JEE PYQ: Motion in a Plane - Question ID c0a4f21bcaa4 (JEE Main 2020)

When a particle moves in a plane with constant acceleration, its motion can be analyzed independently along the $x$ and $y$ axes using the kinematic equations for uniformly accelerated motion. The key formulas are:

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

Here, $\vec{r}_0$ is the initial position, $\vec{v}_0$ is the initial velocity, and $\vec{a}$ is the constant acceleration. Since the motion is in the $x$-$y$ plane, we can write the position vector as: $\vec{r}(t) = x(t) \widehat{i} + y(t) \widehat{j}$

Given that the particle starts from the origin at $t = 0$, we have $\vec{r}_0 = \vec{0}$. The initial velocity is $\vec{v}_0 = 3.0 \widehat{i}$ m/s, and the acceleration is $\vec{a} = 6 \widehat{i} + 4 \widehat{j}$ m/s².

Step 1: Write the position equations for $x(t)$ and $y(t)$.

Using the position formula:

Given:

• $x_0 = 0$, $y_0 = 0$ (starts from origin)
• $v_{0x} = 3$ m/s, $v_{0y} = 0$ m/s (initial velocity only in $x$-direction)
• $a_x = 6$ m/s², $a_y = 4$ m/s²

So,

Step 2: Find the time $t$ when $y(t) = 32$ m.

Set $y(t) = 32$:

Step 3: Compute $x(t)$ at $t = 4$ s.

Substitute $t = 4$ into $x(t)$:

Thus, the $x$-coordinate when $y = 32$ m is $D = 60$ m.

Trap 1: Ignoring the initial velocity in the $x$-direction. Some students mistakenly assume the initial velocity is zero in both directions, leading to incorrect $x(t)$ calculations. Always verify the initial velocity components.

Trap 2: Misapplying the kinematic equations. Students sometimes confuse the position and velocity equations or forget to include the $\frac{1}{2}$ factor in the acceleration term. Double-check the formula before substituting values.

Trap 3: Solving for time incorrectly. When solving $2t^2 = 32$, students might forget to take the square root or consider only the positive root. Since time cannot be negative, $t = 4$ s is the only valid solution.

Exam Tip: Always break 2D motion into $x$ and $y$ components and solve them independently. This simplifies the problem and reduces errors.