Tiga buah vektor satuan, satuan, dan satuan. Tentukan : a. b.

Pertanyaan

Tiga buah vektor $begin mathsize 14px style stack F subscript 1 with rightwards arrow on top equals left parenthesis negative 15 space i with hat on top space plus space 10 space j with hat on top space right parenthesis end style$ satuan, $begin mathsize 14px style stack F subscript 2 with rightwards arrow on top equals left parenthesis 10 space i with hat on top space plus space 5 space j with hat on top space right parenthesis end style$ satuan, dan $begin mathsize 14px style stack F subscript 3 with rightwards arrow on top equals left parenthesis 5 space i with hat on top space minus space 4 space j with hat on top space right parenthesis end style$ satuan. Tentukan :
a. $begin mathsize 14px style stack F subscript 1 with rightwards arrow on top space plus space stack F subscript 2 with rightwards arrow on top space minus space stack F subscript 3 with rightwards arrow on top end style$

b. $begin mathsize 14px style stack F subscript 1 with rightwards arrow on top space minus space stack F subscript 2 with rightwards arrow on top space minus space stack F subscript 3 with rightwards arrow on top end style$

Pembahasan Video:

Pembahasan Soal:

Diketahui :

$undefined$

$stack size 14px F subscript size 14px 2 with size 14px rightwards arrow on top size 14px equals size 14px left parenthesis size 14px 10 size 14px space size 14px i with size 14px hat on top size 14px space size 14px plus size 14px space size 14px 5 size 14px space size 14px j with size 14px hat on top size 14px space size 14px right parenthesis$

$stack size 14px F subscript size 14px 3 with size 14px rightwards arrow on top size 14px equals size 14px left parenthesis size 14px 5 size 14px space size 14px i with size 14px hat on top size 14px space size 14px minus size 14px space size 14px 4 size 14px space size 14px j with size 14px hat on top size 14px space size 14px right parenthesis$

Ditanyakan :
a. $stack F subscript 1 with rightwards arrow on top space plus space stack F subscript 2 with rightwards arrow on top space minus space stack F subscript 3 with rightwards arrow on top$
b. $stack F subscript 1 with rightwards arrow on top space minus space stack F subscript 2 with rightwards arrow on top space minus space stack F subscript 3 with rightwards arrow on top$ .

Pembahasan :

$begin mathsize 14px style stack F subscript 1 with rightwards arrow on top space plus space stack F subscript 2 with rightwards arrow on top space minus space stack F subscript 3 with rightwards arrow on top equals left parenthesis negative 15 space i with hat on top plus 10 space j with hat on top right parenthesis space plus space left parenthesis 10 space i with hat on top plus 5 space j with hat on top right parenthesis space minus space left parenthesis 5 space i with hat on top minus 4 space j with hat on top right parenthesis equals left parenthesis negative 10 space i with hat on top plus 19 space j with hat on top right parenthesis space end style$

$begin mathsize 14px style stack F subscript 1 with rightwards arrow on top space minus space stack F subscript 2 with rightwards arrow on top space minus space stack F subscript 3 with rightwards arrow on top equals left parenthesis negative 15 space i with hat on top space plus space 10 space j with hat on top right parenthesis space minus left parenthesis 10 space i with hat on top space plus space 5 space j with hat on top right parenthesis space minus space left parenthesis 5 space i with hat on top space minus space 4 space j with hat on top right parenthesis equals left parenthesis negative 30 space i with hat on top space plus space 9 space j with hat on top right parenthesis end style$

Jadi $stack bold F subscript bold 1 with bold rightwards arrow on top bold space bold plus bold space stack bold F subscript bold 2 with bold rightwards arrow on top bold space bold minus bold space stack bold F subscript bold 3 with bold rightwards arrow on top bold space bold equals bold space bold left parenthesis bold minus bold 10 bold space bold i with bold hat on top bold plus bold 19 bold space bold j with bold hat on top bold right parenthesis bold space$ dan $stack bold F subscript bold 1 with bold rightwards arrow on top bold space bold minus bold space stack bold F subscript bold 2 with bold rightwards arrow on top bold space bold minus bold space stack bold F subscript bold 3 with bold rightwards arrow on top bold space bold equals bold space bold left parenthesis bold minus bold 30 bold space bold i with bold hat on top bold space bold plus bold space bold 9 bold space bold j with bold hat on top bold right parenthesis$

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 30 Agustus 2021

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Sebuah perahu menyebrangi sungai yang memiliki lebar 250 m dengan kecepatan alirannya 3 m/s. Perahu bergerak dengan arah menyilang dengan kecepatan 2 m/s. Lintasan yang ditempuh perahu sepanjang...

Pembahasan Soal:

Kecepatan aliran air $begin mathsize 14px style v subscript 1 equals 3 space bevelled straight m over straight s end style$

Kecepatan menyilang $begin mathsize 14px style v subscript 2 space equals space 2 space bevelled m over s end style$

Lebar sungai = 250 m

karena arah kedua gerak tegak lurus, maka kecepatan akhir adalah:

$begin mathsize 14px style v equals square root of left parenthesis v subscript 1 squared plus v subscript 2 squared right parenthesis end root v equals 3 comma 6 space bevelled straight m over straight s end style$

panjang lintasan:

$begin mathsize 14px style s equals left parenthesis L over v subscript 1 right parenthesis space v s equals 300 space straight m end style$

Jadi, jawaban yang tepat adalah B.

Roboguru

Perhatikan tiga vektor berikut! Nilai resultan ketiga vektor tersebut adalah....

Pembahasan Soal:

Reesultan vektor dapat dicari menggunakan metode penguraian.

resultan vektor di sumbu x:

$begin mathsize 14px style R with rightwards arrow on top subscript x equals capital sigma F with rightwards arrow on top subscript x equals F with rightwards arrow on top subscript 1 x end subscript plus F with rightwards arrow on top subscript 2 x end subscript plus F with rightwards arrow on top subscript 3 x end subscript space R with rightwards arrow on top subscript x equals 4 space straight N end style$

resultan vektor di sumbu y:

$undefined$

dan resultan vektor:

$begin mathsize 14px style R equals square root of x squared plus y squared end root R equals 4 comma 6 space straight N end style$

Jadi, jawaban yang tepat adalah A.

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Perhatikan kedua vektor kecepatan berikut! Vektor resultan pada sumbu X dan sumbu Y adalah...

Pembahasan Soal:

Diketahui
$stack v subscript 1 with rightwards harpoon with barb upwards on top equals 4 space straight m divided by straight s stack v subscript 2 with rightwards harpoon with barb upwards on top equals 2 space straight m divided by straight s theta subscript 1 equals 30 degree theta subscript 2 equals 60 degree$

Ditanyakan
Vektor resultan pada sumbu X dan sumbu Y

Jawab
Tinjau resultan vektor di masing-masing sumbu.

resultan vektor di sumbu x

$begin mathsize 14px style v with rightwards arrow on top subscript x equals v with rightwards arrow on top subscript 1 space cos space theta space minus space v with rightwards arrow on top subscript 2 space cos space theta v with rightwards arrow on top subscript x equals 4 space cos space 30 to the power of straight omicron minus 2 space cos space 60 to the power of straight omicron v with rightwards arrow on top subscript x equals left parenthesis 2 square root of 3 minus 1 right parenthesis space bevelled straight m over straight s end style$

resultan vektor di sumbu y

$Error converting from MathML to accessible text.$

Dengan demikian, vektor resultan pada sumbu X dan sumbu Y adalah $stack bold v subscript bold x with bold rightwards arrow on top bold equals bold left parenthesis bold 2 square root of bold 3 bold minus bold 1 bold right parenthesis bold space bold m bold divided by bold s bold semicolon$ $stack bold v subscript bold y with bold rightwards arrow on top bold equals bold left parenthesis bold 2 bold minus square root of bold 3 bold right parenthesis bold space bold m bold divided by bold s$ .

Jadi, jawaban yang tepat adalah B.

Roboguru

Perhatikan posisi titik pada koordinat cartesius berikut! Jika 1 kotak mewakili 0,5 meter, nyatakan posisi titik dalam vektor satuan, besar dan arahnya r adalah...

Pembahasan Soal:

Diketahui :

$x equals 3 space kotak cross times 0 comma 5 space equals 1 comma 5 space straight m y equals 2 space kotak cross times 0 comma 5 equals 1 space straight m$

Ditanya : besar dan arah vektor r ?

Jawab :

Sudut yang dibentuk oleh sumbu x negatif dan sumbu y negatif adalah 90°, maka $cos space 90 degree equals 0$ , maka besar vektor r dapat dicari dengan persamaan berikut.

$r equals square root of x squared plus y squared end root r equals square root of 1 comma 5 squared plus 1 squared end root r equals square root of 2 comma 25 plus 1 end root r equals square root of 3 comma 25 end root space straight m$

Untuk menentukan arahnya, kita cari terlebih dahulu sudut vektor r terhadap sumbu x negatif

$table attributes columnalign right center left columnspacing 0px end attributes row cell tan space alpha end cell equals cell y over x end cell row cell tan space alpha end cell equals cell fraction numerator 1 over denominator 1 comma 5 end fraction end cell row alpha equals cell 33 comma 69 degree end cell end table$

Maka arahnya

$theta equals 180 degree plus 33 comma 69 degree theta equals 213 comma 69 degree$

Besar dan arah vektor r adalah $square root of 3 comma 25 end root space straight m$ dan 213,69°.

Jadi, jawaban yang benar adalah B.

Roboguru

Sebuah vektor dituliskan dalam persamaan . Besar dan arah vektornya adalah...

Pembahasan Soal:

Besar vektor:

$begin mathsize 14px style u equals square root of 12 squared plus left parenthesis negative 5 right parenthesis squared end root u equals square root of 169 u equals 13 space satuan end style$

arah vektor:

$begin mathsize 14px style tan space alpha equals y over x tan space alpha equals fraction numerator negative 5 over denominator 12 end fraction alpha equals 157 comma 38 to the power of straight omicron end style$

Jadi, jawaban yang tepat adalah B.

Roboguru