Question

Persamaan gerak suatu partikel dinyatakan oleh fungsi x = ( 5t^3 + 4t^2 + 4 ) meter . Hitunglah :
a.Kecepatan rata - rata dalam selang waktu 0 sekon sampai 2 sekon
b.Kecepatan saat t = 1 sekon
c.Percepatan rata - rata dalam selang waktu 1 sekon sampai 2 sekon
d.Percepatan saat t = 1 sekon

Answer 1

Diket:
x = (5t³ + 4t² + 4) m
maka:
a. v rata2 saat t = 0 s sd t = 2s
r (t=0) = (5.0³ + 4.0² + 4) = 4m
r (t=2) = (5.2³ + 4.2² + 4) = (40 + 16 + 4 ) = 60m
maka:
v = (r2 - r1)/(t2 - t1)
v = (60 - 4 )/(2 - 0)
v = 56/2
v = 28m/s
,
b. kecepatan saat t = 1 s
v = dx/dt (5t³ + 4t² + 4)
v = 15t² + 8t m/s ---> substitusi nilai t = 1
v = 15.1² + 8.1
v = 23 m/s
,
c. percepatan rata2 saat t = 1s sd t = 2s
v (t=1) = 15t² + 8t m/s
v = 15.1² + 8.1 = 23m/s
v (t=2) = 15.2² + 8.2 = 60 + 16 = 76m/s
maka:
a = (v2-v1)/(t2-t1)
a = (76 - 23)/(2-1)
a = 53 m/s²
,
d. percepatan saat t = 1
a = dv/dt (15t² + 8t)
a = 30t + 8 ---> substitusi t = 1
a = 30.1 + 8
a = 38m/s²