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Motion in A Plane question

2009 · Shift 0 · Q75
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Motion in A Plane question

2009 · Shift 0 · Q75

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
A particle has an initial velocity 3i^+4j^3\widehat i + 4\widehat j3i+4j​ and an acceleration of 0.4i^+0.3j^0.4\widehat i + 0.3\widehat j0.4i+0.3j​. Its speed after 10 s is:
  1. A
    727\sqrt 272​ units
  2. B
    7 units
  3. C
    8.5 units
  4. D
    10 units
View written solutionFree

Correct answer: A

  1. Given data

Initial velocity: u⃗=3i^+4j^\vec u = 3\hat i + 4\hat ju=3i^+4j^​

Acceleration: a⃗=0.4i^+0.3j^\vec a = 0.4\hat i + 0.3\hat ja=0.4i^+0.3j^​

Time: t=10 st = 10\,\text{s}t=10s

We need the speed after 101010 s.

  1. Find velocity after 10 s

Using v⃗=u⃗+a⃗t\vec v = \vec u + \vec a tv=u+at

So, 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)v=(3i^+4j^​)+10(0.4i^+0.3j^​)

v⃗=(3i^+4j^)+(4i^+3j^)\vec v = (3\hat i + 4\hat j) + (4\hat i + 3\hat j)v=(3i^+4j^​)+(4i^+3j^​)

v⃗=7i^+7j^\vec v = 7\hat i + 7\hat jv=7i^+7j^​

  1. Find speed

Speed is the magnitude of velocity: ∣v⃗∣=72+72|\vec v| = \sqrt{7^2 + 7^2}∣v∣=72+72​

∣v⃗∣=49+49=98=72|\vec v| = \sqrt{49+49} = \sqrt{98} = 7\sqrt{2}∣v∣=49+49​=98​=72​

  1. Match with options

727\sqrt{2}72​ corresponds to Option A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So, the derived answer agrees with the stored answer.

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