Question

A balloon starts rising from the surface of the Earth. The ascension rate is constant and equal to v_o. Due to the wind the balloon gathers the horizontal velocity component v_x=ay, where a is a constant and y is the height of ascent. Find the total, tangential, and normal accelerations of the balloon.

• A. $a_r=a{v}_0;a_t=\frac{a{v}_0}{\sqrt{1+{\left(\frac{ay}{v_0}\right)}^2}};a_N=a{v}_0$
• B. $a_r=a{v}_0;a_t=a{v}_0;a_N=0$
• C. $a_r=a{v}_0;a_t=\frac{a^{2}y}{\sqrt{1+\frac{a^2{y}^2}{v_0^2}}};a_N=\frac{a{v}_0}{\sqrt{(1+\frac{a^2{y}^2}{v_0^2})}}$
• D. $a_r=\frac{a^{2}y}{\sqrt{1}+\frac{a^2{y}^2}{v_0^2}};a_t=a{v}_0;a_N=a{v}_0$

Answer 1

The correct option is C

$a_r=a{v}_0;a_t=\frac{a^{2}y}{\sqrt{1+\frac{a^2{y}^2}{v_0^2}}};a_N=\frac{a{v}_0}{\sqrt{(1+\frac{a^2{y}^2}{v_0^2})}}$

The path of the balloon will look something like this

After t sec ballon would have gone a height of $v_{0}t$then at that very instance the balloon's $v_x$ will be
$v_x=a{v}_{0}t\Rightarrow a_x=a{v}_0$
$V_y=v_0$
$a_y=0$
$a_r=totalacceleration=a{v}_0.......\left(1\right)$
$v_{tangentical}$ is actually the resultant velocity
$\Rightarrow v_t=\sqrt{v^{2}0+(a{v}_{0}t{)}^2}$
$\Rightarrow v_t=v_0\sqrt{1+a^2{t}^2}$
$a_t=\frac{d{v}_t}{dt}=\frac{v_0{a}^{2}t}{\sqrt{1+a^2{t}^2}}\Rightarrow \frac{a^{2}y}{\sqrt{1+{\left(\frac{ay}{v_0}\right)}^2}}$
Now we know $a_t^2+a_N^2=a_r^2$
$\Rightarrow \frac{v_0^2{a}^4{t}^2}{(1+a^2{t}^2)}+a_N^2=a^2{v}_0^2$
$a_N=\sqrt{v_0^2{a}^4{t}^2-\frac{v_0^2{a}^4{t}^2}{(1+a^2{t}^2)}}$
$a_N=a{v}_0\sqrt{\frac{1+a^2{t}^2-a^2{t}^2}{(1+a^2{t}^2)}}$
$a_N=\frac{a{v}_0}{\sqrt{1+a^2{t}^2}}$
$\Rightarrow \frac{a{v}_0}{\sqrt{1+{\left(\frac{ay}{v_0}\right)}^2}}$