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Motion in A Plane question

2012 · Q155
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Motion in A Plane question

2012 · Q155

NEETPhysicsMotion in A PlaneMCQ+4 / −1
A particle has initial velocity (2i→+3j→)\left( {2\overrightarrow i + 3\overrightarrow j } \right)(2i+3j​) and acceleration (0.3i→+0.2j→)\left( {0.3\overrightarrow i + 0.2\overrightarrow j } \right)(0.3i+0.2j​). The magnitude of velocity after 10 seconds will be
  1. A
    929\sqrt 292​
  2. B
    525\sqrt 252​
  3. C
    5 units
  4. D
    9 units
View written solutionFree

Correct answer: B

P→\overrightarrow P P = vector sum = A→+B→\overrightarrow A + \overrightarrow B A+B

Q→\overrightarrow Q Q​ = Vector differences = A→−B→\overrightarrow A - \overrightarrow B A−B

Since P→\overrightarrow P P and Q→\overrightarrow Q Q​ are perpendicular

∴\therefore∴ P→.Q→=0⇒(A→+B→).(A→−B→)=0\overrightarrow P .\overrightarrow Q = 0 \Rightarrow \left( {\overrightarrow A + \overrightarrow B } \right).\left( {\overrightarrow A - \overrightarrow B } \right) = 0P.Q​=0⇒(A+B).(A−B)=0

⇒A2=B2=∣A→∣=∣B→∣ \Rightarrow {A^2} = {B^2} = \left| {\overrightarrow A } \right| = \left| {\overrightarrow B } \right|⇒A2=B2=​A​=​B​

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