Physics - Calculus Question with Solution | TestHub

The given position function is of the form $x(t)=A\sin (\theta )+B\cos (\theta )$. This expression can be rewritten as $R\sin (\theta +\alpha )$, where $R=\sqrt{A^2+B^2}$. In this case, $A=5$ and $B=12$. So, the maximum displacement $R=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$ m.