JEE Main 2025 Jan 24 Shift 2, Physics Q8: Motion In Two And Three Dimension

JEE Main2025Jan 24, Shift 2Physics

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,

A
$515 m/s$ , making an angle of $tan^{- 1} 4$ with $-$ ve Y axis
B
$515 m/s$ , making an angle of $tan^{- 1} 4$ with $+$ ve X axis
C
$517 m/s$ , making an angle of $tan^{- 1} 4$ with $-$ ve Y axis
D
$517 m/s$ , making an angle of $tan^{- 1} 4$ with $+$ ve X axis

Solution

Differentiate the position with respect to time:

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

At $t = 2 s$ : $v = 20 \hat{i} - 5 \hat{j} m/s$ .

Magnitude:

$∣ v ∣ = 2 0^{2} + (- 5)^{2} = 400 + 25 = 425 = 517 m/s$

Direction: the velocity has $v_{x} = + 20$ (positive x) and $v_{y} = - 5$ (negative y), so it lies in the fourth quadrant, close to the $- y$ axis. Measured from the $- y$ axis:

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

So the velocity makes an angle of $tan^{- 1} 4$ with the $-$ ve Y-axis.

Correct option: (3)

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JEE Main 2025 Jan 24 Shift 2, Physics Q8: Motion In Two And Three Dimension

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