Mohon bantuan nya kawan

[tex]\displaystyle \text{karena }v\text{ maksimum, maka}\\t=-\frac{b}{2a}\\t=-\frac{6}{2(-1)}\\t=3\,\text{s}\\\\v(3)=-3^2+6(3)-5\\v(3)=-9+18-5\\v(3)=4\,\text{m/s}\\\\s=s_0+vt\\s=0+4\cdot3\\\boxed{\boxed{s=12\,\text{m/s}}}[/tex]

V(t) = -t^2 + 6t - 5
Jarak = S(t) = integral (-t^2 + 6t - 5) dt
S(t) = -1/3 t^3 + 6/2 t^2 - 5t + C
S(t) = -1/3 t^3 + 3t^2 - 5t + C ==> saat t = 0, jaraknya = 0
S(0) = -1/3 (0)^3 + 3(0)^2 - 5(0) + C = 0
C = 0
S(t) = -1/3 t^3 + 3t^2 - 5t

V(t) = -t^2 + 6t - 5
Kecepatan tertinggi saat V'(t) = 0
V'(t) = -2t + 6 = 0
=> -2t = -6
=> t = 3 detik
Jarak saat t = 3 detik
S(3) = -1/3 (3)^3 + 3(3)^2 - 5(3)
= -9 + 27 - 15
= 3 m