Calculus-Based Motion: 1D and 2D Kinematics & Projectile Motion

Motion in One Dimension (1D)

Integral Calculus of Motion

Calculus provides a powerful framework for analyzing motion, allowing us to relate acceleration, velocity, and position through integration and differentiation. In one-dimensional motion, these relationships are foundational for solving kinematics problems.

• Acceleration to Velocity: If acceleration as a function of time is known, velocity can be found by integrating acceleration with respect to time.

• Velocity to Position: Similarly, integrating velocity with respect to time yields the position function.

• Special Case (Constant Acceleration): For constant acceleration, the equations simplify to the classic kinematic equations.

Key Equations:

• From acceleration to velocity: $v(t) = v_0 + \int_{0}^{t} a(t') dt'$

• From velocity to position: $x(t) = x_0 + \int_{0}^{t} v(t') dt'$

Example: A rocket car starts at $v_0 = 3.00\ \mathrm{m/s}$ and accelerates according to $a(t) = 9.00\ \mathrm{m/s^3} \cdot t$. Find its velocity at time $t$ and at $t = 2.00\ \mathrm{s}$.

• Integrate acceleration to get velocity: $v(t) = 3.00 + \int_0^t 9.00 t' dt' = 3.00 + 4.50 t^2$

• At $t = 2.00$ s: $v(2.00) = 3.00 + 4.50 \times (2.00)^2 = 3.00 + 18.00 = 21.00\ \mathrm{m/s}$

Example: If the car starts at $x = 0$ at $t = 0$, find its position at $t = 2.00$ s.

• Integrate velocity: $x(t) = \int_0^t [3.00 + 4.50 t'^2] dt' = 3.00 t + 1.50 t^3$

• At $t = 2.00$ s: $x(2.00) = 3.00 \times 2.00 + 1.50 \times (2.00)^3 = 6.00 + 12.00 = 18.00\ \mathrm{m}$

Motion in Two and Three Dimensions (2D/3D)

Vector Representation of Motion

In two or three dimensions, position, velocity, and acceleration are described by vectors. The motion can be analyzed by considering each component separately.

• Position Vector: $\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k}$

• Velocity Vector: $\vec{v}(t) = \frac{d\vec{r}}{dt}$

• Acceleration Vector: $\vec{a}(t) = \frac{d\vec{v}}{dt}$

Example: A Mars rover's position is given by:

• $x = 2.0\ \mathrm{m} - (0.25\ \mathrm{m/s^2}) t^2$

• $y = (1.0\ \mathrm{m/s}) t + (0.025\ \mathrm{m/s^3}) t^3$

To analyze the rover's motion:

• Sketch the path by plotting $x(t)$ and $y(t)$ for $0 \leq t \leq 2.0$ s.

• Find velocity and acceleration by differentiating the position functions.

Key Points:

• Velocity and acceleration vectors change direction and magnitude as the rover moves.

• Graphical analysis helps visualize the motion and interpret the physical meaning of the equations.

Projectile Motion

Projectile motion is a special case of two-dimensional motion where an object moves under the influence of gravity alone, after an initial launch. The horizontal and vertical motions are independent except for the time variable.

• Horizontal Motion: Constant velocity, $a_x = 0$.

• Vertical Motion: Constant acceleration due to gravity, $a_y = -g$.

Equations of Projectile Motion:

• Horizontal position: $x(t) = x_0 + v_{0x} t$

• Vertical position: $y(t) = y_0 + v_{0y} t - \frac{1}{2} g t^2$

Key Concepts:

• Time of flight depends only on vertical motion.

• Range depends on initial velocity and launch angle.

• Maximum height is reached when vertical velocity is zero.

Sample Problem: Horizontal Launch from a Cliff

Problem: An object is thrown horizontally from a cliff with initial speed $v_0 = 10.0\ \mathrm{m/s}$ and takes $t = 4.52$ s to hit the ground. How far from the base does it land? From what height was it thrown?

• Horizontal distance: $x = v_0 t = 10.0 \times 4.52 = 45.2\ \mathrm{m}$

• Vertical distance (height): $y = \frac{1}{2} g t^2 = 0.5 \times 9.80 \times (4.52)^2 = 100\ \mathrm{m}$

Conclusion: The independence of horizontal and vertical motions allows us to solve for range and height separately using the appropriate kinematic equations.

Summary Table: Kinematic Equations for Constant Acceleration

Quantity	Equation	Notes
Velocity (from acceleration)	$v(t) = v_0 + a t$	For constant $a$
Position (from velocity)	$x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$	For constant $a$
Projectile horizontal position	$x(t) = x_0 + v_{0x} t$	$a_x = 0$
Projectile vertical position	$y(t) = y_0 + v_{0y} t - \frac{1}{2} g t^2$	$a_y = -g$

Additional info:

• These notes cover material from "Units, Physical Quantities & Vectors," "Motion Along a Straight Line," and "Motion in Two or Three Dimensions" as outlined in standard calculus-based physics syllabi.

• Images included are directly relevant to the explanation of vector motion, projectile motion, and the graphical analysis of kinematics.