JEE PYQ: Motion in a Plane - Question ID bcb046c45e97 (JEE Main 2024)

In two-dimensional motion under constant acceleration, the position and velocity of a particle at any time $t$ are governed by the kinematic equations:

• Position vector: $\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \tfrac{1}{2} \vec{a} t^2$ where $\vec{r}_0$ is the initial position, $\vec{v}_0$ the initial velocity, and $\vec{a}$ the constant acceleration.
• Velocity vector: $\vec{v}(t) = \vec{v}_0 + \vec{a} t$
• Speed: $v(t) = \bigl|\vec{v}(t)\bigr| = \sqrt{v_x^2 + v_y^2}$

Here the particle starts from the origin ($\vec{r}_0 = \vec{0}$) with initial velocity $\vec{v}_0 = 5\,\hat{i}\,\mathrm{m/s}$ and constant acceleration $\vec{a} = (3\,\hat{i} + 2\,\hat{j})\,\mathrm{m/s^2}$.

1. Write the $x$-coordinate as a function of time: $x(t) = v_{0x}\,t + \tfrac{1}{2}\,a_x\,t^2 = 5\,t + \tfrac{1}{2}\cdot3\,t^2 = 5\,t + 1.5\,t^2$


2. Set $x(t) = 84$ m and solve for $t$: $1.5\,t^2 + 5\,t - 84 = 0$ Multiply by 2 to clear decimals: $3\,t^2 + 10\,t - 168 = 0$ Use the quadratic formula $t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=3$, $b=10$, $c=-168$: $\Delta = 10^2 - 4\cdot3\cdot(-168) = 100 + 2016 = 2116$ $\sqrt{\Delta} = 46$ $t = \frac{-10 \pm 46}{6}$ The positive root is $t = \frac{36}{6} = 6\;\text{s}.$


3. Compute the velocity components at $t = 6$ s: $v_x(t) = v_{0x} + a_x\,t = 5 + 3\cdot6 = 5 + 18 = 23\;\mathrm{m/s}$ $v_y(t) = v_{0y} + a_y\,t = 0 + 2\cdot6 = 12\;\mathrm{m/s}$


4. Find the speed and $\alpha$: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{23^2 + 12^2} = \sqrt{529 + 144} = \sqrt{673}$ Hence $\alpha = 673$.

1. Sign errors in the quadratic equation: Students sometimes drop the negative sign on $c$, leading to wrong roots. 2. Forgetting the $y$-component of velocity: One must compute both $v_x$ and $v_y$ to find the speed. 3. Decimal arithmetic: Clearing fractions early (multiplying by 2) avoids messy decimals. 4. Units and vector notation: Always keep track of units and distinguish between scalar speed and vector velocity.