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Motion in A Straight Line question

2019 · 9 Apr · Shift 2 · Q52
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  5. /2019 · 9 Apr · Shift 2 · Q52

Motion in A Straight Line question

2019 · 9 Apr · Shift 2 · Q52

JEE MainPhysicsMotion in A Straight LineMCQ+4 / −1
The position vector of a particle changes with time according to the relation r→(t)=15t2i^+(4−20t2)j^\overrightarrow r (t) = 15{t^2}\widehat i + (4 - 20{t^2})\widehat jr(t)=15t2i+(4−20t2)j​ What is the magnitude of the acceleration at t = 1 ?
  1. A
    50
  2. B
    25
  3. C
    40
  4. D
    100
View written solutionFree

Correct answer: A

  1. Given position vector

    r⃗(t)=15t2i^+(4−20t2)j^\vec r(t)=15t^2\hat i+(4-20t^2)\hat jr(t)=15t2i^+(4−20t2)j^​

  2. Find velocity by differentiating position w.r.t. time

    v⃗(t)=dr⃗dt=(30t)i^+(−40t)j^\vec v(t)=\frac{d\vec r}{dt}=(30t)\hat i+(-40t)\hat jv(t)=dtdr​=(30t)i^+(−40t)j^​

  3. Find acceleration by differentiating velocity w.r.t. time

    a⃗(t)=dv⃗dt=30i^−40j^\vec a(t)=\frac{d\vec v}{dt}=30\hat i-40\hat ja(t)=dtdv​=30i^−40j^​

    So acceleration is constant, hence at t=1t=1t=1 also:

    a⃗(1)=30i^−40j^\vec a(1)=30\hat i-40\hat ja(1)=30i^−40j^​

  4. Magnitude of acceleration

    ∣a⃗∣=302+(−40)2|\vec a|=\sqrt{30^2+(-40)^2}∣a∣=302+(−40)2​

    =900+1600=\sqrt{900+1600}=900+1600​

    =2500=50=\sqrt{2500}=50=2500​=50

  5. Option check

    • A: 505050 ✅
    • B: 252525 ❌
    • C: 404040 ❌
    • D: 100100100 ❌

Therefore, the correct answer is A.

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