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$$\text{The position vector of a moving body at any instant of time is given as }\vec r=(5t^2\hat i-5t\hat j)\,m.\text{The magnitude and direction of velocity at } t=2\,s \text{ is:}$$
r = 5t²i-5tj. v = 10ti-5j. At t=2: v = 20i-5j. |v| = √(400+25) = √425 = 5√17.
Direction: angle with -ve y-axis: tan θ = 20/5 = 4. So θ = tan⁻¹(4) with -ve y-axis.
The correct answer is Option 4: 5√17 m/s, making angle tan⁻¹4 with -ve Y-axis.
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