JEE PYQ: Motion in a Plane - Question ID b0f8ad374aab (JEE Main 2020)

When a particle moves in a plane with constant acceleration, its motion can be resolved into independent components along the $x$ and $y$ axes. The key kinematic equations for each axis are:

• Position as a function of time: $\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \tfrac{1}{2} \vec{a} t^2$
• Velocity as a function of time: $\vec{v}(t) = \vec{v}_0 + \vec{a} t$

Here $\vec{r}_0$ is the initial position, $\vec{v}_0$ the initial velocity, and $\vec{a}$ the constant acceleration vector. Since the motion is two-dimensional, we treat the $x$ and $y$ components separately.

Given data:

• Initial position: $\vec{r}_0 = \vec{0}$ (origin)
• Initial velocity: $\vec{v}_0 = 5\,\widehat j$ m/s
• Constant acceleration: $\vec{a} = 10\,\widehat i + 4\,\widehat j$ m/s²
• At time $t$, the $x$–coordinate is $20$ m, and the $y$–coordinate is $y_0$.

Step 1 – Write the position vector at time $t$:

Step 2 – Extract the $x$–component and set it equal to $20$ m:

Step 3 – Compute the $y$–coordinate at $t=2$ s:

Conclusion:

1. Mixing components: Students often confuse the $x$ and $y$ parts of the acceleration or initial velocity. Always label each component clearly. 2. Sign errors: Ensure the signs of acceleration and initial velocity match the chosen coordinate axes. 3. Quadratic roots: When solving $5t^2=20$, remember to take only the positive root for time. 4. Unit consistency: Double-check that all units are in meters and seconds to avoid numerical mismatches.