r=5t2i^−5tjv=10i^−5j^v=20i^−5j^ at t=2sec 
tanθ=520=4θ=tan−14 From -veY-axis
The position vector of a moving body at any instant of time is given as r=(5t2i^−5tj^)m. The magnitude and direction of velocity at t=2 s is,
Held on 24 Jan 2025 · Verified 6 Jul 2026.
515 m/s, making an angle of tan−14 with - ve Y axis
515 m/s, making an angle of tan−14 with + ve X axis
517 m/s, making an angle of tan−14 with + ve X axis
517 m/s, making an angle of tan−14 with - ve Y axis
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