The entire trip took 5 and a half hour. How long did Wendy spend on the train?

Problem 3
Kiran drove from Tortula to Cactus, a distance of 250 mi. She increaseed her speed by 10 mi/h for 360-mi trip from cactus to Dry Junction. If the total trip took 11 hr, what was her speed from Tortula to Cactus?

Problem 4
A salesman drives from Ajax to Barrington, a distance of 120 mi, at a steady speed. He then increases his speed by 10 mi/h to drive the 150 mi from Barrington to Collins. If the second leg of his trip took 6 mins more time than the first leg, how fast was he driving between Ajax and Barrington

sidenote:
mi = miles

reply fast please!!

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#2 2008-09-21 16:14:49

Macy
Guest

Re: Problems!!

Problem 1:  C1 ----------X------------------------------C2
             800km/h                                               900km/h

Let X is the point where the jet meet each other
As the jet flies in opposite direction therefore the time they will pass each other be:
2550/(800 + 900) = 1.5h

Problem 2:     D ----------------------X----------------------------------------O
                    bus(40mi/h)                 train (60mi/h)

Total time taken T = 5.5h ; x1 = distance DX; t1 = x1/40 (time spent on bus);
                                         x2 =  distance XO;  t2 = x2/60  (time spent on train) ;

Establish simultaneous equations
       x1 + x2 = 300 (1)
      t1 + t2 = 5.5 (2)

    Sustitute the value of t1 and t2 we have:

      x1 + x2 = 300 (1)
     x1/40 + x2/60 = 5.5 (2)

   Solve simultaneous equations we have: x2 = 240, therefore time spend on train t2 = 240/60 = 4hrs

Problem 3:      T -----------------------------C ---------------------------D 
                            v (mi/h)                         v + 10 (mi/h)

     t1 : time going from T to C = 250/v
     t2 : time goiong from C to D 360/v+10
  As t1 + t2 = 11, therefore:      (250/v) + (360/v + 10) = 11
Rearrange and simplify we have quadratic equation:  11v2 - 500v - 2500 = 0
Solve this quadratic equation we have: v = 50mi/h or v = -4.54 (invalid).
The speed from T to C is 50mi/h

Problem 4:  A -----------120mi------------B----------------150mi-------------------------C
                        v (mi/h)                                v+10 (mi/h)

Time taken from A to B: t1 = 120/v
Time taken from B to C: t2 = 150/(v+10)
As he goes 6mins = 0.1 hr more time in section BC, therefore:
         150/(v+10) - 120/v = 0.1

Rearrange we have the quadratic equation: v2 - 290v + 12000 = 0
Solve we have v = 240mi/h or v = 50 (mi/h)

Hope this help !!! roflol

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