Question

The position vector of the particle at 't' s is given by $\overrightarrow{r} = (t\hat{i} - \frac{t^2}{2}\hat{j})$ m. Then which of the following statements are correct?

• A. The average speed of the particle in the time interval from t = 0 s to t=2s is $\frac{2\sqrt{5} + ln(2 + \sqrt{5})}{4}$ m/s
• B. The magnitude of average velocity of the particle in the time interval from t=0 s to t= 2s is $\sqrt{2}$ m/s
• C. The magnitude of instantaneous acceleration of the particle at t = 2s is $1m/s^2$
• D. The instantaneous speed of the particle at 2s, is $\sqrt{5}$ m/s

Answer 1

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

All options A, B, C, D are correct.

Answer 2

The position vector of the particle is given by $\overrightarrow{r} = (t\hat{i} - \frac{t^2}{2}\hat{j})$ m.

First, let's find the velocity and acceleration vectors. The instantaneous velocity vector is the derivative of the position vector with respect to time: $\overrightarrow{v} = \frac{d\overrightarrow{r}}{dt} = \frac{d}{dt}(t\hat{i} - \frac{t^2}{2}\hat{j}) = (1\hat{i} - \frac{2t}{2}\hat{j}) = (\hat{i} - t\hat{j})$ m/s.

The instantaneous acceleration vector is the derivative of the velocity vector with respect to time: $\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt} = \frac{d}{dt}(\hat{i} - t\hat{j}) = (0\hat{i} - 1\hat{j}) = -\hat{j}$ m/s$^2$.

Now let's evaluate each statement:

A) The average speed of the particle in the time interval from t = 0 s to t=2s is $\frac{2\sqrt{5} + ln(2 + \sqrt{5})}{4}$ m/s Average speed is total distance traveled divided by total time. The instantaneous speed is the magnitude of the velocity vector: $|\overrightarrow{v}| = \sqrt{1^2 + (-t)^2} = \sqrt{1 + t^2}$. The total distance traveled ($S$) from $t=0$ to $t=2$ s is the integral of the instantaneous speed: $S = \int_{0}^{2} |\overrightarrow{v}| dt = \int_{0}^{2} \sqrt{1 + t^2} dt$. Using the standard integral formula $\int \sqrt{a^2 + x^2} dx = \frac{x}{2}\sqrt{a^2 + x^2} + \frac{a^2}{2}\ln|x + \sqrt{a^2 + x^2}|$, with $a=1$ and $x=t$: $S = \left[ \frac{t}{2}\sqrt{1 + t^2} + \frac{1}{2}\ln|t + \sqrt{1 + t^2}| \right]_{0}^{2}$ $S = \left( \frac{2}{2}\sqrt{1 + 2^2} + \frac{1}{2}\ln|2 + \sqrt{1 + 2^2}| \right) - \left( \frac{0}{2}\sqrt{1 + 0^2} + \frac{1}{2}\ln|0 + \sqrt{1 + 0^2}| \right)$ $S = \left( \sqrt{5} + \frac{1}{2}\ln(2 + \sqrt{5}) \right) - \left( 0 + \frac{1}{2}\ln(1) \right)$ Since $\ln(1) = 0$: $S = \sqrt{5} + \frac{1}{2}\ln(2 + \sqrt{5})$ m. The time interval is $\Delta t = 2 - 0 = 2$ s. Average speed $= \frac{S}{\Delta t} = \frac{\sqrt{5} + \frac{1}{2}\ln(2 + \sqrt{5})}{2} = \frac{2\sqrt{5} + \ln(2 + \sqrt{5})}{4}$ m/s. Statement A is correct.

B) The magnitude of average velocity of the particle in the time interval from t=0 s to t= 2s is $\sqrt{2}$ m/s Average velocity is total displacement divided by total time. Position at $t=0$ s: $\overrightarrow{r}(0) = (0\hat{i} - \frac{0^2}{2}\hat{j}) = \vec{0}$ m. Position at $t=2$ s: $\overrightarrow{r}(2) = (2\hat{i} - \frac{2^2}{2}\hat{j}) = (2\hat{i} - 2\hat{j})$ m. Displacement $\Delta \overrightarrow{r} = \overrightarrow{r}(2) - \overrightarrow{r}(0) = (2\hat{i} - 2\hat{j}) - \vec{0} = (2\hat{i} - 2\hat{j})$ m. Time interval $\Delta t = 2 - 0 = 2$ s. Average velocity $\overrightarrow{v}_{avg} = \frac{\Delta \overrightarrow{r}}{\Delta t} = \frac{2\hat{i} - 2\hat{j}}{2} = (\hat{i} - \hat{j})$ m/s. The magnitude of average velocity is $|\overrightarrow{v}_{avg}| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$ m/s. Statement B is correct.

C) The magnitude of instantaneous acceleration of the particle at t = 2s is $1m/s^2$ The acceleration vector is $\overrightarrow{a} = -\hat{j}$ m/s$^2$. The magnitude of acceleration is $|\overrightarrow{a}| = \sqrt{0^2 + (-1)^2} = \sqrt{1} = 1$ m/s$^2$. Since the acceleration is constant, its magnitude is $1$ m/s$^2$ at any time $t$, including $t=2$ s. Statement C is correct.

D) The instantaneous speed of the particle at 2s, is $\sqrt{5}$ m/s The instantaneous speed is $|\overrightarrow{v}| = \sqrt{1 + t^2}$. At $t=2$ s, the instantaneous speed is $|\overrightarrow{v}(2)| = \sqrt{1 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$ m/s. Statement D is correct.

All four statements A, B, C, and D are correct.