JEE PYQ: Motion in a Plane - Question ID 01b44822f4b3 (JEE Main 2019)

To determine the speed of a particle moving in three-dimensional space, we rely on the following fundamental concepts:

• Position Vector: The position of the particle at any time $t$ is given by the vector $\vec{r}(t) = x(t) \hat{i} + y(t) \hat{j} + z(t) \hat{k}$ where $x(t)$, $y(t)$, and $z(t)$ are the time-dependent coordinates.
• Velocity Vector: The velocity of the particle is the time derivative of the position vector: $\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{dx}{dt} \hat{i} + \frac{dy}{dt} \hat{j} + \frac{dz}{dt} \hat{k}$
• Speed: The speed of the particle is the magnitude of the velocity vector: $v = |\vec{v}(t)| = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}$

In this problem, the coordinates are given as: $x = a \cos(\omega t), \quad y = a \sin(\omega t), \quad z = a \omega t$ We will compute the velocity components by differentiating these expressions with respect to time and then find the magnitude of the velocity vector to obtain the speed.

Step 1: Compute the velocity components

Differentiate each coordinate with respect to time $t$:

• $x = a \cos(\omega t)$ $\frac{dx}{dt} = -a \omega \sin(\omega t)$
• $y = a \sin(\omega t)$ $\frac{dy}{dt} = a \omega \cos(\omega t)$
• $z = a \omega t$ $\frac{dz}{dt} = a \omega$

Step 2: Write the velocity vector

The velocity vector is: $\vec{v}(t) = \frac{dx}{dt} \hat{i} + \frac{dy}{dt} \hat{j} + \frac{dz}{dt} \hat{k}$ $\vec{v}(t) = -a \omega \sin(\omega t) \hat{i} + a \omega \cos(\omega t) \hat{j} + a \omega \hat{k}$

Step 3: Compute the magnitude of the velocity vector (speed)

The speed $v$ is: $v = |\vec{v}(t)| = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}$ Substitute the derivatives: $v = \sqrt{(-a \omega \sin(\omega t))^2 + (a \omega \cos(\omega t))^2 + (a \omega)^2}$ Simplify each term: $v = \sqrt{a^2 \omega^2 \sin^2(\omega t) + a^2 \omega^2 \cos^2(\omega t) + a^2 \omega^2}$ Factor out $a^2 \omega^2$: $v = \sqrt{a^2 \omega^2 (\sin^2(\omega t) + \cos^2(\omega t)) + a^2 \omega^2}$ Use the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$: $v = \sqrt{a^2 \omega^2 (1) + a^2 \omega^2} = \sqrt{a^2 \omega^2 + a^2 \omega^2}$ $v = \sqrt{2 a^2 \omega^2} = \sqrt{2} a \omega$

Step 4: Match with the given options

The speed of the particle is $\sqrt{2} a \omega$, which corresponds to option A.

Students often make the following mistakes in this type of problem:

• Forgetting to differentiate the z-coordinate: Some students only consider the $x$ and $y$ components, ignoring the $z$-component, leading to an incorrect speed of $a \omega$ (option B).
• Incorrectly squaring the derivatives: Misapplying the square operation on the derivatives can lead to wrong magnitudes. Always ensure that each component is squared individually before summing.
• Overlooking the trigonometric identity: Failing to use $\sin^2 \theta + \cos^2 \theta = 1$ can complicate the expression unnecessarily. This identity simplifies the calculation significantly.
• Confusing speed with velocity: Speed is a scalar quantity (magnitude of velocity), while velocity is a vector. Ensure you compute the magnitude, not just the components.

Exam Tip: Always double-check that you have included all components of motion (especially in 3D problems) and applied differentiation correctly. Simplify expressions using trigonometric identities to avoid errors.