JEE PYQ: Motion in a Plane - Question ID 3b87b1d57878 (JEE Main 2020)

To find the speed of the second particle as measured in the frame of the first particle, we must compute the relative velocity of the second particle with respect to the first. The key formulas and concepts are:

• Position as a function of time: For any particle, its position vector is given by $\vec{r}(t) = x(t)\,\hat{i} + y(t)\,\hat{j}$.
• Velocity as the time derivative of position: $\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\,\hat{i} + \frac{dy}{dt}\,\hat{j}$.
• Relative velocity: If $\vec{v}_1$ is the velocity of the first particle and $\vec{v}_2$ is the velocity of the second particle, then the velocity of the second particle relative to the first is $\vec{v}_{2/1} = \vec{v}_2 - \vec{v}_1$.
• Speed in the relative frame: The speed is the magnitude of the relative velocity vector: $v_{2/1} = \bigl|\vec{v}_{2/1}\bigr| = \sqrt{(v_{2x} - v_{1x})^2 + (v_{2y} - v_{1y})^2}$.

1. Write down the given position functions:

2. Compute the velocity of the first particle (moving along the x-axis):

3. Compute the velocity of the second particle (moving along the y-axis):

4. Form the relative velocity vector $\vec{v}_{2/1}$:

5. Compute the magnitude of $\vec{v}_{2/1}$ (the relative speed):

1. Forgetting to subtract the first particle’s velocity: Many students compute only the second particle’s speed and stop there, missing the relative‐frame requirement. 2. Sign errors in derivatives: A common mistake is to mis‐differentiate $-3t^2$ or $-8t^3$, leading to wrong velocity components. 3. Mixing up axes: The first particle moves along the x-axis and the second along the y-axis; swapping their velocity components gives an incorrect magnitude.

Exam Tip: Always write down the relative‐velocity formula explicitly and double‐check each derivative before plugging in the time.