1.3 Representing Motion

Representing motion is all about describing how objects move through space and time. We use various tools like diagrams, graphs, and equations to capture the essence of an object's movement.

Understanding motion representation helps us analyze real-world situations and predict outcomes. From simple motion diagrams to complex kinematic equations, these concepts form the foundation for studying more advanced physics topics.

Motion representations

Motion diagrams and descriptions

• Represent motion using various methods like motion diagrams, figures, graphs, equations, and written descriptions 📊
• Motion diagrams depict an object's position at different moments in time, providing a visual representation of its movement
• Figures, such as free-body diagrams, illustrate the forces acting on an object and help analyze its motion
• Graphs show the relationship between kinematic quantities (position, velocity, acceleration) and time, offering insights into an object's behavior
• Equations mathematically describe the motion of an object, relating kinematic quantities and allowing for quantitative analysis
• Narrative descriptions verbally explain an object's motion, often providing context and qualitative information

Kinematic equations

When an object experiences constant acceleration, three fundamental equations can describe its linear motion in one dimension. These equations relate position, velocity, acceleration, and time.

$V_x = V_{xo} + a_xt$

This first equation relates final velocity to initial velocity and acceleration over time. It shows how velocity changes linearly with time under constant acceleration.

$x = x_o + v_{xo}t + \frac{1}{2} a_xt^2$

The second equation describes position as a function of time, showing how an object's position changes due to both its initial velocity and acceleration.

$v^2 = v_{o}^2 + 2a_x (x - x_o)$

The third equation relates velocity directly to position, eliminating time as a variable. This is particularly useful when time information is not available.

These equations apply to motion in any single direction - just substitute the appropriate variables for different dimensions (y for vertical motion, etc.). Remember that these equations only work for constant acceleration scenarios.

Variable Interpretation: Δx is horizontal displacement in meters, Vf is final velocity in meters/second, Vo is initial velocity in meters/second, t is time in seconds, and a is acceleration in m/s/s.

When solving problems, identify which variables you know and which you need to find. Then select the equation that doesn't contain any unknown variables except the one you're solving for.

Acceleration due to gravity

Near Earth's surface, all objects experience a constant downward acceleration due to gravity, regardless of their mass.

$a_g = g = 10 \, \text{m/s}^2$

This value is a simplification used in AP Physics 1, though more precise values ($g = 9.81 \, \text{m/s}^2$ or $g = 9.8 \, \text{m/s}^2$) are also acceptable. This acceleration acts only in the vertical direction and remains constant for objects near Earth's surface.

When solving problems involving falling objects, you can apply the kinematic equations using g as the acceleration value. For objects thrown upward, the acceleration due to gravity still points downward (negative if upward is positive).

Motion graphs

Motion graphs provide visual representations of how position, velocity, and acceleration change over time. These graphs reveal important relationships and allow us to analyze motion patterns at a glance.

Position-time graphs show where an object is at each moment:

• The slope at any point equals instantaneous velocity
• Positive slope means motion in positive direction
• Steeper slopes indicate faster movement
• Horizontal sections mean the object is momentarily stopped

Velocity-time graphs show how fast an object is moving at each moment:

• The slope at any point equals instantaneous acceleration
• Positive slope means the object is speeding up
• Negative slope means the object is slowing down
• The area under the curve equals displacement during that time interval

Acceleration-time graphs show how the rate of velocity change varies:

• The area under the curve equals the change in velocity during that time interval
• Constant acceleration appears as a horizontal line
• Zero acceleration (constant velocity) appears as a line along the time axis

Practice Problem 1: Kinematic Equations

Solution

Let's break this into three phases:

1. Acceleration phase (0-8s)
2. Constant velocity phase (8-20s)
3. Deceleration phase (20-? s)

For phase 1 (acceleration):

• Initial velocity $v_0 = 0$ m/s
• Acceleration $a = 3$ m/s²
• Time $t = 8$ s
• Using $x = x_0 + v_0t + \frac{1}{2}at^2$
• $x_1 = 0 + 0 + \frac{1}{2}(3)(8)^2 = 96$ m
• Final velocity $v_1 = v_0 + at = 0 + 3(8) = 24$ m/s

For phase 2 (constant velocity):

• Velocity $v = 24$ m/s
• Time $t = 12$ s
• Distance $x_2 = v \cdot t = 24 \cdot 12 = 288$ m

For phase 3 (deceleration):

• Initial velocity $v_0 = 24$ m/s
• Final velocity $v_f = 0$ m/s
• Acceleration $a = -2$ m/s²
• Using $v_f^2 = v_0^2 + 2a\Delta x$
• $0 = 24^2 + 2(-2)\Delta x$
• $\Delta x = \frac{24^2}{4} = 144$ m

Total distance = $x_1 + x_2 + x_3 = 96 + 288 + 144 = 528$ m

Practice Problem 2: Motion Graphs

Solution

a) To sketch the velocity-time graph, we need to find the velocity in each segment:

Segment 1 (0-2s):

• Slope of position-time graph = velocity
• Slope = (8m - 0m)/(2s - 0s) = 4 m/s
• Velocity is constant at 4 m/s

Segment 2 (2-4s):

• Horizontal line means zero velocity
• Velocity is constant at 0 m/s

Segment 3 (4-6s):

• Slope = (0m - 8m)/(6s - 4s) = -4 m/s
• Velocity is constant at -4 m/s

The velocity-time graph would show:

• A horizontal line at 4 m/s from t=0s to t=2s
• A horizontal line at 0 m/s from t=2s to t=4s
• A horizontal line at -4 m/s from t=4s to t=6s

b) To calculate the average acceleration during t=4s to t=6s:

Average acceleration = change in velocity / change in time

• Initial velocity at t=4s: 0 m/s
• Final velocity at t=6s: -4 m/s
• Time interval: 2s

Average acceleration = (-4 m/s - 0 m/s) / (2s) = -2 m/s²

The negative sign indicates the acceleration is in the negative direction.

Practice Problem 3: Acceleration Due to Gravity

Solution

a) To find the maximum height: At maximum height, the velocity becomes zero. We can use the equation: $v_f^2 = v_i^2 + 2a\Delta y$

Where:

• $v_f = 0$ m/s (at maximum height)
• $v_i = 30$ m/s
• $a = -10$ m/s² (negative because gravity acts downward)
• $\Delta y$ is the maximum height

Substituting: $0 = 30^2 + 2(-10)\Delta y$ $20\Delta y = 900$ $\Delta y = 45$ m

The maximum height reached is 45 meters.

b) To find the total time in the air: The motion consists of two parts: going up and coming down.

Time to reach maximum height: Using $v_f = v_i + at$ $0 = 30 + (-10)t$ $t = 3$ seconds

Since the acceleration is constant and the initial and final positions are the same (ground level), the time to come down equals the time to go up.

Total time = 3 + 3 = 6 seconds

c) To find the velocity just before hitting the ground: Due to symmetry of the motion, the stone will hit the ground with the same speed as it was thrown up, but in the opposite direction.

Velocity just before hitting the ground = -30 m/s

The negative sign indicates the stone is moving downward.