Question

The position vector of a moving body at any instant of time is given as r = ( 5t²\(\hat{i}\) - 5t\(\hat{j}\)) m. The magnitude and direction of velocity at \( t = 2 \, \text{s} \) are:

• A. \( 5\sqrt{17} \, \text{m/s}, \text{making an angle of } \tan^{-1} \left( \frac{5}{4} \right) \text{ with the } -\hat{y} \text{ axis} \)
• B. \( 5\sqrt{15} \, \text{m/s}, \text{making an angle of } \tan^{-1} \left( \frac{5}{4} \right) \text{ with the } -\hat{y} \text{ axis} \)
• C. \( 5\sqrt{15} \, \text{m/s}, \text{making an angle of } \tan^{-1} \left( \frac{5}{3} \right) \text{ with the } \hat{x} \text{ axis} \)
• D. \( 5\sqrt{17} \, \text{m/s}, \text{making an angle of } \tan^{-1} \left( \frac{5}{4} \right) \text{ with the } +\hat{x} \text{ axis} \)

Hint:

To find the magnitude of velocity, differentiate the position vector with respect to time. To find the direction, use the ratio of the components of the velocity vector along the \( \hat{x} \) and \( \hat{y} \) axes.

Answer 1

The Correct Option is A

The velocity vector is the derivative of the position vector \( \mathbf{r} \) with respect to time: \[ \mathbf{v} = \frac{d}{dt} \left( 5t^2 \hat{i} - 5t \hat{j} \right) = 10t \hat{i} - 5 \hat{j}. \] At \( t = 2 \) s, the velocity is: \[ \mathbf{v} = 20 \hat{i} - 5 \hat{j} \, \text{m/s}. \] The magnitude of the velocity is: \[ |\mathbf{v}| = \sqrt{20^2 + (-5)^2} = \sqrt{400 + 25} = \sqrt{425} = 5\sqrt{17} \, \text{m/s}. \] The direction of the velocity is given by the angle \( \theta \) with the \( -\hat{y} \) axis: \[ \tan \theta = \frac{|\text{component along } \hat{x}|}{|\text{component along } \hat{y}|} = \frac{20}{5} = 4. \] Thus, \( \theta = \tan^{-1}(4) \). 
Final Answer: \( 5\sqrt{17} \, \text{m/s}, \text{making an angle of } \tan^{-1}(4) \text{ with the } -\hat{y} \text{ axis} \).

Answer 2

Step 1: Understand the position vector.
The position vector of the moving body at any instant of time is given by: \[ \vec{r} = ( 5t^2 \hat{i} - 5t \hat{j}) \, \text{m}. \] This represents the position of the body in a 2D plane at any time \( t \), with components in the \( \hat{i} \) (x) and \( \hat{j} \) (y) directions.

Step 2: Differentiate to find the velocity.
Velocity is the time derivative of the position vector. So, differentiate \( \vec{r} \) with respect to time \( t \): \[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt} \left( 5t^2 \hat{i} - 5t \hat{j} \right). \] This gives: \[ \vec{v} = 10t \hat{i} - 5 \hat{j}. \] Thus, the velocity vector as a function of time is \( \vec{v} = (10t \hat{i} - 5 \hat{j}) \, \text{m/s}.

Step 3: Find the velocity at \( t = 2 \, \text{s} \).
Substitute \( t = 2 \, \text{s} \) into the velocity equation: \[ \vec{v}(2) = 10(2) \hat{i} - 5 \hat{j} = 20 \hat{i} - 5 \hat{j} \, \text{m/s}. \] So, the velocity at \( t = 2 \, \text{s} \) is \( \vec{v}(2) = 20 \hat{i} - 5 \hat{j} \, \text{m/s}.

Step 4: Calculate the magnitude of the velocity.
The magnitude of the velocity vector \( \vec{v} \) is given by: \[ |\vec{v}| = \sqrt{(v_x)^2 + (v_y)^2}, \] where \( v_x = 20 \, \text{m/s} \) and \( v_y = -5 \, \text{m/s} \). Substituting the values: \[ |\vec{v}| = \sqrt{(20)^2 + (-5)^2} = \sqrt{400 + 25} = \sqrt{425} = 5\sqrt{17} \, \text{m/s}. \]

Step 5: Calculate the direction of the velocity.
The direction of the velocity vector is given by the angle \( \theta \) with the negative \( \hat{y} \)-axis. We can find this using: \[ \tan \theta = \frac{|v_x|}{|v_y|}. \] Substitute the values \( v_x = 20 \) and \( v_y = -5 \): \[ \tan \theta = \frac{|20|}{|-5|} = \frac{20}{5} = 4. \] Thus, the angle \( \theta \) is: \[ \theta = \tan^{-1}(4). \]

Final Answer:
The magnitude of the velocity at \( t = 2 \, \text{s} \) is \( 5\sqrt{17} \, \text{m/s} \), and the direction is making an angle of \( \tan^{-1} \left( \frac{5}{4} \right) \) with the negative \( \hat{y} \)-axis.