Question

A particle has initial velocity $ ( 2 \vec{i} + 3 \vec{j})$ and acceleartion ($ 0 . 3 \vec{i} + 0 . 2 \vec{j})$ The magnitude of velocity after 10 seconds will be

• A. 9 $ \sqrt 2 $ units
• B. 5 $ \sqrt 2 $ units
• C. 5 units
• D. 9 units

Answer 1

The Correct Option is B

To find the magnitude of the velocity of a particle after 10 seconds, we need to use the kinematic equation for velocity under constant acceleration:

\vec{v} = \vec{u} + \vec{a}t

where:

• \vec{v} is the final velocity vector.
• \vec{u} = (2\vec{i} + 3\vec{j}) is the initial velocity vector.
• \vec{a} = (0.3\vec{i} + 0.2\vec{j}) is the acceleration vector.
• t = 10 seconds is the time duration.

Substituting the known values into the formula, we get:

\vec{v} = (2\vec{i} + 3\vec{j}) + (0.3\vec{i} + 0.2\vec{j}) \times 10

Calculating the second term:

\vec{a}t = (0.3 \times 10)\vec{i} + (0.2 \times 10)\vec{j} = 3\vec{i} + 2\vec{j}

Adding the components:

\vec{v} = (2\vec{i} + 3\vec{j}) + (3\vec{i} + 2\vec{j}) = (2 + 3)\vec{i} + (3 + 2)\vec{j} = 5\vec{i} + 5\vec{j}

The magnitude of the velocity vector is given by:

|\vec{v}| = \sqrt{(5)^2 + (5)^2}

Calculating the magnitude:

|\vec{v}| = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \text{ units}

Thus, the magnitude of the velocity after 10 seconds is 5 \sqrt{2} units, which matches the correct answer provided in the options.