A particle starts from origin… - JEE Main 2024 Physics PYQ

JEE Main 2024 — Physics Mechanics

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A particle starts from origin at $t=0$ with a velocity $5\hat{i}m{s}^{-1}$ and moves in $x-y$ plane under action of a force which produces a constant acceleration of $(3\hat{i}+2\hat{j})m{s}^{-2}$ . If the $x$ -coordinate of the particle at that instant is $84m$ , then the speed of the particle at this time is $\sqrt{\alpha }m{s}^{-1}$ . The value of $\alpha$ is _______.

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Held on 27 Jan 2024 · Verified 6 Jul 2026.

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Solution

Given, ${u}_{x}=5m{s}^{-1},{a}_{x}=3m{s}^{-2},x=84m$

The formula to calculate the velocity of the particle along x-axis is given by

${v}_{x}^{2}={u}_{x}^{2}+2ax...(1)$

From equation (1), it follows that

${v}_{x}^{2}=25+2(3)(84) \Rightarrow {v}_{x}=23m{s}^{-1}$

Also, the velocity of the particle along x-axis can be written as

${v}_{x}={u}_{x}+{a}_{x}t...(2)$

From equation (2), it follows that

$\begin{matrix}t & = & \frac{23-5}{3} \\ & = & 6s\end{matrix}$

Similarly, for the $y-$ component of the velocity, it follows that

$\begin{matrix}{v}_{y} & = & 0+{a}_{y}t \\ & = & 0+2\times (6) \\ & = & 12m{s}^{-1}\end{matrix}$

Hence, the magnitude of the velocity is given by

${v}^{2}={v}_{x}^{2}+{v}_{y}^{2} ={23}^{2}+{12}^{2} =673 v=\sqrt{673}m{s}^{-1}$

Therefore, $\alpha =673$ .

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Solution

Given, ${u}_{x}=5m{s}^{-1},{a}_{x}=3m{s}^{-2},x=84m$

The formula to calculate the velocity of the particle along x-axis is given by

${v}_{x}^{2}={u}_{x}^{2}+2ax...(1)$

From equation (1), it follows that

${v}_{x}^{2}=25+2(3)(84) \Rightarrow {v}_{x}=23m{s}^{-1}$

Also, the velocity of the particle along x-axis can be written as

${v}_{x}={u}_{x}+{a}_{x}t...(2)$

From equation (2), it follows that

$\begin{matrix}t & = & \frac{23-5}{3} \\ & = & 6s\end{matrix}$

Similarly, for the $y-$ component of the velocity, it follows that

$\begin{matrix}{v}_{y} & = & 0+{a}_{y}t \\ & = & 0+2\times (6) \\ & = & 12m{s}^{-1}\end{matrix}$

Hence, the magnitude of the velocity is given by

${v}^{2}={v}_{x}^{2}+{v}_{y}^{2} ={23}^{2}+{12}^{2} =673 v=\sqrt{673}m{s}^{-1}$

Therefore, $\alpha =673$ .

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Solution

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Solution

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JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics