Sebuah bus yabg bratnya 2.5x10^3kg bergerak dgn lju 36km/jam mendekati halte tentukan jarak (m) yg di tempuh bus tersebut sampai halte bila gaya pengereman 5x10

Berat (massa) = 2,5x10³ = 2500 kg
Laju (v) = 36 km/jam = 36x1000/3600 = 10 m/s
Gaya (F) = 5x10³ = 5000 N
Jarak (m) = ...?

F=m.a
5000=2500.v/t
5000=2500.10/t
5000t=25000
t=5 s

v=s/t
10=s/5
s=50 m

= > Massa Benda ( m ) = 2,5 × 10³ Kg = 2,5 × 1000 = 2500 Kg
= > Laju Kendaraan ( v ) = 36 Km / Jam = 36 × 1000 / 3600 Sekon = 36000 m / 3600 Sekon = 10 m / s
= > Gaya ( F ) = 5 × 10³ N = 5 × 1000 = 5000 N
= > F = m . a
5000 N = 2500 Kg . v ÷ t
= 2500 Kg . 10 m / s ÷ t
= 25000 ÷ t
t = 25000 ÷ 5000
= 5 Sekon
= > Jadi , Jarak Tempuh ( s )
v = s ÷ t
10 m / s = s ÷ 5
s = 10 × 5
= 50 m
#SEMOGA_MEMBANTU ^_^