chapter 3 problem 11mcq I am having trouble understanding how this solution came

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chapter 3 problem 11mcq I am having trouble understanding how this solution came to be. Was this diagram divided into shapes and then the area of each shape was calculated to get 2400m? C Chegg Study Guided Sol. X w.chegg.com/homework-helpfogger-exercising-along k-helpogger-eerasing-along-long-straight-road-runs-north-south-chapter-31. h-chapter-3.1 | C | Search OOKS STUDY MORE Find books, solutions, tutors and more CH3 Problem 11MCO show all steps A jogger is exercising along a long, straight road that runs north-south. She starts out heading noth. A graph of uat)is shown below What is the average velocity of the jogger during the 30.0 min? CH3 11CO (a) 1.3 m/s, north (D) 1.7m/s, north (c) 2.1 m/s, north (d) 2.9 m/s, north CH3 1 (m/s) H(min) H t (min) 10.0 20.0 30.0 eb and Windows

Explanation / Answer

yes, the diagram divides into shapes and then took addition of the areas under each segment. we will get the total area of under the curve.

The area under first segment is

               A1 = 1/ 2 ( base ) ( height) = ( 1/ 2) (2 min) ( 1 m.s) = 1x60 m

               A2 = Area of square = side x side = ( 2 min ) ( 1 m/s) = 2 x60 m

               A3 = area of triangle + area of rrectangle

                     = ( 1/ 2) ( base ) ( height ) + ( length ) ( breadth)

                     = (1/2) ( 8 min - 4 min ) ( 3 m/s - 1 m/s ) + ( 8 min - 4 min ) (1 m/s - 0)

                     = (1/2) ( 4min ) (2 m/s) + ( 4 min )( 1 m/s )

                     = ( 8 x 60)

For the segments B to C

               A4 = (12 min - 8 min) (3 m/s - 0)+ ( 1/2) ( 1 m/s) ( 12 - 10 )

                     = 4 (60 s) ( 3 m/s) + 1(60 s) ( 1 m/s)

                      =13 ( 60 ) m

For the segment C D

                 A5 = (18 - 12 min) ( 4 m/s - 0)

                       = (6 ) ( 60 s) ( 4 m/s)

                 A6 = (24 min - 18 min) ( -2 m/s)

                      = 6 ( 60 s ) ( -2 m/s)

                A7 = ( 1/2) ( 28 min - 24 min ) ( 2 m/s)

                      = (1/2) (4 ) ( 60 s) ( 2 m/s)

The total area under the curve is

              A = A1 + A2 + A3 + A4 + A5 + A6 + A7   = 2400 m

Average speed is

          v = 2400 m / 30 min

             = 1.3 m/s

                     

                       

                    

                     =

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