Physics

Chapter 4 Motion in Two and Three Dimensions

Imron Rosyadi

Chapter 4: Motion in Two and Three Dimensions

Overview

• 4.1 Displacement and Velocity Vectors
• 4.2 Acceleration Vector
• 4.3 Projectile Motion
• 4.4 Uniform and Nonuniform Circular Motion
• 4.5 Relative Motion in One and Two Dimensions

4.1 Displacement and Velocity Vectors

Learning Objectives

By the end of this section, you will be able to:

• Calculate position vectors in a multidimensional displacement problem.
• Solve for the displacement in two or three dimensions.
• Calculate the velocity vector given the position vector as a function of time.
• Calculate the average velocity in multiple dimensions.

Today, we’re building on our understanding of 1D motion. The key difference now is that displacement and velocity are vector quantities. This means we need to use vector algebra.

Displacement Vector

Coordinate System

To describe motion in 2D or 3D, we use a coordinate system. A particle at point \(P(x, y, z)\) has coordinates that are functions of time \(t\):

\[ x = x(t) \\ y = y(t) \\ z = z(t) \]

The position vector \(\vec{r}(t)\) from the origin to point \(P\) is:

\[ \vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k} \]

We’ll typically use a right-handed coordinate system, meaning if you curl the fingers of your right hand from the x-axis to the y-axis, your thumb points in the positive z-direction.

Position Vector Visualization

Figure 4.2: A three-dimensional coordinate system with a particle at position \(P(x(t), y(t), z(t))\).

This image helps visualize the position vector \(\vec{r}(t)\) reaching to a point \(P\) in 3D space. Notice how the components \(x(t)\), \(y(t)\), and \(z(t)\) define the coordinates of the point.

Displacement Vector

If a particle is at \(P_1\) with position vector \(\vec{r}(t_1)\) at time \(t_1\), and then at \(P_2\) with position vector \(\vec{r}(t_2)\) at a later time \(t_2\). The displacement vector \(\Delta\vec{r}\) is:

\[ \Delta\vec{r} = \vec{r}(t_2) - \vec{r}(t_1) \]

Note

This is the same concept as 1D displacement, but now we’re subtracting vectors.

Remember that vector subtraction is like adding the negative of a vector. Geometrically, \(\Delta\vec{r}\) points from the initial position \(P_1\) to the final position \(P_2\).

Displacement Vector Visualization

Figure 4.3: The displacement \(\Delta\vec{r} = \vec{r}(t_2) - \vec{r}(t_1)\) is the vector from \(P_1\) to \(P_2\).

Here, you can clearly see the initial position vector \(\vec{r}(t_1)\), the final position vector \(\vec{r}(t_2)\), and the displacement vector \(\Delta\vec{r}\) connecting the two points.

Example: Polar Orbiting Satellite

A satellite orbits Earth at 400 km altitude. Find the displacement from directly over the North Pole to \(-45^\circ\) latitude.

Given:

• Altitude: \(400 \, \text{km}\)
• Earth’s radius: \(R_E = 6370 \, \text{km}\)

Total radius of orbit: \(R = R_E + \text{altitude} = 6370 + 400 = 6770 \, \text{km}\).

Figure 4.4: Position vectors from Earth’s center. Y-axis is North, X-axis is East.

This example shows how to apply the displacement vector concept to a real-world scenario. The key is setting up the coordinate system correctly. The coordinate system is centered on the Earth, with the y-axis pointing North and the x-axis pointing East.

Example: Polar Orbiting Satellite (Solution)

Solution

Initial position (North Pole):
\(\vec{r}(t_1) = 6770 \, \text{km} \, \hat{j}\)

Final position (\(-45^\circ\) latitude):
\(\vec{r}(t_2) = 6770 \, \text{km} \, (\cos(-45^\circ)\hat{i} + \sin(-45^\circ)\hat{j})\)
\(\vec{r}(t_2) = 4787 \, \text{km} \, \hat{i} - 4787 \, \text{km} \, \hat{j}\)

Displacement vector:
\(\Delta\vec{r} = \vec{r}(t_2) - \vec{r}(t_1)\)
\(\Delta\vec{r} = (4787 \, \text{km} \, \hat{i} - 4787 \, \text{km} \, \hat{j}) - (6770 \, \text{km} \, \hat{j})\)
\(\Delta\vec{r} = 4787 \, \text{km} \, \hat{i} - 11557 \, \text{km} \, \hat{j}\)

Magnitude: \(|\Delta\vec{r}| = \sqrt{(4787)^2 + (-11557)^2} \approx 12509 \, \text{km}\)
Direction: \(\theta = \tan^{-1}\left(\frac{-11557}{4787}\right) \approx -67.5^\circ\) (south of east)

We first define the initial and final position vectors using unit vector notation. The North Pole is along the positive y-axis, and -45 degrees latitude is 45 degrees clockwise from the positive y-axis if we’re looking down from above, or just a 45 degree angle with the x-axis in the fourth quadrant. Then, we perform vector subtraction and calculate the magnitude and direction.

Significance of Displacement

Figure 4.5: Displacement vector with components, angle, and magnitude.

Tip

The displacement vector represents the shortest path between two points. Any actual path taken will be longer or equal to the magnitude of the displacement.

This graph shows the displacement vector and its components. It highlights that the satellite did not travel in a straight line, but the displacement vector still points directly from the start to the end point.

Velocity Vector

Instantaneous Velocity

The instantaneous velocity vector \(\vec{v}(t)\) is the derivative of the position function with respect to time:

\[ \vec{v}(t) = \lim_{\Delta t \to 0} \frac{\vec{r}(t+\Delta t) - \vec{r}(t)}{\Delta t} = \frac{d\vec{r}}{dt} \]

The velocity vector is always tangent to the particle’s path.

This definition is a direct extension of 1D instantaneous velocity. The crucial difference is that \(\vec{v}(t)\) is a vector, meaning both its magnitude and direction can change over time.

Velocity Vector Direction

Figure 4.7: As \(\Delta t\) approaches zero, the velocity vector becomes tangent to the path.

This image illustrates how the displacement vector \(\Delta\vec{r}\) becomes tangent to the path as \(\Delta t\) approaches zero, leading to the instantaneous velocity vector.

Velocity Vector Components

Since \(\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k}\), we can write the velocity vector in terms of its components:

\[ \vec{v}(t) = v_x(t)\hat{i} + v_y(t)\hat{j} + v_z(t)\hat{k} \]

where

\[ v_x(t) = \frac{dx(t)}{dt} \\ v_y(t) = \frac{dy(t)}{dt} \\ v_z(t) = \frac{dz(t)}{dt} \]

Each component of the velocity vector is simply the derivative of the corresponding position component. This is a powerful idea: 3D motion can be broken down into three independent 1D motions.

Average Velocity

The average velocity vector \(\vec{v}_{avg}\) for two and three dimensions is:

\[ \vec{v}_{avg} = \frac{\vec{r}(t_2) - \vec{r}(t_1)}{t_2 - t_1} \]

Similar to 1D, average velocity is the total displacement divided by the total time interval. It’s a vector that points in the direction of the displacement.

Example: Calculating the Velocity Vector

A particle’s position function is \(\vec{r}(t) = 2.0t^2\hat{i} + (2.0+3.0t)\hat{j} + 5.0t\hat{k} \, \text{m}\).

1. What is the instantaneous velocity and speed at \(t = 2.0 \, \text{s}\)?
2. What is the average velocity between \(1.0 \, \text{s}\) and \(3.0 \, \text{s}\)?

Example: Calculating the Velocity Vector (Solution)

Solution (a)

Instantaneous velocity function: \(\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = 4.0t\hat{i} + 3.0\hat{j} + 5.0\hat{k} \, \text{m/s}\)

At \(t = 2.0 \, \text{s}\):
\(\vec{v}(2.0 \, \text{s}) = 4.0(2.0)\hat{i} + 3.0\hat{j} + 5.0\hat{k} = 8.0\hat{i} + 3.0\hat{j} + 5.0\hat{k} \, \text{m/s}\)

Speed (magnitude of instantaneous velocity):
\(|\vec{v}(2.0 \, \text{s})| = \sqrt{(8.0)^2 + (3.0)^2 + (5.0)^2} = \sqrt{64 + 9 + 25} = \sqrt{98} \approx 9.9 \, \text{m/s}\)

To find the instantaneous velocity, we differentiate each component of the position vector with respect to time. Then, we plug in the specific time value. The speed is simply the magnitude of this velocity vector.

Example: Calculating the Velocity Vector (Solution)

Solution (b)

Position at \(t_1 = 1.0 \, \text{s}\):
\(\vec{r}(1.0 \, \text{s}) = 2.0(1.0)^2\hat{i} + (2.0+3.0(1.0))\hat{j} + 5.0(1.0)\hat{k} = 2.0\hat{i} + 5.0\hat{j} + 5.0\hat{k} \, \text{m}\)

Position at \(t_2 = 3.0 \, \text{s}\):
\(\vec{r}(3.0 \, \text{s}) = 2.0(3.0)^2\hat{i} + (2.0+3.0(3.0))\hat{j} + 5.0(3.0)\hat{k} = 18.0\hat{i} + 11.0\hat{j} + 15.0\hat{k} \, \text{m}\)

Average velocity:
\(\vec{v}_{avg} = \frac{\vec{r}(3.0 \, \text{s}) - \vec{r}(1.0 \, \text{s})}{3.0 \, \text{s} - 1.0 \, \text{s}}\)
\(\vec{v}_{avg} = \frac{(18.0\hat{i} + 11.0\hat{j} + 15.0\hat{k}) - (2.0\hat{i} + 5.0\hat{j} + 5.0\hat{k})}{2.0 \, \text{s}}\)
\(\vec{v}_{avg} = \frac{16.0\hat{i} + 6.0\hat{j} + 10.0\hat{k}}{2.0} = 8.0\hat{i} + 3.0\hat{j} + 5.0\hat{k} \, \text{m/s}\)

Note

In this specific case, the average velocity equals the instantaneous velocity at \(t = 2.0 \, \text{s}\) because the velocity function is linear. This is generally not the case.

For average velocity, we need the initial and final position vectors. We calculate these by plugging in \(t_1\) and \(t_2\) into the given position function. Then, we apply the average velocity formula.

Independence of Perpendicular Motions

Key Concept:
The motion of an object in two or three dimensions can be divided into separate, independent motions along the perpendicular axes of the coordinate system.

• Motion along the \(x\)-direction does not affect motion along the \(y\) or \(z\) directions.
• This simplifies complex 2D/3D problems into multiple 1D problems.

This is a very fundamental and powerful principle in kinematics. It allows us to break down a seemingly complex problem into simpler parts that we already know how to solve from 1D kinematics. Think of it like separating a vector into its components; each component acts independently.

Independence of Perpendicular Motions (Visual)

Figure 4.8: Two identical balls: one falls from rest, the other thrown horizontally.

• Vertical motion: Identical for both balls, influenced only by gravity.
• Horizontal motion: Constant velocity for the thrown ball (assuming no air resistance).

This strobe photograph beautifully demonstrates the independence of motion. Notice how both balls fall at the same rate vertically, even though one is also moving horizontally. This is because gravity acts only vertically, and in the absence of air resistance, there are no horizontal forces.

4.2 Acceleration Vector

Learning Objectives

By the end of this section, you will be able to:

• Calculate the acceleration vector given the velocity function.
• Describe motion with constant acceleration in three dimensions.
• Use 1D motion equations along perpendicular axes for 2D/3D problems with constant acceleration.
• Express acceleration in unit vector notation.

Just like with velocity, we’re now extending the concept of acceleration to multiple dimensions. It’s still about the rate of change of velocity, but now we’re dealing with vector changes.

Instantaneous Acceleration

The instantaneous acceleration vector \(\vec{a}(t)\) is the derivative of the velocity function with respect to time:

\[ \vec{a}(t) = \lim_{\Delta t \to 0} \frac{\vec{v}(t+\Delta t) - \vec{v}(t)}{\Delta t} = \frac{d\vec{v}(t)}{dt} \]

In terms of components:

\[ \vec{a}(t) = \frac{dv_x(t)}{dt}\hat{i} + \frac{dv_y(t)}{dt}\hat{j} + \frac{dv_z(t)}{dt}\hat{k} \]

Also, as the second derivative of the position function:

\[ \vec{a}(t) = \frac{d^2x(t)}{dt^2}\hat{i} + \frac{d^2y(t)}{dt^2}\hat{j} + \frac{d^2z(t)}{dt^2}\hat{k} \]

The definition of instantaneous acceleration is very similar to instantaneous velocity. It’s the rate at which the velocity vector is changing. This change can be in magnitude (speeding up or slowing down), direction (turning), or both.

Example: Finding an Acceleration Vector

A particle has a velocity of \(\vec{v}(t) = 5.0t\hat{i} + t^2\hat{j} - 2.0t^3\hat{k} \, \text{m/s}\).

1. What is the acceleration function?
2. What is the acceleration vector at \(t = 2.0 \, \text{s}\)? Find its magnitude and direction.

Example: Finding an Acceleration Vector (Solution)

Solution (a)

Acceleration function:
\(\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = \frac{d}{dt}(5.0t)\hat{i} + \frac{d}{dt}(t^2)\hat{j} - \frac{d}{dt}(2.0t^3)\hat{k}\)
\(\vec{a}(t) = 5.0\hat{i} + 2.0t\hat{j} - 6.0t^2\hat{k} \, \text{m/s}^2\)

Solution (b)

At \(t = 2.0 \, \text{s}\):
\(\vec{a}(2.0 \, \text{s}) = 5.0\hat{i} + 2.0(2.0)\hat{j} - 6.0(2.0)^2\hat{k}\)
\(\vec{a}(2.0 \, \text{s}) = 5.0\hat{i} + 4.0\hat{j} - 24.0\hat{k} \, \text{m/s}^2\)

Magnitude:
\(|\vec{a}(2.0 \, \text{s})| = \sqrt{(5.0)^2 + (4.0)^2 + (-24.0)^2} = \sqrt{25 + 16 + 576} = \sqrt{617} \approx 24.8 \, \text{m/s}^2\)

Direction: (difficult to express simply in 3D without specific angles) The vector \(5.0\hat{i} + 4.0\hat{j} - 24.0\hat{k}\) gives the direction.

Notice how the acceleration in this example depends on time, meaning the particle’s rate of change of velocity is not constant. In 3D, describing “direction” usually involves giving the vector components or specific angles with the axes.

Constant Acceleration

For multidimensional motion with constant acceleration, we can treat each dimension independently.

In 2D (xy-plane), constant acceleration \(\vec{a} = a_x\hat{i} + a_y\hat{j}\) means \(a_x\) and \(a_y\) are constants.

Kinematic equations for \(x\)-direction:

\[ x(t) = x_0 + (v_x)_{avg}t \\ v_x(t) = v_{0x} + a_xt \\ x(t) = x_0 + v_{0x}t + \frac{1}{2}a_xt^2 \\ v_x^2(t) = v_{0x}^2 + 2a_x(x - x_0) \]

Kinematic equations for \(y\)-direction:

\[ y(t) = y_0 + (v_y)_{avg}t \\ v_y(t) = v_{0y} + a_yt \\ y(t) = y_0 + v_{0y}t + \frac{1}{2}a_yt^2 \\ v_y^2(t) = v_{0y}^2 + 2a_y(y - y_0) \]

The position and velocity vectors are then:
\(\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}\)
\(\vec{v}(t) = v_x(t)\hat{i} + v_y(t)\hat{j}\)

The beauty of constant acceleration in multiple dimensions is that we can apply our familiar 1D kinematic equations separately to each component of motion. The only common variable linking these independent motions is time. This is a crucial point for solving problems.

Example: A Skier

A skier moves with an acceleration of \(2.1 \, \text{m/s}^2\) down a \(15^\circ\) slope at \(t=0\). Origin at the lodge front, initial position and velocity:
\(\vec{r}(0) = (75.0\hat{i} - 50.0\hat{j}) \, \text{m}\)
\(\vec{v}(0) = (4.1\hat{i} - 1.1\hat{j}) \, \text{m/s}\)

1. What are the \(x\)- and \(y\)-components of the skier’s position and velocity as functions of time?
   
2. What are her position and velocity at \(t = 10.0 \, \text{s}\)?

Figure 4.10: Skier on a slope.

This problem combines constant acceleration with vector components. We need to resolve the acceleration into x and y components based on the slope angle and then use the 1D kinematic equations for each component.

Example: A Skier (Solution Part a)

Solution (a)

Acceleration components:
The acceleration is \(2.1 \, \text{m/s}^2\) down the \(15^\circ\) slope.
\(a_x = (2.1 \, \text{m/s}^2)\cos(15^\circ) \approx 2.0 \, \text{m/s}^2\)
\(a_y = -(2.1 \, \text{m/s}^2)\sin(15^\circ) \approx -0.54 \, \text{m/s}^2\)

Initial conditions:
\(x_0 = 75.0 \, \text{m}\), \(v_{0x} = 4.1 \, \text{m/s}\)
\(y_0 = -50.0 \, \text{m}\), \(v_{0y} = -1.1 \, \text{m/s}\)

Position and velocity as functions of time:

\(x(t) = x_0 + v_{0x}t + \frac{1}{2}a_xt^2 = 75.0 + 4.1t + \frac{1}{2}(2.0)t^2 = 75.0 + 4.1t + 1.0t^2\)
\(v_x(t) = v_{0x} + a_xt = 4.1 + 2.0t\)

\(y(t) = y_0 + v_{0y}t + \frac{1}{2}a_yt^2 = -50.0 - 1.1t + \frac{1}{2}(-0.54)t^2 = -50.0 - 1.1t - 0.27t^2\)
\(v_y(t) = v_{0y} + a_yt = -1.1 - 0.54t\)

It’s crucial to correctly assign the signs for acceleration components based on the chosen coordinate system. Here, \(a_x\) is positive because it points right (east), and \(a_y\) is negative because it points downwards.

Example: A Skier (Solution Part b)

Solution (b)

At \(t = 10.0 \, \text{s}\):

\(x(10.0 \, \text{s}) = 75.0 + 4.1(10.0) + 1.0(10.0)^2 = 75.0 + 41.0 + 100.0 = 216.0 \, \text{m}\)
\(v_x(10.0 \, \text{s}) = 4.1 + 2.0(10.0) = 4.1 + 20.0 = 24.1 \, \text{m/s}\)

\(y(10.0 \, \text{s}) = -50.0 - 1.1(10.0) - 0.27(10.0)^2 = -50.0 - 11.0 - 27.0 = -88.0 \, \text{m}\)
\(v_y(10.0 \, \text{s}) = -1.1 - 0.54(10.0) = -1.1 - 5.4 = -6.5 \, \text{m/s}\)

Position vector:
\(\vec{r}(10.0 \, \text{s}) = (216.0\hat{i} - 88.0\hat{j}) \, \text{m}\)

Velocity vector:
\(\vec{v}(10.0 \, \text{s}) = (24.1\hat{i} - 6.5\hat{j}) \, \text{m/s}\)

Magnitude of velocity:
\(|\vec{v}(10.0 \, \text{s})| = \sqrt{(24.1)^2 + (-6.5)^2} \approx 25.0 \, \text{m/s}\)

By having the equations for \(x(t)\), \(y(t)\), \(v_x(t)\), and \(v_y(t)\), we can find the skier’s state at any future time. The magnitude of the velocity at 10 seconds is roughly 60 mph, quite fast for a skier!

4.3 Projectile Motion

Learning Objectives

By the end of this section, you will be able to:

• Use 1D motion in perpendicular directions to analyze projectile motion.
• Calculate range, time of flight, and maximum height for projectiles on flat ground.
• Find time of flight and impact velocity for projectiles landing at different heights.
• Calculate the trajectory of a projectile.

Projectile motion is a classic application of the independence of perpendicular motions. We’re assuming negligible air resistance, so the only acceleration is due to gravity, which acts purely in the vertical direction.

Projectile Motion Defined

Projectile motion is the motion of an object thrown or projected into the air, subject only to acceleration as a result of gravity.

• Objects are called projectiles.
• Their path is called a trajectory.
• Crucial fact: Motions along perpendicular axes are independent.
  • Horizontal motion (\(x\)-axis): \(a_x = 0\)
  • Vertical motion (\(y\)-axis): \(a_y = -g = -9.8 \, \text{m/s}^2\) (if upward is positive)

This means that horizontally, the projectile moves at a constant velocity, and vertically, it undergoes constant acceleration due to gravity. This separation simplifies the analysis tremendously.

Projectile Displacement (Visual)

Figure 4.11: Total displacement \(\vec{s}\) of a soccer ball, with horizontal (\(\vec{x}\)) and vertical (\(\vec{y}\)) components.

This diagram shows how the total displacement \(\vec{s}\) can be broken down into its horizontal \(\vec{x}\) and vertical \(\vec{y}\) components. This vector breakdown is fundamental to analyzing projectile motion.

Kinematic Equations for Projectile Motion

Using \(a_x = 0\) and \(a_y = -g\):

Horizontal Motion

\[ v_{0x} = v_x \\ x = x_0 + v_xt \]

Vertical Motion

\[ v_y = v_{0y} - gt \\ y = y_0 + v_{0y}t - \frac{1}{2}gt^2 \\ v_y^2 = v_{0y}^2 - 2g(y - y_0) \]

Notice how simple the horizontal motion equations become. The horizontal velocity is constant. For vertical motion, these are the same equations you used for free-fall problems, just with \(a_y = -g\).

Problem-Solving Strategy: Projectile Motion

1. Resolve motion into components: Separate initial velocity \(\vec{v}_0\) into \(v_{0x} = v_0\cos\theta_0\) and \(v_{0y} = v_0\sin\theta_0\).
2. Treat as independent 1D motions: Apply appropriate kinematic equations for horizontal (\(a_x=0\)) and vertical (\(a_y=-g\)) motion.
3. Solve for unknowns: Time \(t\) is the only common variable between horizontal and vertical motions. Solve for \(t\) first if possible.
4. Recombine components: If needed, find total displacement \(\vec{s}\) and velocity \(\vec{v}\) using: \(s = \sqrt{x^2+y^2}\), \(\Phi = \tan^{-1}(y/x)\) \(v = \sqrt{v_x^2+v_y^2}\), \(\theta_v = \tan^{-1}(v_y/v_x)\)

This strategy is your roadmap for solving any projectile motion problem. The most common mistake is mixing up horizontal and vertical quantities. Always keep them separate until the final recombination step.

Projectile Motion Components (Visual)

Figure 4.12: (a) Trajectory; (b) Constant horizontal velocity; (c) Vertical motion under gravity; (d) Total velocity vector.

This figure is fantastic for visualizing the independence of motion. Look closely at how the horizontal velocity vectors are all the same length, while the vertical velocity vectors change, becoming zero at the peak and then increasing downwards.

Example: A Fireworks Projectile

A fireworks shell is launched at \(v_0 = 70.0 \, \text{m/s}\) at \(\theta_0 = 75.0^\circ\) above horizontal. The fuse ignites it at its highest point.

1. Calculate the height at which the shell explodes.
   
2. How much time passes until the explosion?
   
3. What is the horizontal displacement when it explodes?
   
4. What is the total displacement from launch to the highest point?
   

Figure 4.13: Trajectory of a fireworks shell.

This problem asks for several quantities related to the apex of the projectile’s trajectory. Remember that at the apex, the vertical velocity component is zero.

Example: Fireworks Projectile (Solution a & b)

Solution (a): Maximum Height (\(h\))

At the highest point (apex), \(v_y = 0\). Using \(v_y^2 = v_{0y}^2 - 2g(y - y_0)\):
\(0^2 = v_{0y}^2 - 2g(h - 0) \implies h = \frac{v_{0y}^2}{2g}\)

Initial vertical velocity:
\(v_{0y} = v_0\sin\theta_0 = (70.0 \, \text{m/s})\sin(75.0^\circ) \approx 67.6 \, \text{m/s}\)

Maximum height:
\(h = \frac{(67.6 \, \text{m/s})^2}{2(9.80 \, \text{m/s}^2)} \approx 233 \, \text{m}\)

Solution (b): Time to Explosion (\(t\))

At the apex, \(v_y = 0\). Using \(v_y = v_{0y} - gt\):
\(0 = v_{0y} - gt \implies t = \frac{v_{0y}}{g}\)
\(t = \frac{67.6 \, \text{m/s}}{9.80 \, \text{m/s}^2} \approx 6.90 \, \text{s}\)

We use the fact that the vertical velocity is zero at the peak to simplify our kinematic equations and solve for height and time. These are standard calculations for projectile motion.

Example: Fireworks Projectile (Solution c & d)

Solution (c): Horizontal Displacement (\(x\))

Horizontal velocity is constant:
\(v_x = v_0\cos\theta_0 = (70.0 \, \text{m/s})\cos(75.0^\circ) \approx 18.1 \, \text{m/s}\)

Horizontal displacement at time \(t\) from (b):
\(x = x_0 + v_xt = 0 + (18.1 \, \text{m/s})(6.90 \, \text{s}) \approx 125 \, \text{m}\)

Solution (d): Total Displacement (\(\vec{s}\))

Displacement vector:
\(\vec{s} = x\hat{i} + y\hat{j} = (125\hat{i} + 233\hat{j}) \, \text{m}\)

Magnitude:
\(|\vec{s}| = \sqrt{(125)^2 + (233)^2} \approx 264 \, \text{m}\)

Direction:
\(\Phi = \tan^{-1}\left(\frac{233}{125}\right) \approx 61.8^\circ\) above horizontal.

For horizontal displacement, we use the constant horizontal velocity and the time calculated in part (b). For total displacement, we combine the calculated horizontal and vertical components as a vector and then find its magnitude and direction.

Time of Flight, Trajectory, and Range (Level Ground)

For a projectile launched and impacting on a flat horizontal surface (\(y=y_0=0\)):

Time of Flight (\(T_{tof}\))

Setting \(y-y_0 = 0\) in \(y - y_0 = v_{0y}t - \frac{1}{2}gt^2\):
\(0 = (v_0\sin\theta_0)t - \frac{1}{2}gt^2\)
\(t(v_0\sin\theta_0 - \frac{1}{2}gt) = 0\)
Two solutions: \(t=0\) (launch) and \(T_{tof} = \frac{2v_0\sin\theta_0}{g}\)

Trajectory Equation (\(y(x)\))

Eliminate \(t\) from \(x = v_{0x}t\) and \(y = v_{0y}t - \frac{1}{2}gt^2\):
\(t = \frac{x}{v_0\cos\theta_0}\)
\(y = (v_0\sin\theta_0)\left(\frac{x}{v_0\cos\theta_0}\right) - \frac{1}{2}g\left(\frac{x}{v_0\cos\theta_0}\right)^2\)
\(y = (\tan\theta_0)x - \left[\frac{g}{2(v_0\cos\theta_0)^2}\right]x^2\) (equation of a parabola)

These formulas are specifically for scenarios where the projectile lands at the same elevation it was launched from. The trajectory equation shows that projectile motion forms a parabolic path.

Time of Flight, Trajectory, and Range (Level Ground)

Range (\(R\))

The horizontal distance traveled when \(y=0\) (at impact):
Setting \(y=0\) in the trajectory equation gives \(x=0\) (launch) and:
\(0 = (\tan\theta_0)R - \left[\frac{g}{2(v_0\cos\theta_0)^2}\right]R^2\)
\(R = \frac{2v_0^2\sin\theta_0\cos\theta_0}{g}\)
Using trigonometric identity \(2\sin\theta\cos\theta = \sin(2\theta)\):
\(R = \frac{v_0^2\sin(2\theta_0)}{g}\)

Important

The maximum range occurs when \(\sin(2\theta_0) = 1\), which means \(2\theta_0 = 90^\circ\), so \(\theta_0 = 45^\circ\).

The range equation is incredibly useful. It shows that for a given initial speed, the maximum range is achieved at a launch angle of 45 degrees. It also demonstrates that two complementary angles (e.g., 30 and 60 degrees) will yield the same range, assuming the same initial speed and level ground.

Projectile Trajectories on Level Ground (Visual)

Figure 4.15: (a) Greater initial speed \(\vec{v}_0\) gives greater range. (b) Range is max at \(45^\circ\). Complementary angles give same range.

Figure (a) shows how initial speed dramatically affects the range and height. Figure (b) visually confirms that 45 degrees gives the maximum range and that angles like 15 and 75 degrees have the same range but different peak heights.

4.4 Uniform and Nonuniform Circular Motion

Learning Objectives

By the end of this section, you will be able to:

• Solve for the centripetal acceleration of an object in a circular path.
• Use circular motion equations to find position, velocity, and acceleration.
• Explain differences between centripetal and tangential acceleration.
• Evaluate total acceleration in nonuniform circular motion.

Even if an object moves at a constant speed in a circle, its velocity vector is constantly changing direction. This change in direction means there must be an acceleration.

Centripetal Acceleration

Even at constant speed, motion along a curved path (like a circle) requires acceleration because the velocity vector’s direction changes.

The magnitude of this acceleration is:

\[ a_c = \frac{v^2}{r} \]

• \(v\) is the speed of the object.
• \(r\) is the radius of the circular path.
• Its direction is always towards the center of the circle (radial direction).
• This is called centripetal acceleration (“center seeking”).

This is a critical concept. Many students initially find it counter-intuitive that an object moving at a constant speed can be accelerating. Emphasize that acceleration is about changing velocity, not just speed. Since direction is part of velocity, changing direction implies acceleration.

Centripetal Acceleration (Visual)

Figure 4.19: The centripetal acceleration vector \(\vec{a}_c\) points toward the center, perpendicular to the velocity vector \(\vec{v}\).

This image clearly shows that the velocity vector is tangent to the circle, while the centripetal acceleration vector points radially inward. They are always perpendicular in uniform circular motion.

Equations of Motion for Uniform Circular Motion

For a particle moving counterclockwise in a circle of radius \(A\):

Position vector: \(\vec{r}(t) = A\cos(\omega t)\hat{i} + A\sin(\omega t)\hat{j}\)

• \(\omega\) is the angular frequency (radians/second).
• \(\omega = \frac{2\pi}{T}\), where \(T\) is the period.

Velocity vector:
\(\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = -A\omega\sin(\omega t)\hat{i} + A\omega\cos(\omega t)\hat{j}\)
Magnitude of velocity (speed): \(v = |\vec{v}(t)| = A\omega\)

Acceleration vector:
\(\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = -A\omega^2\cos(\omega t)\hat{i} - A\omega^2\sin(\omega t)\hat{j}\)
Notice \(\vec{a}(t) = -\omega^2\vec{r}(t)\)

Magnitude of acceleration: \(a_c = |\vec{a}(t)| = A\omega^2 = \frac{v^2}{A}\)

These equations describe uniform circular motion using trigonometry and derivatives. The angular frequency \(\omega\) dictates how quickly the angle changes. The relationship \(\vec{a}(t) = -\omega^2\vec{r}(t)\) explicitly shows that acceleration is opposite to the position vector, confirming it points towards the center.

Nonuniform Circular Motion

If the speed of the particle changes while it moves in a circle, there is an additional acceleration component: tangential acceleration.

• Tangential acceleration (\(a_T\)): The time rate of change of the magnitude of the velocity (speed). \[ a_T = \frac{d|\vec{v}|}{dt} \] Direction is tangent to the circle, in the direction of motion (speeding up) or opposite (slowing down).

• Centripetal acceleration (\(a_c\)): Still \(a_c = \frac{v^2}{r}\), pointing radially inward.

The total acceleration \(\vec{a}\) is the vector sum of centripetal and tangential accelerations:

\[ \vec{a} = \vec{a}_c + \vec{a}_T \]

Since \(\vec{a}_c\) and \(\vec{a}_T\) are perpendicular, the magnitude of the total acceleration is: \(|\vec{a}| = \sqrt{a_c^2 + a_T^2}\)

This is the more general case. Uniform circular motion is a special case where \(a_T = 0\). In nonuniform circular motion, the object is both changing direction and changing speed. The tangential acceleration tells you about the change in speed, while the centripetal acceleration tells you about the change in direction.

Nonuniform Circular Motion (Visual)

Figure 4.22: The centripetal acceleration (\(\vec{a}_c\)) is radial, tangential acceleration (\(\vec{a}_T\)) is tangential. Total acceleration (\(\vec{a}\)) is their vector sum.

This diagram clearly shows the two perpendicular components of acceleration. The total acceleration vector \(\vec{a}\) will always point inside the curve of the path, but not necessarily directly towards the center if there is tangential acceleration.

4.5 Relative Motion in One and Two Dimensions

Learning Objectives

By the end of this section, you will be able to:

• Explain the concept of reference frames.
• Write position and velocity vector equations for relative motion.
• Draw position and velocity vectors for relative motion.
• Analyze 1D and 2D relative motion problems.

Up until now, we’ve implicitly assumed Earth is our primary reference frame. But what happens when we observe motion from a moving reference frame? This section introduces the tools to handle such situations.

Reference Frames

All motion is described relative to a reference frame.

• When we say a velocity, we implicitly mean velocity with respect to a specific reference frame (e.g., Earth).
• To analyze relative motion, we must define the reference frames involved.

Example: A person walks on a moving train. - Velocity of person relative to the train (\(\vec{v}_{PT}\)) - Velocity of train relative to Earth (\(\vec{v}_{TE}\)) - Velocity of person relative to Earth (\(\vec{v}_{PE}\))

It’s crucial to be explicit about which reference frame you are using. For instance, a person might be stationary relative to the train, but moving at 100 km/h relative to the ground.

Relative Velocity in One Dimension

For a particle \(P\) and two reference frames \(S\) and \(S'\):

Position of \(P\) relative to \(S\): \(\vec{r}_{PS}\)
Position of \(P\) relative to \(S'\): \(\vec{r}_{PS'}\)
Position of \(S'\) relative to \(S\): \(\vec{r}_{S'S}\)

From vector addition (as seen in Figure 4.26, which is omitted for brevity but conceptually links the origins):

\[ \vec{r}_{PS} = \vec{r}_{PS'} + \vec{r}_{S'S} \]

Taking the time derivative, we get the relative velocity equation:

\[ \vec{v}_{PS} = \vec{v}_{PS'} + \vec{v}_{S'S} \]

• The velocity of particle \(P\) relative to frame \(S\) equals its velocity relative to frame \(S'\) plus the velocity of frame \(S'\) relative to frame \(S\).
• Subscript rule: The inner subscripts match (\(\text{S'}\) and \(\text{S'}\)), and the outer subscripts form the left-hand side (\(\text{P}\) and \(\text{S}\)).

This formula is the core of relative motion. The “coupling reference frame” (S’ in this case) acts as an intermediary. Make sure students understand the subscript notation, as it’s key to setting up problems correctly.

Relative Velocity Subscript Rule (Visual)

Figure 4.24: Subscript ordering for relative velocity vector equation.

This visual emphasizes the mnemonic for setting up relative velocity equations. The “inner” subscripts cancel out, leaving the “outer” subscripts for the resultant vector.

Relative Acceleration

Differentiating the relative velocity equation gives the relative acceleration equation:

\[ \vec{a}_{PS} = \vec{a}_{PS'} + \vec{a}_{S'S} \]

If the relative velocity between the two reference frames (\(S'\) and \(S\)) is constant (\(\vec{v}_{S'S} = \text{constant}\)), then \(\vec{a}_{S'S} = 0\).

In this case:

\[ \vec{a}_{PS} = \vec{a}_{PS'} \]

• The acceleration of a particle is the same as measured by two observers moving at a constant velocity relative to each other.

This means that if you’re in a car moving at a constant speed, an object accelerating inside your car will appear to accelerate the same way to an observer standing still outside (assuming both are inertial frames). This principle is fundamental to Newtonian mechanics.

Example: Motion of a Car Relative to a Truck

A truck is traveling south at \(70 \, \text{km/h}\) toward an intersection.
A car is traveling east at \(80 \, \text{km/h}\) toward the intersection.
What is the velocity of the car relative to the truck?

Figure 4.27: Car and truck approaching an intersection.

This is a 2D relative velocity problem. We need to express velocities in vector form with respect to a common ground reference frame and then use the relative velocity formula.

Example: Car Relative to Truck (Solution)

Strategy

1. Define Earth as the common reference frame (E).
2. Write velocities of car (C) and truck (T) with respect to Earth.
3. Use relative velocity equation: \(\vec{v}_{CT} = \vec{v}_{CE} + \vec{v}_{ET}\).

Solution

• Velocity of car with respect to Earth: \(\vec{v}_{CE} = 80 \, \text{km/h} \, \hat{i}\) (east)
• Velocity of truck with respect to Earth: \(\vec{v}_{TE} = -70 \, \text{km/h} \, \hat{j}\) (south)

We need \(\vec{v}_{ET}\), which is the negative of \(\vec{v}_{TE}\):
\(\vec{v}_{ET} = -\vec{v}_{TE} = 70 \, \text{km/h} \, \hat{j}\) (north)

Now, apply the relative velocity equation:
\(\vec{v}_{CT} = \vec{v}_{CE} + \vec{v}_{ET} = (80 \, \text{km/h})\hat{i} + (70 \, \text{km/h})\hat{j}\)

Magnitude:
\(|\vec{v}_{CT}| = \sqrt{(80)^2 + (70)^2} = \sqrt{6400 + 4900} = \sqrt{11300} \approx 106 \, \text{km/h}\)

Direction:
\(\theta = \tan^{-1}\left(\frac{70}{80}\right) \approx 41.2^\circ\) north of east.

The diagram in the next slide will help visualize this vector addition. It’s crucial to correctly identify \(\vec{v}_{ET}\) as the velocity of Earth relative to the truck, which is opposite to the truck’s velocity relative to Earth.

Example: Car Relative to Truck (Visual)

Figure 4.28: Vector diagram for \(\vec{v}_{CT} = \vec{v}_{CE} + \vec{v}_{ET}\).

This diagram makes the vector addition clear. The car’s velocity relative to the truck is the hypotenuse of a right triangle, composed of the car’s velocity relative to Earth and Earth’s velocity relative to the truck.

Key Takeaways

Displacement and Velocity

• Position Vector: \(\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k}\)
• Displacement Vector: \(\Delta\vec{r} = \vec{r}(t_2) - \vec{r}(t_1)\) (shortest path from initial to final position)
• Instantaneous Velocity: \(\vec{v}(t) = \frac{d\vec{r}}{dt}\) (tangent to path)
• Average Velocity: \(\vec{v}_{avg} = \frac{\Delta\vec{r}}{\Delta t}\)
• Independence of Motion: Motions along perpendicular axes are independent.

Key Takeaways

Acceleration

• Instantaneous Acceleration: \(\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d^2\vec{r}}{dt^2}\)
• Constant Acceleration: Kinematic equations apply independently to each component.

Key Takeaways

Projectile Motion (negligible air resistance, constant \(g\))

• Horizontal: \(a_x = 0\), \(v_x = \text{constant}\)
• Vertical: \(a_y = -g\)
• Max Height (level ground): \(h = \frac{v_{0y}^2}{2g}\)
• Time of Flight (level ground): \(T_{tof} = \frac{2v_0\sin\theta_0}{g}\)
• Range (level ground): \(R = \frac{v_0^2\sin(2\theta_0)}{g}\) (max at \(\theta_0 = 45^\circ\))
• Trajectory: Parabolic path \(y = (\tan\theta_0)x - \left[\frac{g}{2(v_0\cos\theta_0)^2}\right]x^2\)

Key Takeaways

Circular Motion

• Centripetal Acceleration: \(a_c = \frac{v^2}{r}\) (always directed towards center, due to change in direction of velocity)
• Nonuniform Circular Motion: Total acceleration \(\vec{a} = \vec{a}_c + \vec{a}_T\)
  • Tangential Acceleration: \(a_T = \frac{d|\vec{v}|}{dt}\) (due to change in speed, tangent to path)

Key Takeaways

Relative Motion

• Relative Velocity: \(\vec{v}_{PS} = \vec{v}_{PS'} + \vec{v}_{S'S}\) (velocity of P relative to S is P relative to S’ plus S’ relative to S)
• Relative Acceleration: If frames move at constant velocity, accelerations are the same: \(\vec{a}_{PS} = \vec{a}_{PS'}\)

Key Equations

Equation	Description
\(\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k}\)	Position vector
\(\Delta\vec{r} = \vec{r}(t_2) - \vec{r}(t_1)\)	Displacement avector
\(\vec{v}(t) = \frac{d\vec{r}}{dt}\)	Instantaneous velocity
\(\vec{a}(t) = \frac{d\vec{v}}{dt}\)	Instantaneous acceleration
\(v_x(t) = v_{0x} + a_xt\) , \(x(t) = x_0 + v_{0x}t + \frac{1}{2}a_xt^2\)	1D kinematic equations (for each component)
\(T_{tof} = \frac{2v_0\sin\theta_0}{g}\)	Time of flight (level ground)
\(h = \frac{v_{0y}^2}{2g}\)	Maximum height (level ground)
\(R = \frac{v_0^2\sin(2\theta_0)}{g}\)	Range (level ground)
\(a_c = \frac{v^2}{r}\)	Centripetal acceleration (magnitude)
\(a_T = \frac{d|\vec{v}|}{dt}\)	Tangential acceleration (magnitude)
\(\vec{a} = \vec{a}_c + \vec{a}_T\)	Total acceleration (nonuniform circular motion)
\(\vec{v}_{PS} = \vec{v}_{PS'} + \vec{v}_{S'S}\)	Relative velocity between frames

Key Terms

Term	Definition
Displacement Vector	The straight-line vector from an object’s initial position to its final position.
Position Vector	A vector from the origin of a coordinate system to an object’s location.
Velocity Vector	The vector quantity that describes an object’s speed and direction of motion.
Acceleration Vector	The vector quantity that describes the rate of change of an object’s velocity.
Projectile Motion	Motion of an object thrown into the air, subject only to gravity (negligible air resistance).
Trajectory	The path followed by a projectile.
Range	The horizontal distance traveled by a projectile launched and impacting on level ground.
Time of Flight	The total time a projectile is in the air.
Uniform Circular Motion	Motion in a circular path at a constant speed.
Centripetal Acceleration	The acceleration of an object moving in a circle, always directed toward the center of the circle.
Nonuniform Circular Motion	Motion in a circular path where the speed is changing.
Tangential Acceleration	The component of acceleration tangent to a circular path, indicating a change in the object’s speed.
Reference Frame	A coordinate system from which physical observations and measurements are made.
Relative Velocity	The velocity of an object as observed from a particular reference frame.