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NEET PHYSICSEasy

A particle has an initial velocity (2i^+3j^)(2\hat{i} + 3\hat{j}) and an acceleration (0.3i^+0.2j^)(0.3\hat{i} + 0.2\hat{j}). The magnitude of velocity after 10 s10 \text{ s} will be:

A

9\sqrt{2} \text{ units}

B

5\sqrt{2} \text{ units}

C

5 \text{ units}

D

9 \text{ units}

Step-by-Step Solution

  1. Formula: For motion with constant acceleration in a plane, the velocity vector v\mathbf{v} at time tt is given by v=v0+at\mathbf{v} = \mathbf{v}_0 + \mathbf{a}t, where v0\mathbf{v}_0 is the initial velocity and a\mathbf{a} is the acceleration .
  2. Given Data:
  • Initial velocity v0=2i^+3j^\mathbf{v}_0 = 2\hat{i} + 3\hat{j}
  • Acceleration a=0.3i^+0.2j^\mathbf{a} = 0.3\hat{i} + 0.2\hat{j}
  • Time t=10 st = 10 \text{ s}
  1. Calculation of Velocity Vector: v=(2i^+3j^)+(0.3i^+0.2j^)×10\mathbf{v} = (2\hat{i} + 3\hat{j}) + (0.3\hat{i} + 0.2\hat{j}) \times 10 v=(2i^+3j^)+(3i^+2j^)\mathbf{v} = (2\hat{i} + 3\hat{j}) + (3\hat{i} + 2\hat{j}) v=(2+3)i^+(3+2)j^=5i^+5j^\mathbf{v} = (2+3)\hat{i} + (3+2)\hat{j} = 5\hat{i} + 5\hat{j}
  2. Calculation of Magnitude: The magnitude of the velocity vector is ∣v∣=vx2+vy2|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}. ∣v∣=52+52=25+25=50=52 units|\mathbf{v}| = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \text{ units}
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