A particle has an initial velocity $(2\hat{i} + 3\hat{j})$ and an acceleration $(0.3\hat{i} + 0.2\hat{j})$. Th

Question

A particle has an initial velocity $(2\hat{i} + 3\hat{j})$ and an acceleration $(0.3\hat{i} + 0.2\hat{j})$. The magnitude of velocity after $10 	ext{ s}$ will be:

Accepted Answer

1. **Formula:** For motion with constant acceleration in a plane, the velocity vector $\mathbf{v}$ at time $t$ is given by $\mathbf{v} = \mathbf{v}_0 + \mathbf{a}t$, where $\mathbf{v}_0$ is the initial velocity and $\mathbf{a}$ is the acceleration .
2. **Given Data:**
   - Initial velocity $\mathbf{v}_0 = 2\hat{i} + 3\hat{j}$
   - Acceleration $\mathbf{a} = 0.3\hat{i} + 0.2\hat{j}$
   - Time $t = 10 	ext{ s}$
3. **Calculation of Velocity Vector:**
   $$ \mathbf{v} = (2\hat{i} + 3\hat{j}) + (0.3\hat{i} + 0.2\hat{j}) 	imes 10 $$
   $$ \mathbf{v} = (2\hat{i} + 3\hat{j}) + (3\hat{i} + 2\hat{j}) $$
   $$ \mathbf{v} = (2+3)\hat{i} + (3+2)\hat{j} = 5\hat{i} + 5\hat{j} $$
4. **Calculation of Magnitude:**
   The magnitude of the velocity vector is $|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}$.
   $$ |\mathbf{v}| = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} 	ext{ units} $$

Step-by-Step Solution

Practice All PHYSICS Questions