AP Physics C Mechanics- 1.5 Motion in Two or Three Dimensions- Study Notes- New Syllabus
AP Physics C Mechanics- 1.5 Motion in Two or Three Dimensions – Study Notes
AP Physics C Mechanics- 1.5 Motion in Two or Three Dimensions – Study Notes – per latest Syllabus.
Key Concepts:
- Motion of an Object in Two or Three Dimensions
- Velocity and Acceleration in Multiple Dimensions (Nonuniform Motion)
- Independence of Motion in Perpendicular Directions
- Projectile Motion
Motion of an Object in Two or Three Dimensions
Many physical motions occur in a plane or in space — such as a projectile’s flight, a planet’s orbit, or an airplane’s trajectory. These can be described using the same kinematic principles applied in one dimension, but extended to two or three dimensions.
Position Vector:
The position of a particle at any time \( \mathrm{t} \) is given by:
\( \mathrm{\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k}} \)
- \( \mathrm{x(t), y(t), z(t)} \): displacement components along the x-, y-, and z-axes.
- The trajectory or path of the motion is the curve traced by the tip of \( \mathrm{\vec{r}(t)} \).
Velocity Vector:
The instantaneous velocity is the time derivative of position:
\( \mathrm{\vec{v}(t) = \dfrac{d\vec{r}}{dt} = \dfrac{dx}{dt}\hat{i} + \dfrac{dy}{dt}\hat{j} + \dfrac{dz}{dt}\hat{k}} \)
- Each component (\( \mathrm{v_x, v_y, v_z} \)) can be treated independently.
- The magnitude of velocity (speed) is \( \mathrm{|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}} \).
Acceleration Vector:
The instantaneous acceleration is the derivative of velocity:
\( \mathrm{\vec{a}(t) = \dfrac{d\vec{v}}{dt} = \dfrac{d^2\vec{r}}{dt^2}} \)
In component form:
\( \mathrm{\vec{a}(t) = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}} \)
- Each acceleration component represents the rate of change of velocity in that direction.
- The total acceleration magnitude is \( \mathrm{|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}} \).
Example
A particle has a constant velocity \( \mathrm{\vec{v} = 4\hat{i} – 1\hat{j} + 2\hat{k} \; (m/s)} \). At time \( \mathrm{t_0 = 0 \, s} \) its position is \( \mathrm{\vec{r}_0 = 1\hat{i} + 2\hat{j} + 3\hat{k} \; (m)} \).
Find:
- (a) The position vector \( \mathrm{\vec{r}(t)} \) at an arbitrary time \( \mathrm{t} \).
- (b) The position at \( \mathrm{t = 5 \, s} \).
- (c) The displacement between \( \mathrm{t = 2 \, s} \) and \( \mathrm{t = 5 \, s} \).
- (d) The speed and show that average velocity equals instantaneous velocity for this motion.
▶️ Answer / Explanation
Given:
- \( \mathrm{\vec{v} = 4\hat{i} – 1\hat{j} + 2\hat{k} \; (m/s)} \) (constant)
- \( \mathrm{\vec{r}_0 = 1\hat{i} + 2\hat{j} + 3\hat{k} \; (m)} \) at \( \mathrm{t_0 = 0 \, s} \)
(a) Position as a function of time:
For constant velocity, integrate velocity (or use \( \mathrm{\vec{r}(t) = \vec{r}_0 + \vec{v}t} \)):
\( \mathrm{\vec{r}(t) = (1 + 4t)\hat{i} + (2 – t)\hat{j} + (3 + 2t)\hat{k} \; (m)} \)
(b) Position at \( \mathrm{t = 5 \, s} \):
\( \mathrm{\vec{r}(5) = (1 + 4(5))\hat{i} + (2 – 5)\hat{j} + (3 + 2(5))\hat{k}} \)
\( \mathrm{\vec{r}(5) = 21\hat{i} – 3\hat{j} + 13\hat{k} \; (m)} \)
(c) Displacement from \( \mathrm{t = 2 \, s} \) to \( \mathrm{t = 5 \, s} \):
Displacement \( \mathrm{\Delta \vec{r} = \vec{r}(5) – \vec{r}(2) = \vec{v}(5 – 2)} \) (since velocity is constant)
First compute \( \mathrm{\vec{r}(2) = (1+8)\hat{i} + (2-2)\hat{j} + (3+4)\hat{k} = 9\hat{i} + 0\hat{j} + 7\hat{k}} \)
Thus \( \mathrm{\Delta \vec{r} = 21\hat{i} – 3\hat{j} + 13\hat{k} – (9\hat{i} + 0\hat{j} + 7\hat{k}) = 12\hat{i} – 3\hat{j} + 6\hat{k}} \)
(Alternatively \( \mathrm{\Delta \vec{r} = \vec{v}\Delta t = (4\hat{i} – 1\hat{j} + 2\hat{k})(3) = 12\hat{i} – 3\hat{j} + 6\hat{k}} \).)
(d) Speed and average vs instantaneous velocity:
Speed (magnitude of velocity):
\( \mathrm{|\vec{v}| = \sqrt{4^2 + (-1)^2 + 2^2} = \sqrt{16 + 1 + 4} = \sqrt{21} = 4.583 \; m/s} \)
Average velocity over any interval \( \mathrm{\Delta t} \) is
\( \mathrm{\vec{v}_{avg} = \dfrac{\Delta \vec{r}}{\Delta t} = \dfrac{\vec{v}\Delta t}{\Delta t} = \vec{v}} \)
Therefore the average velocity equals the instantaneous velocity for uniform (constant) velocity motion.
Velocity and Acceleration in Multiple Dimensions (Nonuniform Motion)
In two- or three-dimensional motion, both velocity and acceleration are vector quantities that may differ in magnitude and direction along each dimension. When the components of velocity or acceleration vary with time, the motion is said to be nonuniform.
Vector Representation:
The instantaneous velocity and acceleration in three-dimensional motion are expressed as:
\( \mathrm{\vec{v}(t) = v_x(t)\hat{i} + v_y(t)\hat{j} + v_z(t)\hat{k}} \)
\( \mathrm{\vec{a}(t) = a_x(t)\hat{i} + a_y(t)\hat{j} + a_z(t)\hat{k}} \)
Each component can be found by differentiating its respective position component:
\( \mathrm{v_x = \dfrac{dx}{dt}}, \quad v_y = \dfrac{dy}{dt}, \quad v_z = \dfrac{dz}{dt}} \)
\( \mathrm{a_x = \dfrac{dv_x}{dt}}, \quad a_y = \dfrac{dv_y}{dt}}, \quad a_z = \dfrac{dv_z}{dt}} \)
Magnitude of Velocity and Acceleration:
\( \mathrm{|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}} \)
\( \mathrm{|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}} \)
- These magnitudes may change with time if any component varies with time.
- Nonuniform motion implies that at least one of the components is time-dependent.
Changing Directions:
- Even if the speed (magnitude of velocity) remains constant, the acceleration can be nonzero if the direction of motion changes (as in circular motion).
- Thus, velocity and acceleration vectors may not always be aligned — acceleration may have both tangential and normal components.
Decomposition of Acceleration:
\( \mathrm{\vec{a} = \vec{a}_t + \vec{a}_n} \)
- \( \mathrm{\vec{a}_t} \): tangential acceleration (changes speed)
- \( \mathrm{\vec{a}_n} \): normal (centripetal) acceleration (changes direction)
Equations of Motion for Variable Acceleration:
If acceleration components are functions of time, then velocity and position must be obtained by integration:
\( \mathrm{\vec{v}(t) = \vec{v}_0 + \displaystyle \int_{t_0}^{t} \vec{a}(t’) \, dt’} \)
\( \mathrm{\vec{r}(t) = \vec{r}_0 + \displaystyle \int_{t_0}^{t} \vec{v}(t’) \, dt’} \)
- These equations are general and apply to all motions, uniform or nonuniform.
Example:
The position of a particle is given by \( \mathrm{\vec{r}(t) = (3t^2)\hat{i} + (4t^3)\hat{j}} \), where \( \mathrm{t} \) is in seconds and position is in meters. Find:
- (a) The velocity vector as a function of time
- (b) The acceleration vector as a function of time
- (c) The velocity and acceleration at \( \mathrm{t = 2 \, s} \)
▶️ Answer / Explanation
Given: \( \mathrm{\vec{r}(t) = 3t^2\hat{i} + 4t^3\hat{j}} \)
(a) Velocity:
Differentiate position with respect to time:
\( \mathrm{\vec{v}(t) = \dfrac{d\vec{r}}{dt} = (6t)\hat{i} + (12t^2)\hat{j}} \)
(b) Acceleration:
Differentiate velocity with respect to time:
\( \mathrm{\vec{a}(t) = \dfrac{d\vec{v}}{dt} = 6\hat{i} + 24t\hat{j}} \)
(c) At \( \mathrm{t = 2 \, s:} \)
\( \mathrm{\vec{v}(2) = (6)(2)\hat{i} + (12)(2)^2\hat{j} = 12\hat{i} + 48\hat{j} \, (m/s)} \)
\( \mathrm{\vec{a}(2) = 6\hat{i} + 24(2)\hat{j} = 6\hat{i} + 48\hat{j} \, (m/s^2)} \)
Magnitudes:
\( \mathrm{|\vec{v}| = \sqrt{12^2 + 48^2} = 49.48 \, m/s} \)
\( \mathrm{|\vec{a}| = \sqrt{6^2 + 48^2} = 48.37 \, m/s^2} \)
Independence of Motion in Perpendicular Directions
In two- or three-dimensional motion, the motion of an object along one axis (for example, the x-direction) is independent of its motion along a perpendicular axis (such as the y-direction). This means that a change in motion along one dimension does not affect the motion along any direction perpendicular to it.
Mathematical Representation:
If an object moves under constant acceleration in two perpendicular directions, its motion can be described by the one-dimensional kinematic equations applied separately to each axis:
\( \mathrm{x = x_0 + v_{0x}t + \dfrac{1}{2}a_x t^2} \)
\( \mathrm{y = y_0 + v_{0y}t + \dfrac{1}{2}a_y t^2} \)
- The motion along the x-axis depends only on \( \mathrm{a_x} \) and \( \mathrm{v_{0x}} \).
- The motion along the y-axis depends only on \( \mathrm{a_y} \) and \( \mathrm{v_{0y}} \).
- Since \( \mathrm{a_x} \) and \( \mathrm{a_y} \) act independently, one motion does not influence the other.
Key Idea:
The principle of independence of motion allows two-dimensional motion (such as projectile motion) to be analyzed as two separate one-dimensional motions — one horizontal and one vertical — connected only through the common time variable \( \mathrm{t} \).
- Horizontal motion → uniform velocity (no horizontal acceleration if air resistance is negligible)
- Vertical motion → uniformly accelerated motion (due to gravity)
Physical Interpretation:
- In projectile motion, gravity acts only vertically, so it affects only the vertical component of velocity, not the horizontal one.
- The object’s horizontal velocity remains constant (if \( \mathrm{a_x = 0} \)), while its vertical velocity changes due to \( \mathrm{a_y = -g} \).
| Direction | Equation of Motion | Influence |
|---|---|---|
| Horizontal (x-direction) | \( \mathrm{x = x_0 + v_{0x}t + \dfrac{1}{2}a_x t^2} \) | Unaffected by vertical motion |
| Vertical (y-direction) | \( \mathrm{y = y_0 + v_{0y}t + \dfrac{1}{2}a_y t^2} \) | Unaffected by horizontal motion |
Example:
A ball is thrown horizontally from a cliff with an initial velocity of \( \mathrm{10 \, m/s} \). It falls under gravity (\( \mathrm{g = 9.8 \, m/s^2} \)). Find:
- (a) The time it takes to fall \( \mathrm{20 \, m} \) vertically.
- (b) The horizontal distance it travels before hitting the ground.
▶️ Answer / Explanation
Given: \( \mathrm{v_{0x} = 10 \, m/s}, \, v_{0y} = 0, \, a_x = 0, \, a_y = -9.8 \, m/s^2, \, \Delta y = -20 \, m} \)
(a) Vertical Motion:
Using \( \mathrm{y = y_0 + v_{0y}t + \dfrac{1}{2}a_y t^2} \):
\( \mathrm{-20 = 0 + 0 – \dfrac{1}{2}(9.8)t^2} \)
\( \mathrm{t^2 = \dfrac{40}{9.8} = 4.08} \Rightarrow \mathrm{t = 2.02 \, s} \)
(b) Horizontal Motion:
Horizontal motion has no acceleration (\( \mathrm{a_x = 0} \)):
\( \mathrm{x = v_{0x}t = 10(2.02) = 20.2 \, m} \)
Projectile Motion — A Special Case of Two-Dimensional Motion
Projectile motion is a special case of two-dimensional motion in which an object moves under the influence of gravity alone, after being projected into the air. In such motion, there is zero acceleration in one dimension (horizontal) and a constant, nonzero acceleration (due to gravity) in the other dimension (vertical).
The motion of a projectile takes place in a vertical plane and can be analyzed as two independent one-dimensional motions:
- Horizontal motion → uniform (constant velocity)
- Vertical motion → uniformly accelerated (constant downward acceleration due to gravity)
These two motions are connected only by the common time variable \( \mathrm{t} \).
Assumptions in Ideal Projectile Motion:
- Air resistance is neglected.
- Acceleration due to gravity \( \mathrm{g} \) is constant and acts vertically downward.
- Motion occurs near Earth’s surface.
Component Analysis of Motion:
Let an object be projected with an initial velocity \( \mathrm{\vec{v}_0} \) making an angle \( \mathrm{\theta} \) with the horizontal.
Horizontal Component:
- Acceleration: \( \mathrm{a_x = 0} \)
- Velocity: \( \mathrm{v_x = v_{0x} = v_0 \cos\theta} \) (constant)
- Displacement: \( \mathrm{x = v_{0x}t = v_0 \cos\theta \, t} \)
Vertical Component:
- Acceleration: \( \mathrm{a_y = -g} \) (downward)
- Velocity: \( \mathrm{v_y = v_{0y} – gt = v_0 \sin\theta – gt} \)
- Displacement: \( \mathrm{y = v_{0y}t – \dfrac{1}{2}gt^2} \)
Trajectory Equation (Path of the Projectile):
Eliminating \( \mathrm{t} \) from the two component equations:
\( \mathrm{y = x\tan\theta – \dfrac{g x^2}{2v_0^2 \cos^2\theta}} \)
- This is the equation of a parabola — showing that a projectile follows a parabolic path.
Important Quantities in Projectile Motion:
| Quantity | Expression | Description |
|---|---|---|
| Time of Flight | \( \mathrm{T = \dfrac{2v_0 \sin\theta}{g}} \) | Total time the projectile remains in the air |
| Maximum Height | \( \mathrm{H = \dfrac{v_0^2 \sin^2\theta}{2g}} \) | Vertical height reached above the point of projection |
| Horizontal Range | \( \mathrm{R = \dfrac{v_0^2 \sin 2\theta}{g}} \) | Horizontal distance traveled before hitting the ground |
Example:
A projectile is launched with an initial speed of \( \mathrm{20 \, m/s} \) at an angle of \( \mathrm{45^\circ} \) above the horizontal. Find:
- (a) Time of flight
- (b) Maximum height
- (c) Horizontal range
- Take \( \mathrm{g = 9.8 \, m/s^2} \).
▶️ Answer / Explanation
Given: \( \mathrm{v_0 = 20 \, m/s}, \, \theta = 45^\circ, \, g = 9.8 \, m/s^2} \)
(a) Time of Flight:
\( \mathrm{T = \dfrac{2v_0 \sin\theta}{g} = \dfrac{2(20)\sin45^\circ}{9.8} = \dfrac{40(0.707)}{9.8} = 2.89 \, s} \)
(b) Maximum Height:
\( \mathrm{H = \dfrac{v_0^2 \sin^2\theta}{2g} = \dfrac{(20)^2(0.707)^2}{2(9.8)} = \dfrac{200}{19.6} = 10.2 \, m} \)
(c) Horizontal Range:
\( \mathrm{R = \dfrac{v_0^2 \sin2\theta}{g} = \dfrac{(20)^2 \sin90^\circ}{9.8} = \dfrac{400}{9.8} = 40.8 \, m} \)