Q.A particle starts from origin at $t = 0$ with a velocity $5.0\,\hat{i}\ \text{m/s}$ and moves in $x$-$y$ plane under action of a force which produces a constant acceleration of $(3.0\,\hat{i} + 2.0\,\hat{j})\ \text{m/s}^{2}$.

Showing the 12 most recent of 17 on this concept.

1. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.The electric potentials at two points A and B are $+60~\mathrm{V}$ and $-30~\mathrm{V}$ respectively. If a particle of mass $20~\mu\mathrm{g}$ and charge $+10~\mu\mathrm{C}$ is released from rest at point A, then the velocity with which the particle reaches the point B is (A) $450~\mathrm{ms}^{-1}$ (B) $150~\mathrm{ms}^{-1}$ (C) $300~\mathrm{ms}^{-1}$ (D) $600~\mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   The particle moves from higher to lower electric potential, gaining kinetic energy equal to the loss in electric potential energy. Using conservation of energy, the velocity at B is $300~\mathrm{m/s}$.

   The key idea here is that a charged particle in an electric field experiences a force that accelerates it, and the work done by the field changes the particle’s kinetic energy. Since the field is conservative, the total mechanical energy (kinetic + electric potential energy) is conserved. The electric potential difference between A and B tells us exactly how much potential energy is converted into kinetic energy — no need to know the path or the field shape.

   Let’s work through it step by step.

   1. Identify the given data and convert to SI units.

      Mass: $m = 20~\mu\mathrm{g} = 20 \times 10^{-9}~\mathrm{kg} = 2 \times 10^{-8}~\mathrm{kg}$.

      Charge: $q = +10~\mu\mathrm{C} = 10 \times 10^{-6}~\mathrm{C} = 1 \times 10^{-5}~\mathrm{C}$.

      Potentials: $V_A = +60~\mathrm{V}$, $V_B = -30~\mathrm{V}$.

      Initial velocity: $u = 0$ (released from rest).

   2. Understand the energy conversion.

      The electric potential energy of a charge $q$ at a point with potential $V$ is $U = qV$.

      As the particle moves from A to B, the change in potential energy is

   $\Delta U = U_B - U_A = q(V_B - V_A).$

   Since $V_B < V_A$ and $q$ is positive, $\Delta U$ is negative — potential energy decreases.

   By conservation of energy, the loss in potential energy equals the gain in kinetic energy:

   $\Delta K = -\Delta U.$

   3. Calculate the potential difference and the energy change.

   $V_B - V_A = (-30) - (+60) = -90~\mathrm{V}.$

   So

   $\Delta U = q(V_B - V_A) = (1 \times 10^{-5})(-90) = -9 \times 10^{-4}~\mathrm{J}.$

   The kinetic energy gained is therefore

   $\Delta K = -\Delta U = 9 \times 10^{-4}~\mathrm{J}.$

   4. Relate kinetic energy to velocity. …

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2. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.A cannon on a cliff 55 m above the ground fires a shell with a velocity of $50\hat{i} + 50\hat{j} \ \mathrm{ms}^{-1}$. The displacement vector of the shell when it hits the ground is (Acceleration due to gravity $= 10 \ \mathrm{ms}^{-2}$) (A) $(550\hat{i} - 55\hat{j}) \ \mathrm{m}$ (B) $(550\hat{i} - 500\hat{j}) \ \mathrm{m}$ (C) $(500\hat{i} - 55\hat{j}) \ \mathrm{m}$ (D) $(500\hat{i} - 550\hat{j}) \ \mathrm{m}$

   ›Reveal solutionSolution

   Time of flight to the ground is $t = 11\ \text{s}$, giving horizontal range $550\ \text{m}$ and a net vertical drop of $55\ \text{m}$: displacement $= (550\hat{i} - 55\hat{j})\ \text{m}$ — option (A).

   The shell is launched from $55\ \text{m}$ above the ground with velocity $\vec{v} = 50\hat{i} + 50\hat{j}\ \text{ms}^{-1}$, so $u_x = 50\ \text{ms}^{-1}$ and $u_y = 50\ \text{ms}^{-1}$, with $g = 10\ \text{ms}^{-2}$ downward.

   Vertical motion (find the time of flight). Taking upward as positive and the launch point as the origin, the ground is at $y = -55\ \text{m}$:

   $y = u_y t - \tfrac{1}{2} g t^2 \implies -55 = 50t - 5t^2$

   $5t^2 - 50t - 55 = 0 \implies t^2 - 10t - 11 = 0 \implies (t-11)(t+1) = 0$

   Taking the physical root, $t = 11\ \text{s}$.

   Horizontal displacement. …

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3. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.A particle of charge 2 C is moving with a velocity of $\left( 3\hat{i} + 4\hat{j} \right)$ ms$^{-1}$ in the presence of magnetic and electric fields. If the magnetic field is $\left( \hat{i} + 2\hat{j} + 3\hat{k} \right)$ T and the electric field is $\left( -2\hat{k} \right)$ NC$^{-1}$, then the Lorentz force on the particle is (A) 50 N (B) 20 N (C) 30 N (D) 40 N

   ›Reveal solutionSolution

   The Lorentz force is the vector sum of electric and magnetic forces on a charged particle. Computing $\vec{F} = q\vec{E} + q(\vec{v} \times \vec{B})$ and finding its magnitude gives 20 N.

   When a charged particle moves through regions containing both electric and magnetic fields, it experiences two distinct forces. The electric field exerts a force along its direction, while the magnetic field exerts a force perpendicular to both the velocity and the field itself. The total electromagnetic force—the Lorentz force—is simply the vector sum of these two contributions.

   The Lorentz force is given by:

   $\vec{F} = q\vec{E} + q(\vec{v} \times \vec{B})$

   where $q$ is the charge, $\vec{E}$ is the electric field, $\vec{v}$ is the velocity, and $\vec{B}$ is the magnetic field.

   Let me work through this systematically.

   1. Identify the given quantities

   • Charge: $q = 2$ C
   • Velocity: $\vec{v} = (3\hat{i} + 4\hat{j})$ m/s
   • Magnetic field: $\vec{B} = (\hat{i} + 2\hat{j} + 3\hat{k})$ T
   • Electric field: $\vec{E} = (-2\hat{k})$ N/C

   2. Calculate the electric force

   The electric force is straightforward:

   $\vec{F}_E = q\vec{E} = 2 \times (-2\hat{k}) = -4\hat{k} \text{ N}$

   3. Calculate the magnetic force using the cross product

   The magnetic force requires $\vec{v} \times \vec{B}$. Using the determinant method:

   $\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 0 \\ 1 & 2 & 3 \end{vmatrix}$

   Expanding along the first row:

   $\vec{v} \times \vec{B} = \hat{i}(4 \cdot 3 - 0 \cdot 2) - \hat{j}(3 \cdot 3 - 0 \cdot 1) + \hat{k}(3 \cdot 2 - 4 \cdot 1)$ …

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4. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.The displacement of a particle in simple harmonic motion is given by $x = A_0 \cos \left( \frac{\pi t}{2} \right)$. The distance travelled by the particle in the interval between $t = 2$ to $t = 5$ seconds and its position at $5^{\text{th}}$ second are (A) $A_0$, and mean position (B) $A_0$, and extreme position (C) $3A_0$, and mean position (D) $3A_0$, and extreme position

   ›Reveal solutionSolution

   The key is to track the particle’s motion over the time interval by finding its positions at the start and end, then summing the absolute distances between turning points. The distance travelled is $3A_0$ and the position at $t=5$ is the mean position, so the correct option is (C).

   We are given $x = A_0 \cos\left(\frac{\pi t}{2}\right)$. This is simple harmonic motion with amplitude $A_0$ and angular frequency $\omega = \frac{\pi}{2}$ rad/s. The period is $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/2} = 4$ seconds. So the particle completes one full oscillation every 4 seconds.

   Why this approach works:

   In SHM, the distance travelled is not simply the difference in displacement — the particle may reverse direction. We must find the positions at key times and sum the absolute distances between consecutive turning points (where velocity = 0) within the interval.

   ---

   1. Find the position at $t = 2$ s

   $x(2) = A_0 \cos\left(\frac{\pi \cdot 2}{2}\right) = A_0 \cos(\pi) = -A_0$

   So at $t=2$, the particle is at the left extreme position ($-A_0$).

   2. Find the position at $t = 5$ s

   $x(5) = A_0 \cos\left(\frac{\pi \cdot 5}{2}\right) = A_0 \cos\left(\frac{5\pi}{2}\right)$

   Since $\frac{5\pi}{2} = 2\pi + \frac{\pi}{2}$, we have $\cos\left(\frac{5\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0$.

   So $x(5) = 0$, which is the mean position.

   3. Determine the motion between $t=2$ and $t=5$

      The period is 4 s, so from $t=2$ to $t=5$ is 3 seconds — three-quarters of a period.

      Let’s list the turning points (where velocity = 0, i.e., $\sin(\pi t/2)=0$):

      • At $t=2$: left extreme ($-A_0$)
      • Next turning point: $t=4$ (since $\cos(\pi \cdot 4/2) = \cos(2\pi) = A_0$), right extreme ($+A_0$)
      • Next turning point: $t=6$ (left extreme again), but our interval ends at $t=5$.

      So the path is:

      • From $t=2$ to $t=4$: moves from left extreme ($-A_0$) to right extreme ($+A_0$). Distance = $2A_0$. …

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5. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.The velocity of a particle having magnitude of $10~\mathrm{ms}^{-1}$ in the direction of $60^\circ$ with positive X-axis is (A) $5\hat{i} - 5\sqrt{3}\hat{j}$ (B) $5\sqrt{3}\hat{i} - 5\hat{j}$ (C) $5\sqrt{3}\hat{i} + 5\hat{j}$ (D) $5\hat{i} + 5\sqrt{3}\hat{j}$

   ›Reveal solutionSolution

   The velocity vector is found by resolving the given magnitude along the X and Y axes using cosine and sine of the angle. The result is $5\hat{i} + 5\sqrt{3}\hat{j}$, which corresponds to option (D).

   Concept & Intuition

   A velocity vector has both magnitude and direction. When we know the magnitude (speed) and the angle it makes with the positive X-axis, we can break it into horizontal (X) and vertical (Y) components using trigonometry. The X-component is magnitude × cos(angle), and the Y-component is magnitude × sin(angle). This works because the vector forms the hypotenuse of a right triangle, with the components as the legs.

   Step-by-step solution

   1. Identify given values

      Magnitude $v = 10\ \mathrm{m/s}$, angle $\theta = 60^\circ$ measured from the positive X-axis.

   2. Find the X-component

      $v_x = v \cos\theta = 10 \cos 60^\circ = 10 \times \frac{1}{2} = 5$

      So the component along $\hat{i}$ is $5$.

   3. Find the Y-component

      $v_y = v \sin\theta = 10 \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$

      So the component along $\hat{j}$ is $5\sqrt{3}$.

   4. Write the vector …

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6. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.An electromagnetic wave has its electric and magnetic fields given by $\vec{E}(t) = \vec{E}_m \sin (kx - \omega t)$ $\vec{B}(t) = \vec{B}_m \sin (kx - \omega t)$ If the direction of $\vec{E}_m$ & $\vec{B}_m$ are in the direction of $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$ respectively, the unit vector that gives the direction of propagation of the wave is (A) $-\hat{k}$ (B) $\hat{k}$ (C) $\hat{i}$ (D) $-\hat{i}$

   ›Reveal solutionSolution

   The Poynting vector $\vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B}$ gives the direction of energy flow. For the given fields, $\vec{E}_m \times \vec{B}_m$ points along $-\hat{k}$, so the wave propagates in the $-\hat{k}$ direction. The correct option is (A).

   The central idea here is that in an electromagnetic wave, the direction of propagation is given by the cross product $\vec{E} \times \vec{B}$. This comes from the Poynting vector, which describes the energy flux density of the wave. For a plane wave, the electric field, magnetic field, and propagation direction are always mutually perpendicular, forming a right-handed triad.

   The problem gives you the directions of the amplitudes $\vec{E}_m$ and $\vec{B}_m$ as vectors. Since the wave is sinusoidal and both fields oscillate in phase, the direction of propagation is simply the direction of $\vec{E}_m \times \vec{B}_m$ (or equivalently, the direction of $\vec{E} \times \vec{B}$ at any instant).

   Let’s work through it step by step.

   1. Write the given direction vectors clearly.

      $\vec{E}_m$ is along $\hat{i} + \hat{j}$.

      $\vec{B}_m$ is along $\hat{i} - \hat{j}$.

      These are not unit vectors yet — they just tell us the direction.

   2. Compute the cross product $\vec{E}_m \times \vec{B}_m$.

      Use the distributive property of the cross product:

   $(\hat{i} + \hat{j}) \times (\hat{i} - \hat{j}) = \hat{i} \times \hat{i} - \hat{i} \times \hat{j} + \hat{j} \times \hat{i} - \hat{j} \times \hat{j}$

   Recall that $\hat{i} \times \hat{i} = 0$, $\hat{j} \times \hat{j} = 0$, $\hat{i} \times \hat{j} = \hat{k}$, and $\hat{j} \times \hat{i} = -\hat{k}$.

   So:

   $= 0 - \hat{k} + (-\hat{k}) - 0 = -2\hat{k}$ …

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7. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.A particle initially at origin starts moving in xy-plane has velocity component $\vec{V} = (6+2t)\hat{i} + (4+2\sqrt{3}t)\hat{j}$ m/s. Acceleration of the particle in m/s$^2$ is [x, y are measured in meters, t in seconds, respectively] (A) $(6+2t)\hat{i} + (4+2\sqrt{3}t)\hat{j}$ (B) $(6+2t)\hat{i} + 2\sqrt{3}\hat{j}$ (C) $2\hat{i} + 2\sqrt{3}\hat{j}$ (D) $2\hat{i} + 2\sqrt{3}\hat{k}$

   ›Reveal solutionSolution

   Acceleration is the time derivative of velocity. Differentiating each component of $\vec{V}$ gives a constant vector, so the answer is $2\hat{i} + 2\sqrt{3}\hat{j}$.

   The key idea here is simple: acceleration is the rate of change of velocity with respect to time. When velocity is given as a function of time, you differentiate component-wise. No need to worry about position, initial conditions, or any other detail — the question only asks for acceleration.

   1. The velocity vector is

   $\vec{V}(t) = (6 + 2t)\,\hat{i} + (4 + 2\sqrt{3}\,t)\,\hat{j}.$

   2. Acceleration is defined as

   $\vec{a}(t) = \frac{d\vec{V}}{dt}.$

   3. Differentiate the $x$-component:

   $\frac{d}{dt}(6 + 2t) = 2.$

   4. Differentiate the $y$-component:

   $\frac{d}{dt}(4 + 2\sqrt{3}\,t) = 2\sqrt{3}.$

   5. The $z$-component is absent (the motion is in the $xy$-plane), so the acceleration vector is $\vec{a} = 2\,\hat{i} + 2\sqrt{3}\,\hat{j}.$ …

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8. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.A bullet is fired at time $t = 0$ with velocity $20~\mathrm{m/s}$ and at an initial angle of $30^\circ$ with the horizontal. The tan angle between the displacement vector and the horizontal after time $0.1~\mathrm{s}$ is (Assume $g = 10~\mathrm{m/s^2}$) (A) $\dfrac{38}{20\sqrt{3}}$ (B) $\dfrac{19}{20\sqrt{3}}$ (C) $\dfrac{19}{20}$ (D) $\dfrac{19\sqrt{3}}{20}$

   ›Reveal solutionSolution

   To find the angle of the displacement vector, we first calculate its horizontal and vertical components using kinematic equations. The tangent of the angle is then the ratio of the vertical displacement to the horizontal displacement. The tan angle is $\boxed{\dfrac{19}{20\sqrt{3}}}$.

   When an object is launched into the air and moves under the influence of gravity alone, we call this projectile motion. The key insight in solving projectile motion problems is to treat the horizontal and vertical components of motion independently. This is because gravity acts only vertically, causing acceleration in that direction, while there is no acceleration horizontally (assuming negligible air resistance).

   The displacement vector, $\vec{r}$, at any time $t$ has components $\Delta x$ (horizontal displacement) and $\Delta y$ (vertical displacement). If $\phi$ is the angle this displacement vector makes with the horizontal, then $\tan \phi = \frac{\Delta y}{\Delta x}$. Our goal is to find these components and then their ratio.

   Here's how we approach the problem:

   1. Decompose the initial velocity:

      The bullet is fired with an initial velocity $v_0 = 20~\mathrm{m/s}$ at an angle $\theta_0 = 30^\circ$ with the horizontal. We need to find its horizontal and vertical components.

      • Horizontal component of initial velocity: $v_{0x} = v_0 \cos \theta_0 = 20 \cos 30^\circ = 20 \left(\frac{\sqrt{3}}{2}\right) = 10\sqrt{3}~\mathrm{m/s}$
      • Vertical component of initial velocity: $v_{0y} = v_0 \sin \theta_0 = 20 \sin 30^\circ = 20 \left(\frac{1}{2}\right) = 10~\mathrm{m/s}$

   2. Calculate horizontal displacement ($\Delta x$):

      In the horizontal direction, there is no acceleration, so the velocity remains constant. The horizontal displacement is simply the product of the horizontal velocity and time.

      $\Delta x = v_{0x} t$

      Given $t = 0.1~\mathrm{s}$:

      $\Delta x = (10\sqrt{3}~\mathrm{m/s}) \times (0.1~\mathrm{s}) = \sqrt{3}~\mathrm{m}$

   3. Calculate vertical displacement ($\Delta y$): …

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9. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.Particle A (which was located at the origin at time t = 0) is moving along the x-axis with a constant speed of 1 m/s. Location of particle B which is moving along the y-axis is given by $y = ct^2$, where $c = 1$ m/s$^2$. Find the speed of particle A relative to particle B at t = 1 sec. (A) $\sqrt{5}$ m/s (B) 2 m/s (C) 1 m/s (D) 0 m/s

   ›Reveal solutionSolution

   The relative speed is found by vector subtraction of velocities, not positions. At $t=1$ s, the relative speed is $\sqrt{5}$ m/s.

   The key idea here is that relative speed is the magnitude of the relative velocity vector. Velocity is the rate of change of position, so we first need the velocity of each particle, then subtract them vectorially. A common mistake is to subtract positions and then differentiate — that works but is slower. The cleanest path is to get velocities directly.

   1. Particle A moves along the $x$-axis at constant speed $1$ m/s, starting from the origin. Its velocity is simply:

   $\vec{v}_A = 1\,\hat{i} \ \text{m/s}$

   (constant, no dependence on time).

   2. Particle B moves along the $y$-axis with position $y = ct^2$, where $c = 1$ m/s$^2$. So its position vector is:

   $\vec{r}_B = (ct^2)\,\hat{j} = t^2\,\hat{j} \ \text{m}$

   The velocity of B is the time derivative:

   $\vec{v}_B = \frac{d}{dt}(t^2)\,\hat{j} = 2t\,\hat{j} \ \text{m/s}$

   At $t = 1$ s, this becomes:

   $\vec{v}_B(1) = 2(1)\,\hat{j} = 2\,\hat{j} \ \text{m/s}$

   3. Relative velocity of A with respect to B is defined as:

   $\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$

   Substituting at $t = 1$ s:

   $\vec{v}_{A/B} = (1\,\hat{i}) - (2\,\hat{j}) = 1\,\hat{i} - 2\,\hat{j} \ \text{m/s}$

   4. Relative speed is the magnitude of this vector: $|\vec{v}_{A/B}| = \sqrt{(1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \ \text{m/s}$ …

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10. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
    Q.A particle is moving in xy - plane as $\vec{x} = (4 + t^2)\hat{i}, \ \vec{y} = \left(2t + \frac{t^2}{2}\right)\hat{j}$ where $\vec{x}$ & $\vec{y}$ are displacements measured along x and y axes respectively in meters and t in seconds. What is the velocity of the particle? (A) $\vec{v} = (4 + t)\hat{i} + (2 + t)\hat{j}$ (B) $\vec{v} = (4 + 2t)\hat{i} + (2 + t)\hat{j}$ (C) $\vec{v} = (4 + 2t)\hat{i} + \left(2 + \frac{t}{2}\right)\hat{j}$ (D) $\vec{v} = (4 + t)\hat{i} + \left(2 + \frac{t}{2}\right)\hat{j}$

    ›Reveal solutionSolution

    The velocity of a particle is found by taking the time derivative of its position vector. By differentiating the given x and y components of displacement, we find the velocity vector. The calculated velocity is $\vec{v} = 2t\hat{i} + (2 + t)\hat{j}$.

    The velocity of a particle describes how its position changes with respect to time. In two or three dimensions, the position of a particle is represented by a position vector, $\vec{r}(t)$, which has components along the x, y, and z axes. If the position vector is given as $\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k}$, then the instantaneous velocity vector, $\vec{v}(t)$, is found by taking the time derivative of each component of the position vector.

    This approach works because differentiation gives us the instantaneous rate of change. For motion along a straight line, velocity is simply $\frac{dx}{dt}$. When motion occurs in a plane (like the xy-plane here), we treat the x and y motions independently, and the overall velocity is the vector sum of these component velocities.

    The velocity vector $\vec{v}(t)$ is the time derivative of the position vector $\vec{r}(t)$:

    $\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} + \frac{dz}{dt}\hat{k}$

    Here's how to find the velocity of the particle:

    1. Identify the position components: The problem states that the displacement along the x-axis is $\vec{x} = (4 + t^2)\hat{i}$ and along the y-axis is $\vec{y} = \left(2t + \frac{t^2}{2}\right)\hat{j}$. This means the x-component of the particle's position is $x(t) = 4 + t^2$ and the y-component is $y(t) = 2t + \frac{t^2}{2}$. Therefore, the position vector of the particle in the xy-plane is:

    $\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} = (4 + t^2)\hat{i} + \left(2t + \frac{t^2}{2}\right)\hat{j}$

    2. Calculate the x-component of velocity: The x-component of velocity, $v_x(t)$, is the time derivative of the x-component of position:

    $v_x(t) = \frac{dx}{dt} = \frac{d}{dt}(4 + t^2)$

    Using the rules of differentiation ($\frac{d}{dt}(c) = 0$ for a constant $c$, and $\frac{d}{dt}(t^n) = nt^{n-1}$):
    $$ v_x(t) = \frac{d}{dt}(4) + \frac{d}{dt}(t^2) = 0 + 2t = 2t $$ …
    

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11. TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQ
    Q.The co-ordinates (x, y) of a moving particle at any time ‘t’ are given by $x = \alpha t^3$ and $y = \beta t^3$. The speed of the particle at time ‘t’ is given by (A) $3t\sqrt{\alpha^2 + \beta^2}$ (B) $3t^2\sqrt{\alpha^2 + \beta^2}$ (C) $t^2\sqrt{\alpha^2 + \beta^2}$ (D) $\sqrt{\alpha^2 + \beta^2}$

    ›Reveal solutionSolution

    Speed is the magnitude of the velocity vector. Differentiate the position components to find velocity, then compute $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$ to get $3t^2\sqrt{\alpha^2 + \beta^2}$.

    Why this approach works

    Speed measures how fast a particle moves along its path, regardless of direction. When position is given as functions of time, velocity is simply the rate of change of position—the derivative with respect to time. Since motion happens in two dimensions here, we have two velocity components, and the speed is the magnitude of that velocity vector.

    The key insight: differentiate each coordinate separately, then use the Pythagorean theorem to find the magnitude.

    Solution

    1. Find the velocity components

       The velocity in the $x$-direction is:

    $v_x = \frac{dx}{dt} = \frac{d}{dt}(\alpha t^3) = 3\alpha t^2$

    The velocity in the $y$-direction is:

    $v_y = \frac{dy}{dt} = \frac{d}{dt}(\beta t^3) = 3\beta t^2$

    2. Compute the speed

       Speed is the magnitude of the velocity vector:

    $v = \sqrt{v_x^2 + v_y^2}$

    Substituting our components:

    $v = \sqrt{(3\alpha t^2)^2 + (3\beta t^2)^2}$

    3. Simplify the expression …

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12. TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
    Q.At $t = 0$, a particle starts moving from origin with velocity $5.0\,\hat{i}$ m/s and it moves in $x-y$ plane due to a force having a constant acceleration of $(2.0\,\hat{i} + 3.0\,\hat{j})$ m/s$^2$. Find the coordinate of the particle at $t = 6$s. (A) $(x = 54\,\text{m},\, y = 66\,\text{m})$ (B) $(x = 66\,\text{m},\, y = 54\,\text{m})$ (C) $(x = 36\,\text{m},\, y = 48\,\text{m})$ (D) $(x = 48\,\text{m},\, y = 36\,\text{m})$

    ›Reveal solutionSolution

    The particle undergoes constant acceleration in 2D, so we apply the kinematic equation $\vec{r} = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2$ separately to each coordinate. At $t = 6$ s, the position is $(66\,\text{m},\, 54\,\text{m})$, which corresponds to option (B).

    The key idea is that motion in perpendicular directions ($x$ and $y$) is independent when acceleration is constant. We can treat each coordinate as a separate 1D problem under constant acceleration, then combine the results.

    ---

    1. Identify the given quantities

       • Initial position: $\vec{r}_0 = (0, 0)$
       • Initial velocity: $\vec{v}_0 = 5.0\,\hat{i}$ m/s, so $v_{0x} = 5.0$ m/s, $v_{0y} = 0$
       • Constant acceleration: $\vec{a} = 2.0\,\hat{i} + 3.0\,\hat{j}$ m/s², so $a_x = 2.0$ m/s², $a_y = 3.0$ m/s²
       • Time: $t = 6$ s

    2. Apply the 2D kinematic equation for position

       The vector form is:

    $\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2$

    Since $\vec{r}_0 = 0$, this simplifies to:

    $\vec{r}(t) = \vec{v}_0 t + \frac{1}{2} \vec{a} t^2$

    3. Find the $x$-coordinate Only the $x$-components matter:

    $x(t) = v_{0x} t + \frac{1}{2} a_x t^2$

    Substitute values:

    $x(6) = (5.0)(6) + \frac{1}{2}(2.0)(6^2) = 30 + \frac{1}{2}(2)(36) = 30 + 36 = 66\ \text{m}$

    4. Find the $y$-coordinate Only the $y$-components matter: $y(t) = v_{0y} t + \frac{1}{2} a_y t^2$ …

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