besar kecepatan suatu partikel yang mengalami perlambatan konstan

straight v subscript 0 equals 30 space straight m divided by straight s straight v subscript straight t space equals space 15 space straight m divided by straight s space rightwards arrow space straight s space equals space 75 space straight m straight v subscript straight t space equals space 0 space straight m divided by straight s space rightwards arrow space straight s space equals space ? straight v subscript straight t squared equals straight v subscript 0 squared plus 2 as left parenthesis 15 right parenthesis squared equals left parenthesis 30 right parenthesis squared plus 2 straight a left parenthesis 75 right parenthesis 225 equals 900 plus 150 straight a 150 straight a space equals space minus 675 straight a equals negative 4 comma 5 space straight m divided by straight s squared Pada space bagian space kedua space colon straight v subscript straight t squared equals straight v subscript 0 squared plus 2 as 0 squared equals left parenthesis 15 right parenthesis squared plus 2 left parenthesis negative 4 comma 5 right parenthesis straight s 0 equals 225 minus 9 straight s 9 straight s space equals space 225 straight s equals 25 space straight m

Jadi, benda berhenti setelah menempuh jarak lagi sebesar C. 25 m

Artikel Terkait:   potensi sumber daya alam singapura

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