For time \(t\gt 0\), the position of an object moving in the \(xy\)-plane is given by the parametric equations \(x(t)=t\cos{\left(\dfrac{t}{2}\right)}\)
and \(y(t)=\sqrt{t^2+2t}\). What is the speed of the object at time \(t=1\) ?
The magnitude or speed is given by:
$$ |v(t)|=\sqrt{v_x^2+v_y^2} $$
\(x(t)=t\cos{\left(\dfrac{t}{2}\right)}\)
$$ v_x=x'(t)=-\frac{t}{2}\sin{\frac{t}{2}}+\cos{\frac{t}{2}} $$
$$ x'(1)\approx 0.638 $$
\(y(t)=\sqrt{t^2+2t}\)
$$ v_y=y'(t)=\frac{2t+2}{2\sqrt{t^2+2t}} $$
$$ y'(1)\approx 1.155 $$
$$ |v(t)|=\sqrt{v_x^2+v_y^2} $$
$$ |v(1)|=\sqrt{(x'(1))^2+(y'(1))^2} $$
$$ \approx \boxed{1.319} $$