The position vector of a moving body at any instant of time is given as overrightarrow r = (5 t 2 hat i -5 t hat j ) m . The magnitude and direction of velocity at t =2 s is,

5 √(17) m / s , making an angle of tan -1 4 with -ve Y axis

5 √(15) m / s , making an angle of tan -1 4 with -ve Y axis

5 √(15) m / s , making an angle of tan -1 4 with +ve X axis

5 √(17) m / s , making an angle of tan -1 4 with +ve X axis

Physics · Motion in one Dimension

JEE Main 2025 — 24 January, Evening Shift — Question 54

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,

Option A:
$5 \sqrt{15} \mathrm{~m} / \mathrm{s}$ , making an angle of $\tan ^{-1} 4$ with -ve Y axis
Option B:
$5 \sqrt{15} \mathrm{~m} / \mathrm{s}$ , making an angle of $\tan ^{-1} 4$ with +ve X axis
Option C:
$5 \sqrt{17} \mathrm{~m} / \mathrm{s}$ , making an angle of $\tan ^{-1} 4$ with -ve Y axis
Correct
Option D:
$5 \sqrt{17} \mathrm{~m} / \mathrm{s}$ , making an angle of $\tan ^{-1} 4$ with +ve X axis

Answer: C

Step-by-step solution

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

$\vec{v}=10 t \hat{i}-5 \hat{j}$

$\vec{v}=20 \hat{i}-5 \hat{j} \quad \text { at } t=2 \mathrm{sec}$

$\tan \theta=\frac{20}{5}=4$

$\theta=\tan ^{-1} 4$

From-veY-axis

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Motion in one Dimension
Topic: Non-Uniformly Accelerated Motion