Question

Решите 2 задание......

Answer 1

$S(t)=3t^2-cos(2t)-10\S'(t)=V(t)\S''(t)=a(t)\\S'(t)=6t+2sin(2t)\V(15)=6*15+2*sin(30)\V(15)=81\S''(t)=6+4cos(2t)\a(15)=6+4cos(30)\a(15)=6+2 sqrt{3}=9,46$