Q123 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Physics

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

  • Correct answer
    A
    $\frac{\sqrt{3} y}{2}, \frac{y}{2}$
  • B
    $\frac{\sqrt{2} y}{3}, \frac{\sqrt{3} y}{2}$
  • C
    $\frac{\sqrt{3} y}{2}, \frac{5 y}{2}$
  • D
    $2 \sqrt{3} y, \frac{11 y}{\sqrt{3}}$

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 $$ a_n=\left|\mathbf{a}_n\right|=\frac{x y}{\sqrt{x^2+y^2 \cdot \frac{3 x^2}{y^2}}}=\frac{x y}{2 x}=\frac{y}{2} $$

Hence, the correct answer is option: 'A'.

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