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, denoted by \( \mathbf{v} \), is derived from the position vector \( \mathbf{r} \) by differentiating 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 time \( t = 2 \) s, the velocity is calculated as: \[ \mathbf{v} = 20 \hat{i} - 5 \hat{j} \, \text{m/s}. \] The magnitude of this velocity is determined by: \[ |\mathbf{v}| = \sqrt{20^2 + (-5)^2} = \sqrt{400 + 25} = \sqrt{425} = 5\sqrt{17} \, \text{m/s}. \] The direction of the velocity is defined by the angle \( \theta \) relative to the \( -\hat{y} \) axis, found using: \[ \tan \theta = \frac{|\text{component along } \hat{x}|}{|\text{component along } \hat{y}|} = \frac{20}{5} = 4. \] Consequently, \( \theta = \tan^{-1}(4) \).
Final Answer: \( 5\sqrt{17} \, \text{m/s}, \text{making an angle of } \tan^{-1}(4) \text{ with the } -\hat{y} \text{ axis} \).