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

Correct answer: D. 5 17 ~m / s , making an angle of ^ -1 4 with - ve Y axis aligned & r =5 t^2 i -5 t_ j & v =10 i -5 j & v =20 i -5 j at t=2 sec aligned aligned & θ=(20)/(5)=4 & θ= ^ -1 4 & From -veY-axis aligned

JEE Main 202524 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Evening Shift
Subject: Physics
Chapter: Motion in Two Dimensions

JEE Main 2025 · 24 January · Evening Shift · Physics · Motion in Two Dimensions

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

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

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

. The magnitude and direction of velocity at

t=2 \mathrm{~s}

is,

Options

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
C $5 \sqrt{17} \mathrm{~m} / \mathrm{s}$ , 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 $Y$ axis

Correct answer

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

Step-by-step solution

\begin{aligned} & \overrightarrow{\mathrm{r}}=5 t^2 \hat{\mathrm{i}}-5 \mathrm{t}_{\mathrm{j}} \\ & \overrightarrow{\mathrm{v}}=10 \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{v}}=20 \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \quad \text{ at } \mathrm{t}=2 \mathrm{sec}\end{aligned}

\begin{aligned} & \tan \theta=\frac{20}{5}=4 \\ & \theta=\tan ^{-1} 4 \\ & \text{ From -veY-axis }\end{aligned}

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