Home › Physics › Mechanics › Kinematics in One, Two, and Three Dimensions
Mechanics
Kinematics in One, Two, and Three Dimensions
Kinematics is the branch of classical mechanics that describes the motion of points, bodies, and systems without consideration of the masses or the forces acting on them. It relies on the mathematical description of position, displacement, velocity, and acceleration as functions of time.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Uniform and uniformly accelerated motion in one and two dimensions, equations of motion, relative velocity, and projectile motion under constant gravitational acceleration.
PGT extension
General curvilinear motion in polar and cylindrical coordinate systems, velocity and acceleration in rotating frames, Coriolis and centrifugal accelerations, and motion under central forces.
Prerequisites
Vector algebra (addition, dot product, cross product, and derivatives of vectors)
Calculus (differentiation and integration of single and multivariable functions)
Fundamental concepts of coordinate systems (Cartesian, polar, and cylindrical)
Displacement, Velocity, and Acceleration
Displacement vector r(t) defines the position of a particle relative to an origin O.
Instantaneous velocity is the time derivative of the position vector: v = dr/dt.
Instantaneous acceleration is the time derivative of the velocity vector: a = dv/d$t = d^{2}r/dt^{2}$.
Motion with Constant Acceleration
When acceleration a is constant, vector equations decouple into orthogonal Cartesian components.
The velocity increases linearly with time: v(t) = v0 + at.
Position varies quadratically with time: r(t) = r0 + v0t + (1/2)at2.
Projectile Motion
Motion in a uniform gravitational field is analyzed by separating horizontal and vertical components.
Horizontal acceleration is zero, maintaining a constant horizontal velocity component (if air resistance is neglected).
Vertical motion is governed by constant downward gravitational acceleration g.
Kinematics in Polar Coordinates
For planar motion, position is given by r = r r̂.
Velocity vector has radial and transverse components: v = ṙ r̂ + r θ̇ θ̂.
Acceleration vector includes centripetal and Coriolis-type inertial terms: a = (r̈ - rθ̇2) r̂ + (rθ̈ + 2ṙθ̇) θ̂.
Derivations
Equation of Path for a Projectile
Conditions: Uniform gravitational field, launch angle θ0 from origin (0,0), neglect air resistance
Write parametric equations for horizontal and vertical positions as functions of time: $x = v_{0}\cos(θ_{0})t$ and $y = v_{0}\sin(θ_{0})t - (1/2)gt^{2}$.
Express time t from the horizontal motion equation: $t = x / (v_{0}\cos(θ_{0}))$.
Substitute the expression for t into the vertical position equation: $y = v_{0}\sin(θ_{0})[x / (v_{0}\cos(θ_{0}))] - (1/2)g[x / (v_{0}\cos(θ_{0}))]^{2}$.
Simplify the trigonometric terms to obtain the parabolic trajectory equation.
v = final velocity, v0 = initial velocity, a = acceleration, t = time m/s
Constant acceleration in 1D
Kinematic equation (Displacement-time)
$$s = v_0 t + \frac{1}{2}at^2$$
s = displacement, v0 = initial velocity, a = acceleration, t = time m
Constant acceleration in 1D
Kinematic equation (Velocity-displacement)
$$v^2 = v_0^2 + 2as$$
v = final velocity, v0 = initial velocity, a = acceleration, s = displacement m2/s2
Constant acceleration in 1D
Maximum height of projectile
$$H = \frac{v_0^2 \sin^2(\theta_0)}{2g}$$
v0 = initial speed, θ0 = projection angle, g = acceleration due to gravity m
Uniform gravitational field, flat earth
Horizontal range of projectile
$$R = \frac{v_0^2 \sin(2\theta_0)}{g}$$
v0 = initial speed, θ0 = projection angle, g = acceleration due to gravity m
Launched and landing at the same vertical level
Centripetal acceleration
$$a_c = \frac{v^2}{r} = \omega^2 r$$
v = tangential speed, r = radius of curvature, ω = angular velocity m/s2
Uniform circular motion
Graphs and experiments
Displacement-Time (s-t) Graph
Draw: A curved line sloping upwards with increasing slope on a Cartesian plane with t on x-axis and s on y-axis.
Interpret: The slope of the tangent at any point on the s-t graph represents the instantaneous velocity of the particle.
Exam use: Direct questions calculating velocity from curve slopes or identifying stationary vs moving states.
Velocity-Time (v-t) Graph
Draw: A straight line with a positive constant slope starting from a non-zero intercept on the v-axis.
Interpret: The slope of the v-t graph gives acceleration, and the area under the curve represents total displacement.
Exam use: Extensively used to find displacement under varying acceleration and to derive kinematic equations graphically.
Acceleration-Time (a-t) Graph
Draw: A horizontal straight line parallel to the time axis for uniform acceleration.
Interpret: The area under the a-t graph over a given time interval represents the change in velocity (Δv).
Exam use: Testing cumulative velocity changes when acceleration is a function of time.
Solved numericals
TGT core
A particle moves along a straight line with an initial velocity of 5 m/s and a constant acceleration of 2 m/s2. Find its displacement and final velocity after 4 seconds.
Identify given parameters: $u = 5 m/s$, $a = 2 m/s^{2}$, t = 4 s.
Use displacement formula: $s = ut + (1/2)at^{2}$. Substitute values: $s = (5)(4) + (1/2)(2)(4)^{2}$.
Calculate displacement: $s = 20 + 16 = 36 m$.
Use velocity formula: $v = u + at$. Substitute values: $v = 5 + (2)(4) = 13 m/s$.
Answer: Displacement = 36 m, Final Velocit$y = 13 m/s$
PGT extension
The position of a particle moving in a plane is given by r(t) = 3t2î + 4t2ĵ. Find the magnitude of its total acceleration at t = 2 s.
Differentiate position vector to find velocity: v(t) = dr/d$t = (6t)î + (8t)ĵ$.
Observe that acceleration is independent of time and constant: a = 6î + 8ĵ m/s2.
Calculate the magnitude of acceleration: |a| = √(62 + 82) = √(36 + 64) = √$100 = 10 m/s^{2}$.
Answer: 10 m/s2
TGT core
A projectile is fired with an initial velocity of 20 m/s at an angle of 30° to the horizontal. Calculate the maximum height reached. (Take $g = 10 m/s^{2})$
Macroscopic point-mass approximation: Objects are treated as point particles, ignoring rotational kinematics about their internal center of mass unless explicitly specified.
Galilean relativity: Kinematic equations assume inertial reference frames unless fictitious forces (e.g., centrifugal, Coriolis) are explicitly incorporated for non-inertial frames.
Common misconceptions
Common error: At the maximum height of a projectile, both velocity and acceleration are zero.
Correct: At maximum height, only the vertical component of velocity is zero; the horizontal velocity remains non-zero, and acceleration is still constant downward equal to g.
Gravity acts continuously throughout the flight, and horizontal motion has no opposing force in the absence of air resistance.
Common error: If the speed of a particle is constant, its acceleration must be zero.
Correct: A particle moving at constant speed in a curved path experiences non-zero centripetal acceleration.
Acceleration is the rate of change of the velocity vector, which includes changes in direction as well as magnitude.
Quick revision
Velocity is the time derivative of position; acceleration is the time derivative of velocity.
In projectile motion, horizontal velocity remains constant while vertical motion is under uniform acceleration g.
Polar acceleration includes both centripetal (rθ̇2) and Coriolis (2ṙθ̇) components.
Must remember
Instantaneous velocity is the time derivative of the position vector: $\mathbf{v} = d\mathbf{r}/dt$.
For constant acceleration, position varies quadratically with time: $\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2}\mathbf{a}t^2$.
At the maximum height of a projectile, the vertical component of velocity is zero, but the acceleration is non-zero and equal to $g$ downward.
In polar coordinates, the acceleration vector includes centripetal and Coriolis-type inertial terms: $\mathbf{a} = (\ddot{r} - r\dot{\theta}^2)\hat{r} + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\theta}$.
Maximum range in projectile motion on a flat surface is achieved at a projection angle of $45^\circ$.
Glossary
Position VectorA vector $\mathbf{r}(t)$ that defines the position of a particle relative to an origin $O$.
Instantaneous AccelerationThe time derivative of the velocity vector: $\mathbf{a} = d\mathbf{v}/dt = d^2\mathbf{r}/dt^2$.
Horizontal RangeThe total horizontal distance traveled by a projectile, given by $R = \frac{v_0^2 \sin(2\theta_0)}{g}$ for level ground.
Centripetal AccelerationThe component of acceleration directed toward the center of curvature in circular motion, given by $a_c = v^2/r = \omega^2 r$.
Cartesian vs. Polar Kinematics
Feature
Cartesian Coordinates
Polar Coordinates
Position Representation
$\mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k}$
$\mathbf{r} = r\hat{r}$
Velocity Components
$v_x = dx/dt$, $v_y = dy/dt$
$v_r = \dot{r}$, $v_\theta = r\dot{\theta}$
Acceleration Terms
Direct time derivatives of components
Includes centripetal ($-r\dot{\theta}^2$) and Coriolis ($2\dot{r}\dot{\theta}$) terms
Practice Quiz
Choose an option to reveal the explanation.
1. What is the time derivative of the position vector $\mathbf{r}(t)$?
Explanation: Instantaneous velocity is defined as the first time derivative of the position vector: $\mathbf{v} = d\mathbf{r}/dt$.
2. Which of the following kinematic equations represents displacement-time relation under constant acceleration in 1D?
Explanation: The displacement-time relation for constant acceleration is given by $s = v_0 t + \frac{1}{2}at^2$.
3. At the maximum height of a projectile launched from level ground, what is true about its velocity and acceleration?
Explanation: At maximum height, the vertical component of velocity becomes zero while the horizontal velocity remains constant. Acceleration remains constant at $g$ downwards.
4. What is the formula for the maximum height $H$ of a projectile with initial speed $v_0$ and angle $\theta_0$?
Explanation: The maximum height attained by a projectile is given by $H = \frac{v_0^2 \sin^2(\theta_0)}{2g}$.
5. What is the centripetal acceleration of a particle moving with tangential speed $v$ in a circle of radius $r$?
Explanation: Centripetal acceleration is expressed as $a_c = \frac{v^2}{r} = \omega^2 r$.
6. In planar polar coordinates, what is the radial component of the velocity vector $\mathbf{v}$?
Explanation: The velocity vector in polar coordinates is $\mathbf{v} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}$, making the radial component $\dot{r}$.
7. Which term in the polar acceleration vector corresponds to the centripetal acceleration component?
Explanation: The term $-r\dot{\theta}^2 \hat{r}$ in the polar acceleration formula represents the centripetal (or radial inward) acceleration.
8. If the speed of a particle is constant while moving along a curved path, what can be said about its acceleration?
Explanation: Even if speed is constant, curving motion involves a continuous change in the direction of the velocity vector, yielding a non-zero centripetal acceleration.
9. What is the horizontal range of a projectile launched with speed $v_0$ at an angle of $45^\circ$ on flat ground?
Explanation: The range formula is $R = \frac{v_0^2 \sin(2\theta_0)}{g}$. For $\theta_0 = 45^\circ$, $\sin(90^\circ) = 1$, giving $R = \frac{v_0^2}{g}$.
10. What represents the time derivative of the velocity vector?
Explanation: Instantaneous acceleration is defined as $\mathbf{a} = d\mathbf{v}/dt$.
11. In projectile motion (neglecting air resistance), what is the nature of the horizontal component of acceleration?
Explanation: In the absence of air resistance, there are no horizontal forces acting on the projectile, making the horizontal acceleration zero.
12. What is the unit of the velocity-displacement relation $v^2 = v_0^2 + 2as$?
Explanation: Velocity squared has units of $(\text{m/s})^2 = \text{m}^2/\text{s}^2$, which matches the term $2as$ ($(\text{m/s}^2)\times\text{m} = \text{m}^2/\text{s}^2$).
13. Which term in polar coordinates represents the Coriolis-type acceleration component?
Explanation: The transverse acceleration term contains $2\dot{r}\dot{\theta}$, which is the Coriolis-type inertial acceleration component.
14. If a particle's position is given by $\mathbf{r}(t) = 3t \hat{i} + 4t^2 \hat{j}$, what is its acceleration vector?
15. Under what condition is the equation $v = v_0 + at$ valid?
Explanation: The kinematic equations like $v = v_0 + at$ are derived specifically assuming constant acceleration in one dimension.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A particle moves in the $xy$-plane with position vector $\mathbf{r}(t) = (A \cos \omega t)\hat{i} + (A \sin \omega t)\hat{j}$. The trajectory of the particle is:
Explanation: Given $x = A \cos \omega t$ and $y = A \sin \omega t$, squaring and adding yields $x^2 + y^2 = A^2(\cos^2 \omega t + \sin^2 \omega t) = A^2$, which represents a circle of radius $A$.
2. The position of a particle is given by $\mathbf{r} = t^3 \hat{i} - 2t^2 \hat{j} + 5\hat{k}$. What is the magnitude of its acceleration at $t = 1\text{ s}$?
3. A projectile is thrown with an initial speed $v_0$ at an angle $\theta_0$ to the horizontal. At what angle is the horizontal range equal to the maximum height?
Explanation: Equating range $R = \frac{v_0^2 \sin(2\theta_0)}{g}$ and height $H = \frac{v_0^2 \sin^2(\theta_0)}{2g}$ gives $\frac{2v_0^2 \sin\theta_0 \cos\theta_0}{g} = \frac{v_0^2 \sin^2\theta_0}{2g}$, leading to $4\cos\theta_0 = \sin\theta_0$, so $\tan\theta_0 = 4$, or $\theta_0 = \tan^{-1}(4)$.
4. In polar coordinates, if a particle moves along a circular path of constant radius $R$ with constant angular velocity $\omega$, its acceleration vector is:
Explanation: For constant radius ($r=R$, $\dot{r}=\ddot{r}=0$) and constant angular velocity ($\dot{\theta}=\omega$, $\ddot{\theta}=0$), the polar acceleration formula $\mathbf{a} = (\ddot{r} - r\dot{\theta}^2)\hat{r} + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\theta}$ reduces to $-R\omega^2 \hat{r}$.
5. A body is projected such that its kinetic energy at maximum height is 3/4th of its initial kinetic energy. The angle of projection is:
Explanation: At maximum height, vertical velocity is zero, so $K_{\max\,h} = \frac{1}{2}m(v_0 \cos\theta_0)^2 = K_0 \cos^2\theta_0$. Given $K_{\max\,h} = \frac{3}{4} K_0$, we have $\cos^2\theta_0 = \frac{3}{4}$, so $\cos\theta_0 = \frac{\sqrt{3}}{2}$, which means $\theta_0 = 30^\circ$.
6. The velocity of a particle moving along a straight line is given by $v = 3t^2 - 2t$. Find the displacement of the particle from $t = 0$ to $t = 2\text{ s}$.
Attempt these questions under a ten-minute limit using a separate timer.
1. What is the time derivative of the velocity vector called?
Explanation: Acceleration is defined as the time derivative of velocity: $\mathbf{a} = d\mathbf{v}/dt$.
2. Which of the following equations is NOT one of the standard constant acceleration kinematic formulas in 1D?
Explanation: The formula $v = v_0 t + a^2 t^2$ is dimensionally incorrect and physically meaningless.
3. In uniform circular motion, the angle between the velocity vector and the acceleration vector is always:
Explanation: In uniform circular motion, velocity is tangential and centripetal acceleration is radially inward, making the angle between them $90^\circ$.
4. If a projectile has a time of flight $T$, what is the time taken to reach its maximum height?
Explanation: Due to the symmetry of projectile motion on flat ground, the time to reach maximum height is exactly half of the total time of flight, i.e., $T/2$.
5. What is the radial acceleration component for a particle moving in polar coordinates when $\dot{r} = 0$ and $\dot{\theta} = \text{constant}$?
Explanation: The radial acceleration component is $(\ddot{r} - r\dot{\theta}^2)$. With $\ddot{r}=0$, it simplifies to $-r\dot{\theta}^2$.
6. A stone is dropped from a tower of height $h$. With what speed does it hit the ground? (Neglect air resistance)
Explanation: Using $v^2 = v_0^2 + 2as$ with $v_0 = 0$, $a = g$, and $s = h$, we get $v^2 = 2gh$, so $v = \sqrt{2gh}$.
7. Which coordinate system is most convenient for analyzing motion under central forces?
Explanation: Central forces are directed toward a single point, making polar (or spherical) coordinates ideal due to natural symmetry.
8. What is the physical significance of the area under an acceleration-time graph?
Explanation: Since $a = dv/dt$, integrating $a$ with respect to time gives $\int a dt = \Delta v$, representing the change in velocity.
9. If a particle's position vector is $\mathbf{r} = t^2\hat{i} + t\hat{j}$, its trajectory in the $xy$-plane is:
Explanation: Here $x = t^2$ and $y = t$. Substituting $t = y$ into $x$ gives $x = y^2$, which is the equation of a parabola.
10. What is the maximum horizontal range for a given initial velocity $v_0$ in projectile motion?
Explanation: Range $R = \frac{v_0^2 \sin(2\theta_0)}{g}$ is maximized when $\sin(2\theta_0) = 1$, giving $\theta_0 = 45^\circ$ and $R_{\max} = \frac{v_0^2}{g}$.