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

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

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

(B) $5 \sqrt{15} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with +ve X axis

(D) $5 \sqrt{17} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with +ve X axis

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Kinematics

The position vector of a moving body at any instant of time is given as

r = (5 t^{2} \hat{i} - 5 t \hat{j}) m

. The magnitude and direction of velocity at

t = 2 s

is,

Answer: (C)

Given

r = 5 t^{2} \hat{i} - 5 t \hat{j}

(in metres), velocity is obtained by differentiating with respect to time:

v = \frac{d r}{d t} = 10 t \hat{i} - 5 \hat{j} .

t = 2 s

v = 20 \hat{i} - 5 \hat{j} m/s .

Magnitude:

∣ v ∣ = 2 0^{2} + 5^{2} = 400 + 25 = 425 = 517 m/s .

Direction: the x-component is

+ 20

and the y-component is

- 5

, so the velocity vector lies below the x-axis, tilted toward the negative y-direction. Measuring the angle

θ

from the −y axis,

tan θ = \frac{v _{x}}{∣ v _{y} ∣} = \frac{20}{5} = 4 ⟹ θ = tan^{- 1} 4.

The velocity is

517 m/s

, making an angle

tan^{- 1} 4

with the −ve Y axis.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option