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goldenfox [79]
3 years ago
6

Kecepatan sebuah benda dinyatakan dengan persamaan vx=8t+2 vy=12+2t, jika posisi awal pastikel (2,5), maka posisi benda pada saa

t t=4s adalah?
Physics
1 answer:
Vlad1618 [11]3 years ago
8 0
X = 2
y = 5

persamaan 1 : vx = 8t + 2 , x = 2 dan t = 0 karena posisi awal
                      2v = 8(0) + 2
                      2v = 2
                       v1 = 1

persamaan 2 : vy = 12 + 2t , y = 5 dan t = 0 karena posisi awal
                     5v = 12 + 2(0)
                     5v = 12
                     v2 = 2,4
masukin pada persamaan t = 4 dengan menggunakan v1 (v pada x) dan v2 (v pada y)
vx = 8t + 2 , t = 4 dan v = 1
1x = 8(4) + 2
x = 32 + 2
x = 34

vy = 12 + 2t , t = 4 dan v = 2,4
2,4y = 12 + 2(4)
2,4y = 20
y = 8,33

maka (x,y) = (34 ; 8,33) pada t = 4

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