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NEET PHYSICSEasy

A particle has an initial velocity (3i^+4j^)(3\hat{i} + 4\hat{j}) and an acceleration (0.4i^+0.3j^)(0.4\hat{i} + 0.3\hat{j}). Its speed after 10 s10 \text{ s} is:

A

7 units

B

72 units7\sqrt{2} \text{ units}

C

8 units

D

82 units8\sqrt{2} \text{ units}

Step-by-Step Solution

  1. Formula: For motion with constant acceleration, the final velocity vector v\mathbf{v} is given by v=u+at\mathbf{v} = \mathbf{u} + \mathbf{a}t .
  2. Given Data:
  • Initial velocity u=3i^+4j^\mathbf{u} = 3\hat{i} + 4\hat{j}
  • Acceleration a=0.4i^+0.3j^\mathbf{a} = 0.4\hat{i} + 0.3\hat{j}
  • Time t=10 st = 10 \text{ s}
  1. Calculate Velocity Vector: v=(3i^+4j^)+(0.4i^+0.3j^)×10\mathbf{v} = (3\hat{i} + 4\hat{j}) + (0.4\hat{i} + 0.3\hat{j}) \times 10 v=(3i^+4j^)+(4i^+3j^)\mathbf{v} = (3\hat{i} + 4\hat{j}) + (4\hat{i} + 3\hat{j}) v=(3+4)i^+(4+3)j^=7i^+7j^\mathbf{v} = (3+4)\hat{i} + (4+3)\hat{j} = 7\hat{i} + 7\hat{j}
  2. Calculate Speed (Magnitude): The speed is the magnitude of the velocity vector: ∣v∣=vx2+vy2=72+72=49+49=98=72 units|\mathbf{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{7^2 + 7^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2} \text{ units}
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