Question

The velocity of a moving particle at any instant is $\hat{i} + \hat{j}$. The magnitude and direction of the velocity of the particle are 

• A. √2 units and 45° with the x-axis
• B. 2 units and 30° with the z-axis
• C. 2 units and 45° with the x-axis
• D. √2 units and 60° with the y-axis
• E. 2 units and 60° with the x-axis

Answer 1

The Correct Option is A

The velocity of a particle is given as \( \hat{i} + \hat{j} \), which means the velocity vector has components of 1 unit in both the \( x \)-direction and \( y \)-direction. So the velocity vector is:

\[ \vec{v} = \hat{i} + \hat{j} \] The magnitude of the velocity vector \( |\vec{v}| \) is given by the formula for the magnitude of a vector: \[ |\vec{v}| = \sqrt{(1)^2 + (1)^2} = \sqrt{2} \] So, the magnitude of the velocity is \( \sqrt{2} \) units. The direction of the velocity vector is given by the angle \( \theta \) it makes with the \( x \)-axis. The angle \( \theta \) can be calculated using the formula: \[ \tan(\theta) = \frac{\text{component in the } y \text{-direction}}{\text{component in the } x \text{-direction}} = \frac{1}{1} = 1 \] Thus, \[ \theta = \tan^{-1}(1) = 45^\circ \] Therefore, the magnitude of the velocity is \( \sqrt{2} \) units and the direction is \( 45^\circ \) with respect to the \( x \)-axis.

Correct Answer:

Correct Answer: (A) \( \sqrt{2} \) units and 45° with the x-axis