in problem number in chapter 3 of Essentials of College Physics. Aroller coaster

Question

in problem number in chapter 3 of Essentials of College Physics. Aroller coaster moves 200ft horizontally at an angle of 30.0 degreesabove the horizontal. Then 135 ft at 30.0 degree above thehorizontal. Then 135 ft at 40.0 degrees below the horizontal. Findthe displacement from starting point (O) to the end (C).  
the problem was converted into scale as follows: 50 ft= 1cm 135 ft= 2.75 cm
200 ft= 4 cm
    the solution given from point O to point C is equal to 8.4 cm(scale)= 420 ft.
I dont understand quite well how you get the 8.4 cm. theway i tried to solve it that gave me 8.4 cm as an answer was:   
  2.75*Cos (30)= 2.3   2.75* Cos(40)= 2.1

then i added them both to the 4cm   4+2.3+2.1= 8.4 is thisstrategy correct?
the problem was converted into scale as follows: 50 ft= 1cm 135 ft= 2.75 cm
200 ft= 4 cm
    the solution given from point O to point C is equal to 8.4 cm(scale)= 420 ft.
I dont understand quite well how you get the 8.4 cm. theway i tried to solve it that gave me 8.4 cm as an answer was:   
  2.75*Cos (30)= 2.3   2.75* Cos(40)= 2.1

then i added them both to the 4cm   4+2.3+2.1= 8.4 is thisstrategy correct? 200 ft= 4 cm
    the solution given from point O to point C is equal to 8.4 cm(scale)= 420 ft.
I dont understand quite well how you get the 8.4 cm. theway i tried to solve it that gave me 8.4 cm as an answer was:   
  2.75*Cos (30)= 2.3   2.75* Cos(40)= 2.1

then i added them both to the 4cm   4+2.3+2.1= 8.4 is thisstrategy correct?     the solution given from point O to point C is equal to 8.4 cm(scale)= 420 ft.
I dont understand quite well how you get the 8.4 cm. theway i tried to solve it that gave me 8.4 cm as an answer was:   
  2.75*Cos (30)= 2.3   2.75* Cos(40)= 2.1

then i added them both to the 4cm   4+2.3+2.1= 8.4 is thisstrategy correct?

Explanation / Answer

Given : 50 ft= 1cm 135 ft= 2.75 cm
200 ft= 4 cm     Here it is a horizontal component The agles ( 1 ) = 30o                 ( 2 ) = 40o    a = 135 ft = 2.75 cm     acos1=  2.75*Cos (30)= 2.3       a cos2= 2.75* Cos(40)= 2.1 So finally we can add thevalues    4cm+2.3cm +2.1cm = 8.4 cm    You are right I hope it helps you
50 ft= 1cm 135 ft= 2.75 cm
200 ft= 4 cm     Here it is a horizontal component The agles ( 1 ) = 30o                 ( 2 ) = 40o    a = 135 ft = 2.75 cm    a = 135 ft = 2.75 cm     acos1=  2.75*Cos (30)= 2.3       a cos2= 2.75* Cos(40)= 2.1 So finally we can add thevalues    4cm+2.3cm +2.1cm = 8.4 cm    You are right I hope it helps you
      a cos2= 2.75* Cos(40)= 2.1 So finally we can add thevalues    4cm+2.3cm +2.1cm = 8.4 cm    You are right I hope it helps you

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