The position vector of a moving body at any instant of time is given as r→=(5t2i^−5t^)m. The magnitude and direction of velocity at t=2s,is,

5 17 m / s , making an angle of tan − 1 ⁡ 4 with –ve Y axis

The position vector of a moving body at any instant of time is given as r→=(5t2i^−5t^)m. The magnitude and direction of velocity at t=2s,is,

5 17 m / s , making an angle of tan − 1 ⁡ 4 with –ve Y axis

5 15 m / s , making an angle of tan − 1 ⁡ 4 +ve Y axis

5 17 m / s , making an angle of tan − 1 ⁡ 4 +ve Y axis

5 17 m / s , making an angle of tan − 1 ⁡ 4 with –ve Y axis

5 15 m / s , making an angle of tan − 1 ⁡ 4 with –ve Y axis

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The position vector of a moving body at any instant of time is given as $\vec{r} = (5 t^{2} \hat{i} - 5 \hat{t}) m .$ The magnitude and direction of velocity at $t = 2 s,$ is,

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$5 \sqrt{15} m / s,$ making an angle of $\tan^{- 1} 4$ +ve Y axis

$5 \sqrt{17} m / s,$ making an angle of $\tan^{- 1} 4$ +ve Y axis

$5 \sqrt{17} m / s,$ making an angle of $\tan^{- 1} 4$ with –ve Y axis

$5 \sqrt{15} m / s,$ making an angle of $\tan^{- 1} 4$ with –ve Y axis

answer is C.

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Detailed Solution

$\begin{aligned} \vec{r} = 5 t^{2} \hat{i} - 5 \hat{t} \\ \vec{v} = 10 t \hat{i} - 5 \hat{j} \\ \vec{v} = 20 \hat{i} - 5 \hat{j} at t = 2 \sec \end{aligned}$
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$\begin{aligned} \tan θ = \frac{20}{5} = 4 \\ θ = \tan^{- 1} 4 \\ From-veY-axis \end{aligned}$