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Benda yang bergerak lurus memiliki persamaan kecepatan v ( t ) = ( 3 − 6 t ) i + ( 4 + 8 t ) j ​ . Jika kecepatan awal benda 0 m/s, makavektor posisi benda pada saat t 1 = 1 s dan t 2 = 2 s adalah ....

Benda yang bergerak lurus memiliki persamaan kecepatan . Jika kecepatan awal benda 0 m/s, maka vektor posisi benda pada saat t1 = 1 s dan t2 = 2 s adalah ....

  1. begin mathsize 14px style r with bar on top subscript 1 equals 6 j with hat on top space straight m space dan space r with bar on top subscript 2 equals negative 12 i with hat on top plus 24 j with hat on top space straight m end style

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  5. begin mathsize 14px style r with bar on top subscript 1 equals 8 j with hat on top space straight m space dan space r with bar on top subscript 2 equals negative 24 i with hat on top plus 24 j with hat on top space straight m end style

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Pembahasan

Terlebih dahulu tentukan persamaan posisinya. Vektor posisi awalketika t 1 = 1 s : Vektor posisi akhir ketika t 2 = 2 s :

Terlebih dahulu tentukan persamaan posisinya.

begin mathsize 14px style top enclose r left parenthesis t right parenthesis equals top enclose r subscript o plus integral top enclose v left parenthesis t right parenthesis space dt top enclose r left parenthesis t right parenthesis equals 0 plus integral left parenthesis left parenthesis 3 minus 6 t right parenthesis i with hat on top plus left parenthesis 4 plus 8 t right parenthesis j with hat on top right parenthesis dt top enclose r left parenthesis t right parenthesis equals left parenthesis 3 t minus 3 t squared right parenthesis i with hat on top plus left parenthesis 4 t plus 4 t squared right parenthesis j with hat on top space straight m end style

Vektor posisi awal ketika t1 = 1 s :

begin mathsize 14px style r with bar on top left parenthesis t subscript 1 right parenthesis equals left parenthesis 3 t subscript 1 minus 3 t subscript 1 squared right parenthesis i with hat on top plus left parenthesis 4 t subscript 1 plus 4 t subscript 1 squared right parenthesis j with hat on top r with bar on top left parenthesis 1 right parenthesis equals left parenthesis 3 left parenthesis 1 right parenthesis minus 3 left parenthesis 1 right parenthesis squared right parenthesis i with hat on top plus left parenthesis 4 left parenthesis 1 right parenthesis plus 4 left parenthesis 1 right parenthesis squared right parenthesis j with hat on top r with bar on top subscript 1 equals 8 j with hat on top space straight m end style

Vektor posisi akhir ketika t2 = 2 s :

begin mathsize 14px style r with bar on top left parenthesis t subscript 2 right parenthesis equals left parenthesis 3 t subscript 2 minus 3 t subscript 2 squared right parenthesis i with hat on top plus left parenthesis 4 t subscript 2 plus 4 t subscript 2 squared right parenthesis j with hat on top r with bar on top left parenthesis 2 right parenthesis equals left parenthesis 3 left parenthesis 2 right parenthesis minus 3 left parenthesis 2 right parenthesis squared right parenthesis i with hat on top plus left parenthesis 4 left parenthesis 2 right parenthesis plus 4 left parenthesis 2 right parenthesis squared right parenthesis j with hat on top r with bar on top subscript 2 equals negative 6 i with hat on top plus 24 j with hat on top space straight m end style

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Percepatan motor yang sedang berjalan dapat disampaikan dalam persamaan a ( t ) = − 2 t + 3 t 2 m / s 2 . Jika kecepatan motor saat t = 2 s adalah 12 m/s, maka kecepatan motor saat t = 3 s adalah ….

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