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Motion in A Straight Line question

2023 · Q134
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Motion in A Straight Line question

2023 · Q134

NEETPhysicsMotion in A Straight LineMCQ+4 / −1

The position of a particle is given by

r⃗(t)=4ti^+2t2j^+5k^\vec{r}(t)=4 t \hat{i}+2 t^2 \hat{j}+5 \hat{k}r(t)=4ti^+2t2j^​+5k^

where t\mathrm{t}t is in seconds and r\mathrm{r}r in metre. Find the magnitude and direction of velocity v(t)v(t)v(t), at t=1 st=1 \mathrm{~s}t=1 s, with respect to x\mathrm{x}x-axis

  1. A
    42 ms−1,45∘4 \sqrt{2} \mathrm{~ms}^{-1}, 45^{\circ}42​ ms−1,45∘
  2. B
    42 ms−1,60∘4 \sqrt{2} \mathrm{~ms}^{-1}, 60^{\circ}42​ ms−1,60∘
  3. C
    32 ms−1,30∘3 \sqrt{2} \mathrm{~ms}^{-1}, 30^{\circ}32​ ms−1,30∘
  4. D
    32 ms−1,45∘3 \sqrt{2} \mathrm{~ms}^{-1}, 45^{\circ}32​ ms−1,45∘
View written solutionFree

Correct answer: A

$$\begin{aligned} & \overrightarrow{\mathrm{V}}=\frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}=4 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+0 \hat{\mathrm{k}} \\ & \text { at } \mathrm{t}=1 \mathrm{sec} \\ & \overrightarrow{\mathrm{V}}=4 \hat{\mathrm{i}}+4(1) \hat{\mathrm{j}} \\ & |\overrightarrow{\mathrm{V}}|=\sqrt{4^2+4^2}=4 \sqrt{2} \\ & \tan \alpha=\frac{4}{4}=1 \\ & \alpha=45^{\circ}\end{aligned}$$

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