Question

The velocity of a particle having a magnitude of 10 ms^-1 in the direction of 60° with positive X-axis is

• A. 5i - 5√3j
• B. 5√3i - 5j
• C. 5√3i + 5j
• D. 5i + 5√3j

Answer 1

The Correct Option is D

To solve the problem, we need to resolve the velocity vector of a particle moving at 10 m/s at an angle of $60^\circ$ with respect to the positive X-axis into its vector components.

1. Understanding Vector Components:
Any vector $V$ with magnitude and direction $\theta$ can be resolved as:
$ \vec{V} = V \cos\theta \, \hat{i} + V \sin\theta \, \hat{j} $
Here,
$V = 10 \, \text{m/s}$,
$\theta = 60^\circ$

2. Applying Trigonometric Values:
We know that:
$ \cos(60^\circ) = \frac{1}{2} $
$ \sin(60^\circ) = \frac{\sqrt{3}}{2} $
Substitute into the component form:
$ \vec{V} = 10 \cdot \frac{1}{2} \, \hat{i} + 10 \cdot \frac{\sqrt{3}}{2} \, \hat{j} = 5 \, \hat{i} + 5\sqrt{3} \, \hat{j} $

Final Answer:
The velocity vector of the particle is $ 5 \hat{i} + 5\sqrt{3} \hat{j} $.

Answer 2

The correct option is: (D): 5i + 5√3j .

The given velocity vector can be represented as a combination of its components along the X-axis and Y-axis. The velocity's X-component (Vx) is given by the magnitude (10 m/s) multiplied by the cosine of the angle (60°), and the Y-component (Vy) is given by the magnitude multiplied by the sine of the angle.

Vx = 10 m/s * cos(60°) = 10 * 0.5 = 5 m/s Vy = 10 m/s * sin(60°) = 10 * (√3 / 2) = 5√3 m/s

So, the velocity vector is 5i + 5√3j, which corresponds to 5 m/s in the positive X-direction and 5√3 m/s in the positive Y-direction.