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NEET PHYSICSEasy

A particle has initial velocity (3i^+4j^)(3\hat{i}+4\hat{j}) and has 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

7\sqrt{2} units

C

8.5 units

D

10 units

Step-by-Step Solution

  1. Identify Given Values:
  • Initial velocity u⃗=3i^+4j^\vec{u} = 3\hat{i} + 4\hat{j}
  • Acceleration a⃗=0.4i^+0.3j^\vec{a} = 0.4\hat{i} + 0.3\hat{j}
  • Time t=10 st = 10 \text{ s}
  1. Apply Kinematic Equation: For motion with constant acceleration in a plane, the velocity vector at time tt is given by v⃗=u⃗+a⃗t\vec{v} = \vec{u} + \vec{a}t .
  2. Calculate Final Velocity Vector: v⃗=(3i^+4j^)+10(0.4i^+0.3j^)\vec{v} = (3\hat{i} + 4\hat{j}) + 10(0.4\hat{i} + 0.3\hat{j}) v⃗=(3i^+4j^)+(4i^+3j^)\vec{v} = (3\hat{i} + 4\hat{j}) + (4\hat{i} + 3\hat{j}) v⃗=(3+4)i^+(4+3)j^=7i^+7j^\vec{v} = (3+4)\hat{i} + (4+3)\hat{j} = 7\hat{i} + 7\hat{j}
  3. Calculate Speed: Speed is the magnitude of the velocity vector. ∣v⃗∣=vx2+vy2=72+72=49+49=98=72 units|\vec{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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