Answer:
Hello,
Step-by-step explanation:
[tex]v_x(t)=ln(t-2)\ \Longrightarrow\ a_x(t)=\dfrac{1}{t-2} \\\\v_y(t)=e^{2t-t^2}\ \Longrightarrow\ a_y(t)=e^{2t-t^2}*(2-2t)\\\\t=3\ i\ suppose\ \\\\a_x(3)=\dfrac{1}{3-2}=1\\\\a_y(t)=e^{6-9}*(2-6)=-\dfrac{4}{e^3}[/tex]