Chapter 4 - Motion in Two and Three Dimensions

Chapter 4: Motion in Two and Three Dimensions

Main Ideas

• Two- and three-dimensional motion can be treated as separate one-dimensional problems. Analyze motion independently along the (x), (y), and, when needed, (z) axes. Apply the familiar one-dimensional kinematic equations to each axis.

• Position, velocity, and acceleration are vectors. In two dimensions, for example, [ \vec v=v_x\hat i+v_y\hat j,\qquad |\vec v|=\sqrt{v_x^2+v_y^2}. ] In three dimensions, add the corresponding (z)-component and calculate the magnitude using all three squared components.

• Time links the directions. The elapsed time is the same for motion along each axis, even though the position, velocity, and acceleration components can differ.

• A change in direction is a change in velocity. Therefore, an object can accelerate even if its speed stays constant.

Projectile Motion: Two-Column Method

For an object launched at an angle, separate the knowns and unknowns into (x)- and (y)-direction columns, then use the one-dimensional equations in each column.

1. Resolve the Initial Velocity into Components

For launch speed (v_0) at angle (\theta) above the horizontal:

[ v_{0x}=v_0\cos\theta,\qquad v_{0y}=v_0\sin\theta. ]

2. Identify the Acceleration in Each Direction

For the idealized projectile motion discussed in the video, air resistance is ignored:

• (a_x=0): no horizontal acceleration.
• (a_y=-9.8\ \text{m/s}^2): gravity acts downward, taking upward as positive.

3. Use the Appropriate One-Dimensional Equations

Useful forms include:

[ \Delta x=v_{0x}\Delta t+\tfrac12a_x(\Delta t)^2, ]

[ v_f=v_0+a\Delta t, ]

[ a\Delta x=\tfrac12(v_f^2-v_0^2). ]

Apply these separately to the horizontal and vertical components. Choose an equation based on which quantities are known and which are sought.

4. Use the Apex to Find Maximum Height or Ascent Time

At the highest point, the vertical velocity is zero, although the horizontal velocity generally is not. Use (v_{fy}=0) in a vertical-motion equation to find the maximum height or the time to reach the apex.

5. Find Total Flight Time and Range

If the projectile lands at the same height from which it was launched, its ascent and descent times are equal. Double the time to reach the apex to obtain the total flight time. Then use that time in the horizontal-motion equation to find the range.

Worked Example: Discus Throw

The lecturer applies this method to a discus thrown at (26\ \text{m/s}) at (45^\circ), assuming it lands at its launch height and neglecting air resistance.

• Components of the initial velocity: [ v_{0x}=26\cos45^\circ\approx18.4\ \text{m/s}, \qquad v_{0y}=26\sin45^\circ\approx18.4\ \text{m/s}. ]

• Maximum height: At the apex, (v_{fy}=0). Using the vertical kinematic equation gives a height of about 17.3 m.

• Time to apex: Using the vertical velocity equation gives about 1.88 s.
• Total flight time: By symmetry, the total is approximately (2(1.88)=) 3.76 s.
• Horizontal range: With (a_x=0), the range is (v_{0x}\Delta t), or about 69.2 m.

The velocity immediately before impact is not zero; the object has not yet stopped. At the apex, only its vertical velocity is zero.

Uniform Circular Motion

• In uniform circular motion, an object’s speed can stay constant while its velocity changes direction.
• The velocity at any point is tangent to the circle.
• Because the velocity changes, the object accelerates. The acceleration is centripetal, directed toward the circle’s center, with magnitude [ a_c=\frac{v^2}{r}, ] where (v) is speed and (r) is the circle’s radius.

Relative Motion and Galilean Addition of Velocities

When objects are described from different reference frames, their velocities are related by vector addition:

[ \vec v=\vec v’+\vec V_0. ]

Here:

• (\vec v) is the object’s velocity relative to the stationary frame.
• (\vec v’) is its velocity relative to the moving frame.
• (\vec V_0) is the velocity of the moving frame relative to the stationary frame.

The boat-and-current example illustrates the rule: a boat’s velocity relative to the shore combines its velocity relative to the water with the water’s velocity relative to the shore. For the example’s perpendicular components—(18\ \text{km/h}) across the water and (4.8\ \text{km/h}) with the current—the resultant speed is about 18.6 km/h, at about 15° from the boat’s heading.

The video ends by posing a related question: where should the boat be steered to reach the point directly opposite its starting position?

Speakers and Sources Featured

• Speaker: One instructor/narrator, explaining the chapter and worked examples.
• Sources mentioned: The course textbook, Physics for Engineers and Scientists by Ohanian and Markery, including its Chapter 4 material and Example 7 (discus throw). The instructor also refers to Example 1 and Review Task 4.1.