Question

The total, tangential and normal accelerations of the balloon.

Answer 1

For finding the tangential and normal accelerations, we require an expression for the speed as a function of height y
$v_y=v_0$ and $v_x=ky$
$\therefore v=\sqrt{v_x^2+v_y^2}$
$=\sqrt{v_0^2+k^2{y}^2}$
Therefore, tangential acceleration,
$a_t=\frac{dv}{dt}=\frac{k^{2}y}{\sqrt{v_0^2+k^2{y}^2}}\frac{dy}{dt}$
$=\frac{k^{2}y{v}_0}{\sqrt{v_0^2+k^2{y}^2}}$
or $a_t=\frac{k^{2}y}{\sqrt{1+k^2{y}^2{v}_0^2}}$
Now,
the total acceleration is, $a=\sqrt{{\left(\frac{d{v}_y}{dt}\right)}^2+{\left(\frac{d{v}_x}{dt}\right)}^2}$
$=\frac{d{v}_x}{dt}=k\frac{dy}{dt}=k{v}_0$
Normal acceleration, $a_n=\sqrt{a^2-a_t^2}$
$=\frac{k{v}_0}{\sqrt{1+{\left(\frac{ky}{v_0}\right)}^2}}$