dari keadaan diam sebuah mobil mencapai 100km/jam sesudah 10 detik pada percepatan konstan.Mobil ditahan kecepatannya selama 10 detik dan kemudian direm sampai

dalam soal mobil mengalami tiga proses
1 . GLBB dipercepat (glbb1)
2. GLB (glb)
3. GLBB diperlambat (glbb2)

v1 = 0 m/s
v2 = 100 km/jam = 250/9 m/s
v3 = 0 m/s

a.)
GLBB diperlambat
v3 = v2 + at
0 = 250/9 + (-5)t
5t = 250/9
t = 50/9 s

t total = t glbb1 + t glb + t glbb2
t total = 10 + 10 + 50/9
t total = 90/9 + 90/9 + 50/9
t total = 230/9
t total = 25,5556 s
t total = 26 s

b.)
GLBB dipercepat
v2 = v1 + at
250/9 = 0 + a10
a = 25/9 m/s²

s = v1t + 1/2 at² .
s = 0 + 1/2 . 25/9 . 100
s = 1250/9 m
s = 138,8889 m
s = 139 m

GLB
s = v2 t
s = 250/9 . 10
s = 2500/9 m
s = 277,7778 m
s = 278 m

GLBB diperlambat
v3² = v2² + 2as
0 = (250/9)² + 2.-5 .s
0 = 62500/81 - 10s
10s = 62500/81
s = 6250/81
s = 77,1605 m
s = 77 m

s total = sglbb1 + sglb + sglbb2
s total = 139 + 278 + 77
s total = 494 m