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NEET PHYSICSMOTION IN A PLANEEasy

Question

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 u=3i^+4j^\vec{u} = 3\hat{i} + 4\hat{j}
  • Acceleration a=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 v=u+at\vec{v} = \vec{u} + \vec{a}t .
  2. Calculate Final Velocity Vector: v=(3i^+4j^)+10(0.4i^+0.3j^)\vec{v} = (3\hat{i} + 4\hat{j}) + 10(0.4\hat{i} + 0.3\hat{j}) v=(3i^+4j^)+(4i^+3j^)\vec{v} = (3\hat{i} + 4\hat{j}) + (4\hat{i} + 3\hat{j}) v=(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. v=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}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOTION IN A PLANE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOTION IN A PLANEparticleinitialvelocityhatihatjacceleration

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