Concept Card: Motion in a Plane & 2D Kinematics
1. Position and Displacement Vectors in Two Dimensions (2D):
In a Cartesian coordinate plane ($x-y$), the position of a particle at time $t$ is represented by the position vector $\vec{r}(t)$ directed from origin $O$ to point $P(x, y)$:
- Position Vector: $\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}$
Magnitude: $r = |\vec{r}| = \sqrt{x^2 + y^2}$
Direction with $+x$ axis: $\theta = \arctan\left(\frac{y}{x}\right)$ - Displacement Vector: For motion from $P_1(x_1, y_1)$ at $t_1$ to $P_2(x_2, y_2)$ at $t_2$:
$\Delta\vec{r} = \vec{r}_2 – \vec{r}_1 = (x_2 – x_1)\hat{i} + (y_2 – y_1)\hat{j} = \Delta x\hat{i} + \Delta y\hat{j}$
Magnitude: $|\Delta\vec{r}| = \sqrt{(\Delta x)^2 + (\Delta y)^2}$
2. Velocity Vector in 2D:
- Average Velocity: $\vec{v}_{\text{avg}} = \frac{\Delta\vec{r}}{\Delta t} = \frac{\Delta x}{\Delta t}\hat{i} + \frac{\Delta y}{\Delta t}\hat{j} = v_{x,\text{avg}}\hat{i} + v_{y,\text{avg}}\hat{j}$
- Instantaneous Velocity: The time derivative of position:
$\vec{v} = \lim_{\Delta t \to 0} \frac{\Delta\vec{r}}{\Delta t} = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} = v_x\hat{i} + v_y\hat{j}$ - Speed (Magnitude): $v = |\vec{v}| = \sqrt{v_x^2 + v_y^2}$
- Direction: $\tan\theta = \frac{v_y}{v_x} \implies \theta = \arctan\left(\frac{v_y}{v_x}\right)$
Crucial Geometric Property: The instantaneous velocity vector $\vec{v}$ is always tangent to the particle’s trajectory at every point along its path.
3. Acceleration Vector in 2D:
- Instantaneous Acceleration: $\vec{a} = \frac{d\vec{v}}{dt} = \frac{dv_x}{dt}\hat{i} + \frac{dv_y}{dt}\hat{j} = a_x\hat{i} + a_y\hat{j} = \frac{d^2 x}{dt^2}\hat{i} + \frac{d^2 y}{dt^2}\hat{j}$
- Magnitude: $a = |\vec{a}| = \sqrt{a_x^2 + a_y^2}$
- Resolution into Tangential and Normal Components:
Any 2D acceleration can be resolved into two mutually perpendicular geometric components:- Tangential Acceleration ($a_t$): Component parallel to velocity; responsible for changing the magnitude of velocity (speed):
$a_t = \frac{dv}{dt} = \frac{\vec{a}\cdot\vec{v}}{v}$ - Normal / Centripetal Acceleration ($a_n$): Component perpendicular to velocity; responsible for changing the direction of velocity (curvature of path):
$a_n = \sqrt{a^2 – a_t^2} = \frac{v^2}{R_c}$
where $R_c = \frac{v^2}{a_n} = \frac{v^3}{|\vec{v}\times\vec{a}|}$ is the radius of curvature of the trajectory at that point.
- Tangential Acceleration ($a_t$): Component parallel to velocity; responsible for changing the magnitude of velocity (speed):
4. Independence of Perpendicular Coordinates (Motion Superposition):
A two-dimensional motion with constant acceleration $\vec{a} = a_x\hat{i} + a_y\hat{j}$ can be completely separated into two independent one-dimensional motions along the orthogonal $x$ and $y$ axes:
| Kinematic Equation | X-Component | Y-Component |
|---|---|---|
| Velocity-Time | $v_x = u_x + a_x t$ | $v_y = u_y + a_y t$ |
| Position-Time | $x = x_0 + u_x t + \frac{1}{2}a_x t^2$ | $y = y_0 + u_y t + \frac{1}{2}a_y t^2$ |
| Velocity-Position | $v_x^2 = u_x^2 + 2a_x \Delta x$ | $v_y^2 = u_y^2 + 2a_y \Delta y$ |
5. Equation of Trajectory (Path Equation):
The equation of the path $y = f(x)$ describes the geometric curve of motion in space. It is found by eliminating the time parameter $t$ between $x(t)$ and $y(t)$.
– If $\vec{a}$ is constant and parallel to $\vec{u}$, the trajectory is a straight line.
– If $\vec{a}$ is constant and non-parallel to $\vec{u}$, the trajectory is a parabola.
Solved Examples
Example 1 (2D Kinematics with Constant Acceleration):
A particle starts from the origin at $t = 0$ with initial velocity $\vec{u} = 5.0\hat{i}\text{ m/s}$ and accelerates in the $x-y$ plane with constant acceleration $\vec{a} = (3.0\hat{i} + 2.0\hat{j})\text{ m/s}^2$.
(a) At what instant is the $y$-coordinate of the particle equal to $16\text{ m}$?
(b) What is the $x$-coordinate of the particle at this time?
(c) What is the speed of the particle at this instant?
Solution:
Given: $x_0 = 0, y_0 = 0$; $u_x = 5.0\text{ m/s}, u_y = 0$; $a_x = 3.0\text{ m/s}^2, a_y = 2.0\text{ m/s}^2$.
(a) For motion along $y$-axis:
$y(t) = u_y t + \frac{1}{2}a_y t^2 \implies 16 = 0 + \frac{1}{2}(2.0)t^2 \implies t^2 = 16 \implies t = 4.0\text{ seconds}$.
(b) For motion along $x$-axis at $t = 4.0\text{ s}$:
$x(t) = u_x t + \frac{1}{2}a_x t^2 = 5.0(4.0) + \frac{1}{2}(3.0)(4.0^2) = 20.0 + 1.5(16) = 20.0 + 24.0 = 44.0\text{ meters}$.
(c) Velocity components at $t = 4.0\text{ s}$:
$v_x = u_x + a_x t = 5.0 + 3.0(4.0) = 17.0\text{ m/s}$
$v_y = u_y + a_y t = 0 + 2.0(4.0) = 8.0\text{ m/s}$
Speed $v = \sqrt{v_x^2 + v_y^2} = \sqrt{17.0^2 + 8.0^2} = \sqrt{289 + 64} = \sqrt{353} \approx 18.79\text{ m/s}$.
Example 2 (Parametric Trajectory & Central Acceleration):
The position vector of a particle moving in the $x-y$ plane is given by:
$\vec{r}(t) = a\cos(\omega t)\hat{i} + b\sin(\omega t)\hat{j}$
where $a, b$, and $\omega$ are positive constants ($a \neq b$).
(a) Determine the Cartesian equation of the trajectory $y = f(x)$.
(b) Prove that the acceleration $\vec{a}$ is always directed towards the origin and is proportional to the position vector.
(c) Find the radius of curvature of the path at the vertex $(a, 0)$.
Solution:
(a) Parametric equations: $x = a\cos(\omega t) \implies \cos(\omega t) = \frac{x}{a}$, and $y = b\sin(\omega t) \implies \sin(\omega t) = \frac{y}{b}$.
Using the trigonometric identity $\cos^2(\omega t) + \sin^2(\omega t) = 1$:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
The trajectory is an ellipse centered at the origin with semi-major axis $a$ and semi-minor axis $b$.
(b) Differentiating position to find velocity and acceleration:
$\vec{v}(t) = \frac{d\vec{r}}{dt} = -a\omega\sin(\omega t)\hat{i} + b\omega\cos(\omega t)\hat{j}$
$\vec{a}(t) = \frac{d\vec{v}}{dt} = -a\omega^2\cos(\omega t)\hat{i} – b\omega^2\sin(\omega t)\hat{j} = -\omega^2 [a\cos(\omega t)\hat{i} + b\sin(\omega t)\hat{j}] = -\omega^2\vec{r}(t)$.
Since $\vec{a} = -\omega^2\vec{r}$, the acceleration vector is always directed opposite to the position vector (i.e., towards the origin $O$) and its magnitude is directly proportional to the distance from the origin ($a = \omega^2 r$).
(c) At point $(a, 0)$, $\omega t = 0$:
Velocity: $\vec{v} = 0\hat{i} + b\omega\hat{j} \implies v = b\omega$ (directed along $+y$).
Acceleration: $\vec{a} = -a\omega^2\hat{i} + 0\hat{j} \implies a = a\omega^2$ (directed along $-x$).
Since $\vec{v} \perp \vec{a}$, the entire acceleration is normal/centripetal: $a_n = a\omega^2$.
Radius of curvature: $R_c = \frac{v^2}{a_n} = \frac{(b\omega)^2}{a\omega^2} = \frac{b^2}{a}$.
Example 3 (Tangential and Normal Acceleration):
A particle moves in the $x-y$ plane according to $x(t) = c t$ and $y(t) = c t(1 – \alpha t)$, where $c$ and $\alpha$ are positive constants.
(a) Find the trajectory equation $y(x)$.
(b) Find the velocity and acceleration vectors at $t = 0$.
(c) Determine the tangential acceleration $a_t$ and normal acceleration $a_n$ at $t = 0$.
Solution:
(a) From $x = c t \implies t = \frac{x}{c}$.
Substitute into $y$: $y = c\left(\frac{x}{c}\right)\left[1 – \alpha\left(\frac{x}{c}\right)\right] = x – \frac{\alpha}{c}x^2$.
The trajectory is an inverted parabola.
(b) Differentiating:
$v_x = \frac{dx}{dt} = c, \quad v_y = \frac{dy}{dt} = c – 2c\alpha t$
$\vec{v}(t) = c\hat{i} + c(1 – 2\alpha t)\hat{j} \implies \vec{v}(0) = c\hat{i} + c\hat{j}$
$a_x = \frac{dv_x}{dt} = 0, \quad a_y = \frac{dv_y}{dt} = -2c\alpha$
$\vec{a}(t) = -2c\alpha\hat{j} \implies \vec{a}(0) = -2c\alpha\hat{j}$.
(c) At $t = 0$:
Speed: $v = |\vec{v}(0)| = \sqrt{c^2 + c^2} = c\sqrt{2}$.
$\vec{a}\cdot\vec{v} = (0)(c) + (-2c\alpha)(c) = -2c^2\alpha$.
Tangential acceleration: $a_t = \frac{\vec{a}\cdot\vec{v}}{v} = \frac{-2c^2\alpha}{c\sqrt{2}} = -\sqrt{2}c\alpha$.
Total acceleration magnitude: $a = |-2c\alpha| = 2c\alpha$.
Normal acceleration: $a_n = \sqrt{a^2 – a_t^2} = \sqrt{(2c\alpha)^2 – (-\sqrt{2}c\alpha)^2} = \sqrt{4c^2\alpha^2 – 2c^2\alpha^2} = \sqrt{2c^2\alpha^2} = \sqrt{2}c\alpha$.
Example 4 (Calculus in 2D Kinematics):
The velocity of a particle moving in the $x-y$ plane is given by $\vec{v}(t) = (6t\hat{i} + 3t^2\hat{j})\text{ m/s}$. At $t = 0$, the particle is located at $\vec{r}(0) = (2\hat{i} – 5\hat{j})\text{ m}$.
(a) Find the position vector $\vec{r}(t)$ and acceleration vector $\vec{a}(t)$.
(b) Determine the Cartesian equation of the trajectory $y(x)$.
(c) Find the magnitude of acceleration when the $x$-coordinate is $14\text{ m}$.
Solution:
(a) Integrating velocity to obtain position:
$x(t) = x(0) + \int_{0}^{t} v_x\,dt = 2 + \int_{0}^{t} 6t\,dt = 2 + 3t^2$
$y(t) = y(0) + \int_{0}^{t} v_y\,dt = -5 + \int_{0}^{t} 3t^2\,dt = -5 + t^3$
$\vec{r}(t) = (2 + 3t^2)\hat{i} + (t^3 – 5)\hat{j}\text{ m}$.
Differentiating velocity to obtain acceleration:
$\vec{a}(t) = \frac{d\vec{v}}{dt} = 6\hat{i} + 6t\hat{j}\text{ m/s}^2$.
(b) From $x(t) = 2 + 3t^2 \implies 3t^2 = x – 2 \implies t^2 = \frac{x – 2}{3} \implies t = \left(\frac{x – 2}{3}\right)^{1/2}$.
Substitute into $y(t)$:
$y(x) = \left(\frac{x – 2}{3}\right)^{3/2} – 5$.
(c) When $x = 14\text{ m}$:
$14 = 2 + 3t^2 \implies 3t^2 = 12 \implies t^2 = 4 \implies t = 2.0\text{ s}$.
At $t = 2.0\text{ s}$:
$\vec{a}(2) = 6\hat{i} + 6(2)\hat{j} = 6\hat{i} + 12\hat{j}\text{ m/s}^2$.
Magnitude: $a = \sqrt{6^2 + 12^2} = \sqrt{36 + 144} = \sqrt{180} = 6\sqrt{5}\text{ m/s}^2 \approx 13.42\text{ m/s}^2$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A particle moves in the $x-y$ plane such that its coordinates are given by $x = 3t^2$ and $y = 4t^2$. The trajectory of the particle is:
(A) A circle
(B) An ellipse
(C) A straight line passing through the origin
(D) A parabola
Problem 2 (JEE Main – Single Correct):
A particle moves in the $x-y$ plane according to $\vec{r}(t) = A(\cos\omega t\hat{i} + \sin\omega t\hat{j})$. The angle between the velocity vector $\vec{v}$ and the acceleration vector $\vec{a}$ is:
(A) $0^\circ$
(B) $45^\circ$
(C) $90^\circ$
(D) $180^\circ$
Problem 3 (JEE Main – Single Correct):
A body starts from the origin at $t = 0$ with an initial velocity $\vec{u} = 3\hat{i}\text{ m/s}$ and experiences a constant acceleration $\vec{a} = 4\hat{j}\text{ m/s}^2$. The speed of the body at $t = 2\text{ s}$ is:
(A) $5\text{ m/s}$
(B) $\sqrt{73}\text{ m/s}$
(C) $8\text{ m/s}$
(D) $11\text{ m/s}$
Problem 4 (JEE Main – Single Correct):
The coordinates of a moving particle at any time $t$ are given by $x = \alpha t^3$ and $y = \beta t^3$. The instantaneous speed of the particle at time $t$ is:
(A) $3t^2 \sqrt{\alpha^2 + \beta^2}$
(B) $t^2 \sqrt{\alpha^2 + \beta^2}$
(C) $\sqrt{\alpha^2 + \beta^2}$
(D) $3t \sqrt{\alpha + \beta}$
Problem 5 (JEE Main – Single Correct):
A particle moves in a plane such that its velocity vector is $\vec{v} = 4\hat{i} + 3\hat{j}\text{ m/s}$ and its acceleration vector is $\vec{a} = 2\hat{i} + 6\hat{j}\text{ m/s}^2$. The tangential component of acceleration $a_t$ is:
(A) $2.6\text{ m/s}^2$
(B) $5.2\text{ m/s}^2$
(C) $6.0\text{ m/s}^2$
(D) $4.0\text{ m/s}^2$
Problem 6 (JEE Advanced – One or More Correct):
A particle moves in the $x-y$ plane according to the position vector $\vec{r}(t) = a\cos\omega t\hat{i} + a\sin\omega t\hat{j}$. Which of the following statements is/are correct?
(A) The velocity vector $\vec{v}$ is perpendicular to the position vector $\vec{r}$ at all times ($\vec{v}\cdot\vec{r} = 0$).
(B) The speed of the particle is constant and equal to $a\omega$.
(C) The acceleration vector is directed radially inward towards the origin.
(D) The magnitude of acceleration is $a\omega^2$.
Problem 7 (JEE Advanced – One or More Correct):
The trajectory of a particle in the $x-y$ plane is given by $y = \sqrt{3}x – \frac{g}{2}x^2$, and its horizontal velocity component is constant at $v_x = 10\text{ m/s}$. Which of the following statements is/are TRUE?
(A) The particle was launched at an angle of $60^\circ$ to the horizontal.
(B) The initial speed of the particle is $20\text{ m/s}$.
(C) The acceleration of the particle is directed purely along the $-y$ axis with magnitude $100g$.
(D) The vertical component of velocity at $x = 0$ is $10\sqrt{3}\text{ m/s}$.
Problem 8 (JEE Advanced – One or More Correct):
For a particle executing two-dimensional motion in a plane with acceleration $\vec{a}$ and velocity $\vec{v}$:
(A) If $\vec{a}\cdot\vec{v} > 0$, the speed of the particle increases.
(B) If $\vec{a}\cdot\vec{v} < 0$, the speed of the particle decreases.
(C) If $\vec{a}\cdot\vec{v} = 0$, the speed remains constant.
(D) If $\vec{a} \times \vec{v} = \vec{0}$, the path must be a straight line.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
The position vector of a particle moving in the $x-y$ plane is given by $\vec{r}(t) = 3t\hat{i} + 2t^2\hat{j}\text{ m}$. The speed of the particle in $\text{m/s}$ at $t = 1.0\text{ s}$ is $v$. Find $v$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A particle starts from rest at the origin at $t = 0$ and moves in the $x-y$ plane with constant acceleration $\vec{a} = (3\hat{i} + 4\hat{j})\text{ m/s}^2$. The distance of the particle from the origin at $t = 2\text{ s}$ in meters is $D$. Find $D$.
Solutions & Explanations
Answer Key Summary:
1. (C) | 2. (C) | 3. (B) | 4. (A) | 5. (B) | 6. (A, B, C, D) | 7. (A, B, D) | 8. (A, B, C, D) | 9. 5 | 10. 10
Solution 1:
$t^2 = \frac{x}{3} \implies y = 4t^2 = 4\left(\frac{x}{3}\right) = \frac{4}{3}x$.
This is the equation of a straight line $y = mx$ passing through the origin with slope $m = 4/3$.
Correct Answer: (C)
Solution 2:
$\vec{r} = A\cos\omega t\hat{i} + A\sin\omega t\hat{j}$ represents uniform circular motion.
$\vec{v} = -A\omega\sin\omega t\hat{i} + A\omega\cos\omega t\hat{j}$ (tangential).
$\vec{a} = -\omega^2\vec{r}$ (radial, directed towards center).
In uniform circular motion, the tangential velocity is perpendicular to centripetal acceleration ($\vec{v}\cdot\vec{a} = 0$). Hence angle is $90^\circ$.
Correct Answer: (C)
Solution 3:
$v_x = u_x + a_x t = 3 + 0 = 3\text{ m/s}$.
$v_y = u_y + a_y t = 0 + 4(2) = 8\text{ m/s}$.
Speed $v = \sqrt{v_x^2 + v_y^2} = \sqrt{3^2 + 8^2} = \sqrt{9 + 64} = \sqrt{73}\text{ m/s}$.
Correct Answer: (B)
Solution 4:
$v_x = \frac{dx}{dt} = 3\alpha t^2, \quad v_y = \frac{dy}{dt} = 3\beta t^2$.
$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(3\alpha t^2)^2 + (3\beta t^2)^2} = \sqrt{9t^4(\alpha^2 + \beta^2)} = 3t^2 \sqrt{\alpha^2 + \beta^2}$.
Correct Answer: (A)
Solution 5:
Speed $v = \sqrt{4^2 + 3^2} = 5\text{ m/s}$.
Dot product: $\vec{a}\cdot\vec{v} = (2)(4) + (6)(3) = 8 + 18 = 26\text{ m}^2/\text{s}^3$.
Tangential acceleration: $a_t = \frac{\vec{a}\cdot\vec{v}}{v} = \frac{26}{5} = 5.2\text{ m/s}^2$.
Correct Answer: (B)
Solution 6:
– (A) $\vec{v}\cdot\vec{r} = -a^2\omega\sin\omega t\cos\omega t + a^2\omega\sin\omega t\cos\omega t = 0$ (True).
– (B) $v = \sqrt{(-a\omega\sin\omega t)^2 + (a\omega\cos\omega t)^2} = a\omega$ (True).
– (C) $\vec{a} = -\omega^2\vec{r}$ points inward to origin (True).
– (D) $|\vec{a}| = \omega^2|\vec{r}| = a\omega^2$ (True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
Standard trajectory equation: $y = x\tan\theta – \frac{g x^2}{2u^2\cos^2\theta}$.
Comparing: $\tan\theta = \sqrt{3} \implies \theta = 60^\circ$ (A is True).
$u_x = u\cos(60^\circ) = 10 \implies u(1/2) = 10 \implies u = 20\text{ m/s}$ (B is True).
$u_y = u\sin(60^\circ) = 20(\sqrt{3}/2) = 10\sqrt{3}\text{ m/s}$ (D is True).
In standard projectile motion, $a_y = -g$ (not $100g$, so C is False).
Correct Answer: (A, B, D)
Solution 8:
– $\frac{d(v^2)}{dt} = 2v\frac{dv}{dt} = 2(\vec{a}\cdot\vec{v})$. Hence $\vec{a}\cdot\vec{v} > 0 \implies \text{speed increases}$ (A is True).
– $\vec{a}\cdot\vec{v} < 0 \implies \text{speed decreases}$ (B is True).
– $\vec{a}\cdot\vec{v} = 0 \implies \text{speed is constant}$ (C is True).
– $\vec{a} \times \vec{v} = 0 \implies \vec{a} \parallel \vec{v}$, so acceleration is strictly along the line of motion, tracing a straight line (D is True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 9:
$\vec{v}(t) = \frac{d\vec{r}}{dt} = 3\hat{i} + 4t\hat{j}\text{ m/s}$.
At $t = 1.0\text{ s}$: $\vec{v}(1) = 3\hat{i} + 4\hat{j}\text{ m/s}$.
Speed $v = \sqrt{3^2 + 4^2} = 5\text{ m/s}$.
Correct Answer: 5
Solution 10:
$\vec{s} = \vec{u}t + \frac{1}{2}\vec{a}t^2 = 0 + \frac{1}{2}(3\hat{i} + 4\hat{j})(2^2) = 2(3\hat{i} + 4\hat{j}) = 6\hat{i} + 8\hat{j}\text{ m}$.
Distance from origin: $D = |\vec{s}| = \sqrt{6^2 + 8^2} = 10\text{ meters}$.
Correct Answer: 10