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JEE PYQ: Motion in a Plane - Question ID 721f057aa15d (JEE Main 2022)

ID: 721f057aa15dJEE Main 2022Single Correct MCQ

At time $t=0$ a particle starts travelling from a height $7 \hat{z} \mathrm{~cm}$ in a plane keeping z coordinate constant. At any instant of time it's position along the $\hat{x}$ and $\hat{y}$ directions are defined as $3 \mathrm{t}$ and $5 \mathrm{t}^{3}$ respectively. At t = 1s acceleration of the particle will be

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Step-by-step Explanation

Core Formula & Concept:

In two-dimensional motion (here the $xy$ -plane, since $z$ is constant), the acceleration is the second time-derivative of the position vector. Let the position vector at time $t$ be $\vec{r}(t) = x(t)\,\hat{x} + y(t)\,\hat{y} + z(t)\,\hat{z}.$ Then

Velocity $\vec{v}(t) = \dfrac{d\vec{r}}{dt} = \dot{x}\,\hat{x} + \dot{y}\,\hat{y} + \dot{z}\,\hat{z}.$
Acceleration $\vec{a}(t) = \dfrac{d\vec{v}}{dt} = \ddot{x}\,\hat{x} + \ddot{y}\,\hat{y} + \ddot{z}\,\hat{z}.$

Since

z(t)

is constant,

\dot{z}=0

and

\ddot{z}=0

, so the acceleration has no

\hat{z}

component.

Step-by-Step Derivation: 1. Write the given position functions:

x(t) = 3\,t,\qquad y(t) = 5\,t^{3},\qquad z(t) = 7.

2. Compute the first derivatives (velocity components):

\dot{x} = \frac{d}{dt}(3t) = 3,\qquad \dot{y} = \frac{d}{dt}(5t^{3}) = 15\,t^{2},\qquad \dot{z} = 0.

3. Compute the second derivatives (acceleration components):

\ddot{x} = \frac{d}{dt}(3) = 0,\qquad \ddot{y} = \frac{d}{dt}(15\,t^{2}) = 30\,t,\qquad \ddot{z} = 0.

4. Evaluate at $t=1$ s:

\ddot{x}(1) = 0,\qquad \ddot{y}(1) = 30\cdot1 = 30,\qquad \ddot{z}(1) = 0.

5. Form the acceleration vector at $t=1$ s:

\vec{a}(1) = 0\,\hat{x} + 30\,\hat{y} + 0\,\hat{z} = 30\,\hat{y}.

6. Match with the given options: Option B is exactly

30\,\hat{y}

. Common Traps & Exam Tip:

Trap 1: Forgetting that $z$ is constant and including a $\hat{z}$ term in the acceleration. Trap 2: Stopping at the first derivative (velocity) instead of computing the second derivative. Trap 3: Misreading the question and evaluating at $t=0$ instead of $t=1$ s. Exam Tip: Always write down the position functions explicitly, then differentiate twice. Double-check the time at which you evaluate.

JEE PYQ: Motion in a Plane - Question ID 721f057aa15d (JEE Main 2022)

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Step-by-step Explanation

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