In physics, physical quantities are classified into two categories based on their directional properties: scalars and vectors .
To describe motion in a plane, a reference point $O$ (origin) is chosen.
Two vectors $\vec{A}$ and $\vec{B}$ are equal if and only if they have the same magnitude and the same direction. Graphically, if vector $\vec{B}$ is shifted parallel to itself until its tail coincides with the tail of $\vec{A}$, their tips must also coincide.
Multiplying a vector $\vec{A}$ by a positive real number $\lambda$ yields a vector whose magnitude is scaled by $\lambda$ while maintaining the same direction: $|\lambda \vec{A}| = \lambda |\vec{A}|$ for $\lambda > 0$. Multiplying by a negative number $-\lambda$ reverses the direction of the vector while scaling its magnitude by $\lambda$.
Vectors do not follow ordinary algebraic addition. Instead, they are added geometrically:
Vector addition is both commutative ($\vec{A} + \vec{B} = \vec{B} + \vec{A}$) and associative ($(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C})$).
The difference between two vectors $\vec{A}$ and $\vec{B}$ is defined as the sum of $\vec{A}$ and the negative of $\vec{B}$:
$$\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$$
A null vector (or zero vector) $\vec{0}$ is a vector with zero magnitude and an unspecified direction. It arises when subtracting a vector from itself ($\vec{A} - \vec{A} = \vec{0}$) or multiplying any vector by the scalar zero ($0 \vec{A} = \vec{0}$).
To simplify vector operations, vectors are resolved into perpendicular components along coordinate axes using unit vectors .
Unit vectors are dimensionless vectors of magnitude 1 pointing along specific coordinate axes. In a right-handed Cartesian system, they are denoted by $\hat{i}$, $\hat{j}$, and $\hat{k}$ along the $x$, $y$, and $z$ axes respectively, where $|\hat{i}| = |\hat{j}| = |\hat{k}| = 1$.
Any vector $\vec{A}$ in the $x$-$y$ plane can be uniquely resolved into component vectors along the axes:
$$\vec{A} = A_x \hat{i} + A_y \hat{j}$$
Using trigonometry, the components are related to the magnitude $A$ and angle $\theta$ with the $x$-axis by:
Analytical addition is highly accurate and involves summing the respective components of the vectors. For $\vec{R} = \vec{A} + \vec{B}$:
Motion in a plane involves position, velocity, and acceleration vectors varying with time.
| Kinematic Variable | Vector Definition | Component Form |
|---|---|---|
| Position | $\vec{r}(t)$ | $x(t)\hat{i} + y(t)\hat{j}$ |
| Velocity | $\vec{v} = \frac{d\vec{r}}{dt}$ | $v_x \hat{i} + v_y \hat{j} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j}$ |
| Acceleration | $\vec{a} = \frac{d\vec{v}}{dt}$ | $a_x \hat{i} + a_y \hat{j} = \frac{dv_x}{dt}\hat{i} + \frac{dv_y}{dt}\hat{j}$ |
The direction of the instantaneous velocity vector $\vec{v}$ at any point on the path of an object is always tangential to the path at that point.
When acceleration $\vec{a}$ is constant, the equations of motion along perpendicular axes are completely independent. Thus, 2D motion can be treated as two simultaneous, independent 1D motions:
Mathematical Problem Statement: A particle starts from the origin at $t = 0$ with an initial velocity of $5.0 \hat{i}$ m/s and moves in the $x$-$y$ plane under a force producing a constant acceleration of $(3.0\hat{i} + 2.0\hat{j})$ m/sĀ². Find the $y$-coordinate of the particle at the instant its $x$-coordinate is 84 m, and determine its speed at this time.
Step-by-Step Algorithmic Solution Trace:
- Using the position equation $\vec{r}(t) = \vec{v}_0 t + \frac{1}{2}\vec{a}t^2$, resolve into components: $x(t) = 5.0t + 1.5t^2$ and $y(t) = 1.0t^2$.
- Set $x(t) = 84$ m: $1.5t^2 + 5.0t - 84 = 0$. Solving the quadratic equation yields $t = 6$ s.
- Substitute $t = 6$ s into the $y$-coordinate equation: $y(6) = 1.0(6)^2 = 36$ m.
- Differentiate position to find velocity: $\vec{v}(t) = (5.0 + 3.0t)\hat{i} + 2.0t\hat{j}$. At $t = 6$ s, $\vec{v} = 23.0\hat{i} + 12.0\hat{j}$ m/s.
- Calculate speed: $v = \sqrt{23.0^2 + 12.0^2} \approx 26$ m/s.
Final Evaluation Output: $y = 36$ m , $v \approx 26$ m/s
A projectile is an object in flight after being projected into the air. Its motion is a combination of horizontal motion with constant velocity (zero acceleration) and vertical motion with constant downward acceleration due to gravity ($a_y = -g$).
By eliminating time $t$ from the horizontal and vertical position equations, we obtain the equation of the path (trajectory):
$$y = (\tan \theta_0) x - \frac{g}{2(v_0 \cos \theta_0)^2} x^2$$
Since this equation is of the form $y = ax + bx^2$, the path of a projectile is a parabola .
| Parameter | Physical Meaning | Mathematical Formula |
|---|---|---|
| Time of Maximum Height ($t_m$) | Time taken to reach the peak where $v_y = 0$ | $t_m = \frac{v_0 \sin \theta_0}{g}$ |
| Time of Flight ($T_f$) | Total time the projectile remains in the air | $T_f = \frac{2 v_0 \sin \theta_0}{g}$ |
| Maximum Height ($h_m$) | The peak vertical displacement reached | $h_m = \frac{(v_0 \sin \theta_0)^2}{2g}$ |
| Horizontal Range ($R$) | Total horizontal distance covered when $y = 0$ | $R = \frac{v_0^2 \sin 2\theta_0}{g}$ |
The horizontal range is maximum when $\theta_0 = 45^\circ$, giving $R_{max} = \frac{v_0^2}{g}$. For any two complementary angles of projection ($\theta_0$ and $90^\circ - \theta_0$), the horizontal ranges are equal.
Historical Chronicle: Galileo Galilei
Mathematical Problem Statement: A cricket ball is thrown at a speed of 28 m/s in a direction 30Ā° above the horizontal. Calculate (a) the maximum height, (b) the time taken to return to the same level, and (c) the horizontal distance from the thrower to the point where the ball returns to the same level.
Step-by-Step Algorithmic Solution Trace:
- Identify given values: $v_0 = 28$ m/s, $\theta_0 = 30^\circ$, $g = 9.8$ m/sĀ².
- Calculate maximum height: $h_m = \frac{(28 \sin 30^\circ)^2}{2 \times 9.8} = \frac{14^2}{19.6} = 10.0$ m.
- Calculate time of flight: $T_f = \frac{2 \times 28 \times \sin 30^\circ}{9.8} = \frac{28}{9.8} \approx 2.9$ s.
- Calculate horizontal range: $R = \frac{28^2 \sin 60^\circ}{9.8} = \frac{784 \times 0.866}{9.8} \approx 69$ m.
Final Evaluation Output: $h_m = 10.0$ m , $T_f = 2.9$ s , $R = 69$ m
When an object follows a circular path at a constant speed, its motion is called uniform circular motion . Although speed is constant, the velocity vector continuously changes direction, resulting in an acceleration directed towards the center of the circle.
The acceleration of an object in uniform circular motion is always directed towards the center of the circular path and is called centripetal acceleration ($a_c$):
$$a_c = \frac{v^2}{R}$$
Let the object turn through an angle $\Delta \theta$ in time $\Delta t$. The rate of change of angular displacement is the angular speed ($\omega$):
$$\omega = \frac{\Delta \theta}{\Delta t}$$
The linear speed $v$ is related to angular speed by $v = \omega R$. Thus, centripetal acceleration can also be expressed as:
$$a_c = \omega^2 R$$
If $T$ is the time period of one complete revolution and $\nu$ is the frequency ($\nu = 1/T$), then:
Historical Chronicle: Christiaan Huygens