A particle moves in the $x y$-plane with velocity $\mathbf{v}=x \hat{\mathbf{i}}+y t \hat{\mathbf{j}}$. At…
A particle moves in the $x y$-plane with velocity $\mathbf{v}=x \hat{\mathbf{i}}+y t \hat{\mathbf{j}}$. At $t=\frac{x \sqrt{3}}{y}$, the
$\frac{\sqrt{3} y}{2}, \frac{y}{2}$
$\frac{\sqrt{2} y}{3}, \frac{\sqrt{3} y}{2}$
$\frac{\sqrt{3} y}{2}, \frac{5 y}{2}$
$2 \sqrt{3} y, \frac{11 y}{\sqrt{3}}$
Solution
Given, velocity of particle is $\mathbf{v}=x \hat{\mathbf{i}}+y t \hat{\mathbf{j}}$
So, $|\mathbf{v}|=v=\sqrt{x^2+y^2 t^2}$
Magnitude of tangential acceleration is
$
a_t=\frac{d v}{d t}=\frac{0+2 t y^2}{2 \sqrt{x^2+y^2 t^2}}
$
or $\quad a_t=\frac{t y^2}{\sqrt{x^2+y^2 t^2}}$
Now, substituting $t=\frac{x \sqrt{3}}{y}$, we get
$
a_t=\frac{\frac{x \sqrt{3}}{y} \times y^2}{\sqrt{x^2+y^2 \times\left(\frac{x^2 \times 3}{y^2}\right)}}
$
$
\Rightarrow \quad a_t=\frac{\sqrt{3} y}{2}
$
Also, total acceleration of particle is $\mathbf{a}=\frac{d v}{d t}$
$
\begin{array}{ll}
\Rightarrow & \mathbf{a}=\frac{d}{d t}(x \hat{\mathbf{i}}+y t \hat{\mathbf{j}}) \\
\Rightarrow & \mathbf{a}=y \hat{\mathbf{j}}
\end{array}
$
or magnitude of total acceleration is
$
a=|\mathbf{a}|=y
$
Hence, normal acceleration is
$
\mathbf{a}_n=\mathbf{a}_{\text {total }}-\mathbf{a}_{\text {tangential }}
$
So, $a_n=\left|\mathbf{a}_n\right|=$ magnitude of normal acceleration
$
\begin{aligned}
& =\sqrt{a^2-a_t^2}=\sqrt{y^2-\left(\frac{y^4 t^2}{x^2+y^2 t^2}\right)} \\
& =\sqrt{\frac{x^2 y^2}{x^2+y^2 t^2}}=\frac{x y}{\sqrt{x^2+y^2 t^2}}
\end{aligned}
$
Now with $t=\frac{x \sqrt{3}}{y}$, we get
$
a_n=\left|\mathbf{a}_n\right|=\frac{x y}{\sqrt{x^2+y^2 \cdot \frac{3 x^2}{y^2}}}=\frac{x y}{2 x}=\frac{y}{2}
$