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Sebuah benda bergerak pada bidang xy dengan kecepa...

Sebuah benda bergerak pada bidang xy dengan kecepatan v subscript x left parenthesis t right parenthesis equals t minus 2 dan v subscript y left parenthesis t right parenthesis equals 3 t plus 1. Jika diketahui t = 0 ketika benda berada di x subscript 0 equals negative 1 m dan y subscript 0 equals 2 space m, pada saat t = 2 detik

  1. y = 0 m dan besar percepatan square root of 10 space m divided by s squared

  2. y = 0 m dan besar percepatan square root of 8 space m divided by s squared

  3. y = 10 m dan besar percepatan square root of 10 space m divided by s squared

  4. x = 0 m dan besar percepatan square root of 8 space m divided by s squared

  5. y = 10 m dan besar percepatan square root of 8 space m divided by s squared

Jawaban:

x open parentheses t close parentheses equals x subscript 0 plus integral v subscript x space d t equals x subscript 0 plus 1 half t squared minus 2 t  y open parentheses t close parentheses equals y subscript 0 plus integral v subscript y space d t equals y subscript 0 plus 3 over 2 t squared plus t  S a a t space t space equals space 2 space d e t i k comma space m a k a space colon  x open parentheses 2 close parentheses equals negative 1 plus 1 half open parentheses 2 close parentheses squared minus 2 open parentheses 2 close parentheses equals negative 1 plus 2 minus 4 equals negative 3 space m  y open parentheses 2 close parentheses equals 2 plus 3 over 2 open parentheses 2 close parentheses squared plus open parentheses 2 close parentheses equals 2 plus 6 plus 2 equals 10 space m  P e r c e p a tan space a d a l a h space t u r u n a n space p e r t a m a space d a r i space f u n g s i space k e c e p a tan space t e r h a d a p space w a k t u comma space m a k a space colon  a equals fraction numerator d v over denominator d t end fraction  a subscript x equals fraction numerator d v subscript x over denominator d t end fraction equals fraction numerator d over denominator d t end fraction open parentheses t minus 2 close parentheses equals 1 space m divided by s squared  a subscript y equals fraction numerator d v subscript y over denominator d t end fraction equals fraction numerator d over denominator d t end fraction open parentheses 3 t plus 1 close parentheses equals 3 space m divided by s squared  a subscript R equals square root of a subscript x superscript 2 plus a subscript y superscript 2 end root  a subscript R equals square root of open parentheses 1 close parentheses squared plus open parentheses 3 close parentheses squared end root  a subscript R equals square root of 10 space m divided by s squared

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