Verified Solution • 2025–26 Session moderate

QNEET-2023-Ph-320: NEET 2023 Physics - Class 11 NEET-2023-Ph-320 Solution: The position of a particle is given by \( {r}(t) = 4t {i} +…

The position of a particle is given by r⃗(t)=4ti^+2t2j^+5k^\vec{r}(t) = 4t\hat{i} + 2t^2\hat{j} + 5\hat{k} where t is in seconds and r in metre. Find the magnitude and direction of velocity v⃗(t)\vec{v}(t) at t=1 s with respect to x-axis 42 ms−145∘4\sqrt{2}\ \text{ms}^{-1} 45^\circ 42 ms−160∘4\sqrt{2}\ \text{ms}^{-1} 60^\circ 32 ms−130∘3\sqrt{2}\ \text{ms}^{-1} 30^\circ 32 ms−145∘3\sqrt{2}\ \text{ms}^{-1} 45^\circ

Multiple Choice Options

  • A 42 ms−1,45∘4 \sqrt{2} \mathrm{~ms}^{-1}, 45^{\circ} 4 2 _   ms _ 1 , 4 5 _
  • B 42 ms−1,60∘4 \sqrt{2} \mathrm{~ms}^{-1}, 60^{\circ} 4 2 _   ms _ 1 , 6 0 _
  • C 32 ms−1,30∘3 \sqrt{2} \mathrm{~ms}^{-1}, 30^{\circ} 3 2 _   ms _ 1 , 3 0 _
  • D 32 ms−1,45∘3 \sqrt{2} \mathrm{~ms}^{-1}, 45^{\circ} 3 2 _   ms _ 1 , 4 5 _

Step-by-Step Verified Solution

V _ = d r _ dt = 4 i ^ + 4 j ^ + 0 k ^  at  t = 1 sec V _ = 4 i ^ + 4 ( 1 ) j ^ | V _ | = 4 2 + 4 2 = 4 2 tan _ _ = 4 4 = 1 _ = 45 _

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