Question

A particle undergoes curvilinear motion in xy horizontal plane. Its position vector is given as $\overrightarrow{r}$ = (t²î + tj) m (mass of particle m = 1 kg.) Choose the correct statement(s) :

• A. The tangential acceleration of particle at t = 1 sec is $\frac{4}{\sqrt{5}}$ m/s²
• B. The centripetal acceleration of particle at t = 1 sec is $\sqrt{\frac{4}{5}}$ m/s²
• C. The radius of curvature of the curve at t = 1 sec is R = ($\frac{5\sqrt{5}}{2}$) m
• D. The radius of curvature of curve at t = 1 sec is R = ($\frac{5\sqrt{3}}{2}$) m

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

The particle's position vector is given by $\overrightarrow{r} = (t^2\hat{i} + t\hat{j})$ m.

1. Velocity Vector ($\overrightarrow{v}$): The velocity vector is the first derivative of the position vector with respect to time: $\overrightarrow{v} = \frac{d\overrightarrow{r}}{dt} = \frac{d}{dt}(t^2\hat{i} + t\hat{j}) = (2t\hat{i} + \hat{j})$ m/s.

2. Acceleration Vector ($\overrightarrow{a}$): The acceleration vector is the first derivative of the velocity vector with respect to time: $\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt} = \frac{d}{dt}(2t\hat{i} + \hat{j}) = 2\hat{i}$ m/s².

3. Evaluate at t = 1 sec: At $t = 1$ sec:

   • Velocity: $\overrightarrow{v}(1) = (2(1)\hat{i} + \hat{j}) = (2\hat{i} + \hat{j})$ m/s.
   • Speed: $v = |\overrightarrow{v}(1)| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$ m/s.
   • Acceleration: $\overrightarrow{a}(1) = 2\hat{i}$ m/s².
   • Magnitude of acceleration: $a = |\overrightarrow{a}(1)| = \sqrt{2^2 + 0^2} = 2$ m/s².

4. Tangential Acceleration ($a_t$): The tangential acceleration is the component of acceleration along the velocity vector. It can be calculated as $a_t = \frac{\overrightarrow{a} \cdot \overrightarrow{v}}{|\overrightarrow{v}|}$. At $t = 1$ sec: $a_t = \frac{(2\hat{i}) \cdot (2\hat{i} + \hat{j})}{\sqrt{5}} = \frac{(2)(2) + (0)(1)}{\sqrt{5}} = \frac{4}{\sqrt{5}}$ m/s². Alternatively, $a_t = \frac{dv}{dt}$. $v = \sqrt{(2t)^2 + 1^2} = \sqrt{4t^2 + 1}$. $\frac{dv}{dt} = \frac{1}{2\sqrt{4t^2 + 1}} \cdot (8t) = \frac{4t}{\sqrt{4t^2 + 1}}$. At $t = 1$ sec, $a_t = \frac{4(1)}{\sqrt{4(1)^2 + 1}} = \frac{4}{\sqrt{5}}$ m/s².

5. Centripetal Acceleration ($a_c$): The centripetal acceleration is the component of acceleration perpendicular to the velocity vector. The magnitude of total acceleration is related to tangential and centripetal accelerations by $a^2 = a_t^2 + a_c^2$. So, $a_c = \sqrt{a^2 - a_t^2}$. At $t = 1$ sec: $a_c = \sqrt{2^2 - \left(\frac{4}{\sqrt{5}}\right)^2} = \sqrt{4 - \frac{16}{5}} = \sqrt{\frac{20 - 16}{5}} = \sqrt{\frac{4}{5}}$ m/s².

6. Radius of Curvature (R): The centripetal acceleration is also given by the formula $a_c = \frac{v^2}{R}$. Therefore, the radius of curvature $R = \frac{v^2}{a_c}$. At $t = 1$ sec: $R = \frac{(\sqrt{5})^2}{\sqrt{\frac{4}{5}}} = \frac{5}{\frac{2}{\sqrt{5}}} = \frac{5\sqrt{5}}{2}$ m.

All statements (A), (B), and (C) are correct.