Д А Н О :

$V_0 = 0 \\ t_1 = 3 c \\ t_2 = 4c \\ S(t_2) - S(t_1) = 7 m \\ t_3 = 10 c\\ - \\ V(t3) =? \\ S(t3) =? \\ - \\S = V_0t+\frac{at^2}{2} \\V_0=0 \\ S(t_2)-S(t_1)= \frac{at_2\-^2}{2}- \frac{at_1\-^2}{2} = a* \frac{t_2\-^2- t_1\-^2}{2} \\ a = \frac{2(S(t_2)-S(t_1))}{t_2\-^2- t_1\-^2} \\ a= \frac{2*7}{4^2-3^2}=2m/c^2 \\ \\ S(t_3)=V_0t_3 + at_3\-^2/2=at_3\-^2/2 \\ V(t_3) = V_0+at_3=at_3 \\ S(10) = 2*10^2/2=100 \ m \\ V(10) = 2*10=20 \ m/c$

ОТВЕТ: 100 м; 20 м/с