Chapter 2: Describing Motion - Kinematics in One Dimension

Describing Motion: Kinematics in One Dimension

Descriptors of Motion

  • HOW FAR: Magnitude only matters (Distance - Scalar).

  • HOW FAST: Direction and Magnitude matter (Velocity - Vector).

  • HOW FASTER: "HOW FAST" Changing (Acceleration).

2-1 Reference Frames and Displacement

  • Any measurement of position, distance, or speed must be made with respect to a reference frame.

  • Displacement: How far the object is from its starting point, regardless of the path taken (VECTOR quantity).

  • Distance Traveled: Measured along the actual path (SCALAR quantity).

Displacement Formula:

  • \Delta x = x2 - x1

    • Where \Delta x is the displacement, x2 is the final position, and x1 is the initial position.

Distance Formula:

  • X{total} = |\Delta x1| + |\Delta x_2|

Example:

  • If an object moves 70 m East and then 30 m West:

    • X_{total} = 70 m + 30 m = 100 m

    • \Delta x = 40 m (East)

Concept Tests:

  • Walking the Dog: You and your dog have the same displacement if you start and end at the same positions, but the dog travels a greater distance due to side trips.

  • Odometer: Measures distance, not displacement. If you return home, the odometer records the total miles traveled, not zero.

2-2 Average Velocity

  • Speed: How far an object travels in a given time interval (SCALAR).

  • Velocity: Includes directional information (VECTOR).

  • Average Velocity: \text{average velocity} = \frac{\text{displacement}}{\text{time elapsed}} = \frac{\Delta x}{\Delta t}

2-3 Instantaneous Velocity

  • The instantaneous velocity is the average velocity, in the limit as the time interval becomes infinitesimally short: v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}

Concept Test:

  • If the average velocity is non-zero over some time interval, it doesn't necessarily mean the instantaneous velocity is never zero during that interval. Example: stopping for lunch during a trip.

2-4 Acceleration

  • Acceleration: The rate of change of velocity.

  • Average Acceleration: \text{average acceleration} = \frac{\text{change of velocity}}{\text{time elapsed}} = \frac{\Delta v}{\Delta t}

Positive vs. Negative Acceleration:

  • Negative Acceleration: Acceleration in the negative direction.

  • Deceleration: Occurs when acceleration is opposite in direction to the velocity.

Example:

  • A car traveling in the +x direction at 18 m/s slows to 8 m/s in 5 s.

    • Change in speed: -10 m/s (indicates a decrease).

    • Change in velocity: -10 m/s (a vector, pointing in the negative x-direction).

    • Acceleration: \frac{-10 m/s}{5 s} = -2 m/s^2

Instantaneous Acceleration:

  • The instantaneous acceleration is the average acceleration, in the limit as the time interval becomes infinitesimally short: a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t}

Our Motion Descriptors

SCALAR

VECTOR

X_{tot} (Distance)

\Delta x (Displacement)

Time Rate of Change

Speed

Velocity

Time Rate of Change

Acceleration

Acceleration

2-5 Motion at Constant Acceleration

  • Average velocity: \bar{v} = \frac{x - x_0}{t}

  • Constant acceleration: a = \frac{v - v_0}{t}

Equations of Motion (Constant Acceleration):

  • v = v_0 + at

  • x = x0 + v0t + \frac{1}{2}at^2

  • v^2 = v0^2 + 2a(x - x0)

2-6 Solving Problems

  1. Read the problem carefully.

  2. Identify the object(s) and time interval.

  3. Draw a diagram and choose coordinate axes.

  4. List known and unknown quantities.

  5. Determine the applicable physics and choose appropriate equations.

  6. Solve algebraically and check dimensions.

  7. Calculate the solution with significant figures.

  8. Assess if the result is reasonable.

  9. Check the units.

Example 1:

  • A car travels at 15 m/s and decreases its speed by 2 m/s each second. How far will it go before stopping, and how much time will it take?

Example 2:

  • Car A travels at a constant 15 m/s. Car B starts from rest 20 m behind Car A and accelerates at -1.5 m/s². How far from Car A's starting position do they collide?

    • Collision Condition: xA = xB

2-7 Falling Objects

  • Near the Earth's surface, all objects experience approximately the same acceleration due to gravity, assuming negligible air resistance.

  • Acceleration due to gravity: g = 9.80 m/s^2 (downwards).

Equations for Falling Objects:

  • v = v_0 - gt

  • y = y0 + v0t - \frac{1}{2}gt^2

  • v^2 = v0^2 - 2g(y - y0)

Examples:

  • A banana thrown upwards with an initial speed of 12 m/s reaches its peak in 1.2 s.

Key Discoveries:

  • At the peak, v = 0.

  • Acceleration at the peak is -9.80 m/s^2.

  • What goes up must come down in equal amounts of time.

  • Launch Speed = Landing Speed (if initial and final heights are the same).

Concept Test:

  • When throwing a ball straight up, at the highest point, v = 0, but a \neq 0.

Additional Example:

  • A plantain dropped past a window takes 0.30 seconds to pass the 1.5 m long window.

    • The speed of the plantain as it first passes the top of the window.

    • The speed as it passes the bottom of the window.

    • The height above the top of the window Jorge dropped it from.

2-8 Graphical Analysis of Linear Motion

  • The slope of an x vs. t curve is equal to velocity.

  • The instantaneous velocity is tangent to the x vs. t curve at each point.

  • The area beneath a v vs. t curve is equal to displacement.

  • The slope of a v vs. t graph is equal to acceleration.

  • Acceleration and Displacement are vectors, whose directions are given by their signs (+ / –) in 1D problems.

  • The y-intercept of a v vs. t graph is initial velocity \v_0.

Example:

  • What is the average velocity for the whole trip? Displacement between t = 0 and t = 30.0 s:

Mother eagle and eaglet example

  • A mother eagle is training her baby eaglet to fly. She drops the eaglet from a 30.0 m high nest, and once the eaglet has fallen 14.5 m, she drops from the nest to catch the eaglet when it is 1.0 m above the ground.
    a) Determine the velocity of the eaglet when it has fallen 14.5 m.
    b) Determine the time it takes the eaglet to fall the additional 14.5 m, at which point it will be exactly 1.0 m from the ground.
    c) Determine the acceleration that mother eagle must have in order to catch the eaglet 1.0 m from the ground, as described in the problem above.

Summary of Chapter 2

  • Kinematics describes how objects move with respect to a reference frame.

  • Displacement is the change in position.

  • Average speed is distance traveled divided by time; average velocity is displacement divided by time.

  • Instantaneous velocity is the limit as the time interval becomes infinitesimally short.

  • Average acceleration is the change in velocity divided by time.

  • Instantaneous acceleration is the limit as the time interval becomes infinitesimally small.

  • Equations of motion for constant acceleration require different sets of quantities.

  • Objects falling near Earth experience a gravitational acceleration of g = -9.80 m/s^2.