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JEE Main 2019 (9 Apr, Shift 2) · Velocity, Acceleration & Variable Motion

Velocity, Acceleration & Variable Motion question from JEE Main 2019 (9 Apr, Shift 2)

Q30JEE Main 2019 · 9 APR · SHIFT 2Easy

Velocity, Acceleration & Variable Motion

The position vector of a particle changes with time according to the relation

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

What is the magnitude of the acceleration at t = 1?

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Answer: (a)

  1. v⃗=dr⃗dt=30t i^−40t j^\vec{v} = \dfrac{d\vec{r}}{dt} = 30t\,\hat{i} - 40t\,\hat{j}
  2. a⃗=dv⃗dt=30i^−40j^\vec{a} = \dfrac{d\vec{v}}{dt} = 30\hat{i} - 40\hat{j} (constant)
  3. ∣a⃗∣=302+402=50|\vec{a}| = \sqrt{30^2 + 40^2} = 50
  4. So the answer is (a) 50.
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