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Sebuah objek berjalan sepanjang suatu garis koordinat menurut percepatan α (dalam centimeter per detik) dengan kecepatan awal v 0 ​ (dalam centimeter per detik) dan jarak s 0 ​ (dalam centimeter). Tentukan kecepatan v beserta jarak berarah s setelah 2 detik. c. α = 3 2 t + 1 ​ , v 0 ​ = 0 , s 0 ​ = 10

Sebuah objek berjalan sepanjang suatu garis koordinat menurut percepatan (dalam centimeter per detik) dengan kecepatan awal (dalam centimeter per detik) dan jarak (dalam centimeter). Tentukan kecepatan  beserta jarak berarah  setelah 2 detik.

c.  ,   

  1. ... 

  2. ... 

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Pembahasan

a. Kecepatan setelah detik Misal Karena , maka Sehingga Kecepatan setelah detik b. Jarak setelah 2 detik Misal Sehingga Karena Sehingga Jarak setelah 2 detik Jadi kecepatan setelah 2 detik dan jarak setelah 2 detik

a. Kecepatan setelah undefined detik

v equals integral a blank d t v equals integral cube root of 2 t plus 1 end root d t

Misal

open parentheses 2 t plus 1 close parentheses equals u 2 d t equals d u d t equals fraction numerator d u over denominator 2 end fraction v equals integral cube root of u fraction numerator d u over denominator 2 end fraction v equals 1 half integral u to the power of 1 third end exponent d u v equals 1 half open parentheses fraction numerator 1 over denominator 1 third plus 1 end fraction close parentheses u to the power of 1 third plus 1 end exponent plus C v equals 1 half fraction numerator 1 over denominator 4 over 3 end fraction u to the power of 4 over 3 end exponent plus C v equals 1 half 3 over 4 u to the power of 4 over 3 end exponent plus C v equals 3 over 8 u to the power of 4 over 3 end exponent plus C v equals 3 over 8 open parentheses 2 t plus 1 close parentheses to the power of 4 over 3 end exponent plus C

Karena v subscript o equals 0, maka

v equals 3 over 8 open parentheses 2 t plus 1 close parentheses to the power of 4 over 3 end exponent plus C 0 equals 3 over 8 open parentheses 2 left parenthesis 0 right parenthesis plus 1 close parentheses to the power of 4 over 3 end exponent plus C 0 equals 3 over 8 open parentheses 1 close parentheses to the power of 4 over 3 end exponent plus C 0 equals 3 over 8 plus C C equals negative 3 over 8

Sehingga

v equals 3 over 8 open parentheses 2 t plus 1 close parentheses to the power of 4 over 3 end exponent minus 3 over 8

Kecepatan setelah 2 detik

v equals 3 over 8 open parentheses 2 t plus 1 close parentheses to the power of 4 over 3 end exponent minus 3 over 8 v equals 3 over 8 open parentheses 2 left parenthesis 2 right parenthesis plus 1 close parentheses to the power of 4 over 3 end exponent minus 3 over 8 v equals 3 over 8 open parentheses 5 close parentheses to the power of 4 over 3 end exponent minus 3 over 8 v equals 3 over 8 open parentheses 5 cube root of 5 minus 1 close parentheses cm over detik

b. Jarak setelah 2 detik

s equals integral v blank d t s equals integral open parentheses 3 over 8 open parentheses 2 t plus 1 close parentheses to the power of 4 over 3 end exponent minus 3 over 8 close parentheses d t

Misal

open parentheses 2 t plus 1 close parentheses equals u 2 d t equals d u d t equals fraction numerator d u over denominator 2 end fraction

Sehingga

s equals integral 3 over 8 u to the power of 4 over 3 end exponent fraction numerator d u over denominator 2 end fraction blank minus integral 3 over 8 d t s equals 3 over 8 open parentheses 1 half integral u to the power of 4 over 3 end exponent d u minus integral d t blank close parentheses s equals 3 over 8 open parentheses 1 half fraction numerator 1 over denominator left parenthesis 4 over 3 plus 1 right parenthesis end fraction u to the power of 4 over 3 plus 1 end exponent minus t plus C close parentheses s equals 3 over 8 open parentheses 1 half fraction numerator 1 over denominator 7 over 3 end fraction u to the power of 7 over 3 end exponent minus t right parenthesis close parentheses plus C s equals 3 over 8 open parentheses 1 half 3 over 7 u to the power of 7 over 3 end exponent minus t close parentheses plus C s equals 3 over 8 open parentheses 3 over 14 open parentheses 2 t plus 1 close parentheses to the power of 7 over 3 end exponent minus t close parentheses plus C

Karena s subscript 0 equals 10

s equals 3 over 8 open parentheses 3 over 14 open parentheses 2 t plus 1 close parentheses to the power of 7 over 3 end exponent minus t close parentheses plus C 10 equals 3 over 8 open parentheses 3 over 14 open parentheses 2 left parenthesis 0 right parenthesis plus 1 close parentheses to the power of 7 over 3 end exponent minus open parentheses 0 close parentheses close parentheses plus C 10 equals 3 over 8 open parentheses 3 over 14 open parentheses 1 close parentheses to the power of 7 over 3 end exponent minus open parentheses 0 close parentheses close parentheses plus C 10 equals 3 over 8 open parentheses 3 over 14 minus 0 close parentheses plus C 10 equals 3 over 8 open parentheses 3 over 14 close parentheses plus C 10 equals 9 over 112 plus C 10 minus 9 over 112 equals C 1111 over 112 equals C

Sehingga

s equals 3 over 8 open parentheses 3 over 14 open parentheses 2 t plus 1 close parentheses to the power of 7 over 3 end exponent minus t close parentheses plus 1111 over 112

Jarak setelah 2 detik

s equals 3 over 8 open parentheses 3 over 14 open parentheses 2 t plus 1 close parentheses to the power of 7 over 3 end exponent minus t close parentheses plus 1111 over 112 s equals 3 over 8 open parentheses 3 over 14 open parentheses 2 left parenthesis 2 right parenthesis plus 1 close parentheses to the power of 7 over 3 end exponent minus 2 close parentheses plus 1111 over 112 s equals 3 over 8 open parentheses 3 over 14 open parentheses 4 plus 1 close parentheses to the power of 7 over 3 end exponent minus 2 close parentheses plus 1111 over 112 s equals 3 over 8 open parentheses 3 over 14 open parentheses 5 close parentheses to the power of 7 over 3 end exponent minus 2 close parentheses plus 1111 over 112 s equals 3 over 8 open parentheses 3 over 14 left parenthesis 25 cube root of 5 minus 2 close parentheses plus 1111 over 112 blank cm

Jadi kecepatan setelah 2 detik 3 over 8 open parentheses 5 cube root of 5 minus 1 close parentheses cm over detik dan jarak setelah 2 detik     3 over 8 open parentheses 3 over 14 left parenthesis 25 cube root of 5 minus 2 close parentheses plus 1111 over 112 blank cm

 

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