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, (a) 5√(17) m/s, making an angle of tan ⁻¹ 4 with - ve Y axis (b) 5√(15) m/s, making an angle of tan ⁻¹ 4 with + ve X axis (c) 5√(17) m/s, making an angle of tan ⁻¹ 4 with + ve X axis (d) 5√(15) m/s, making an angle of tan ⁻¹ 4 with - ve Y axis

Answer: (a). v = (d r)/(dt) = 10t i - 5 j; at t = 2 s, v = 20 i - 5 j | v| = √(400 + 25) = √(425) = 5√(17) m/s Angle with -y axis: tan φ = 20/5 = 4 So the answer is (a) 5√(17) m/s at tan ⁻¹4 with - ve Y axis.

The position vector of a moving body at any instant of time is given… | JEE Main 2025 (24 Jan, Shift 2)

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Motion in a Plane (2-D Kinematics)

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

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Answer: (a)

$\vec v = \frac{d\vec r}{dt} = 10t\,\hat i - 5\,\hat j$ ; at $t = 2$ s, $\vec v = 20\hat i - 5\hat j$
$|\vec v| = \sqrt{400 + 25} = \sqrt{425} = 5\sqrt{17}$ m/s
Angle with $-y$ axis: $\tan\phi = \frac{20}{5} = 4$
So the answer is (a) $5\sqrt{17}$ m/s at $\tan^{-1}4$ with $-$ ve Y axis.

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