Comprehensive Study Guide: Acceleration and Nonuniform Motion

Fundamentals of Motion and Nonuniform Motion Diagrams

Uniform Motion:
- An object in uniform motion moves along a straight line with an unchanging velocity.
- Uniform motion feels smooth; if an observer closes their eyes, it feels as if there is no motion at all.
Nonuniform Motion:
- Motion in which an object's velocity is changing over time.
- Nonuniform motion causes a feeling of being pushed or pulled (e.g., moving around a sharp curve or traveling up and down a roller coaster hill).
- Examples of Nonuniform Motion along a Straight Line:
- A ball rolling down a steep hill.
- A car applying brakes to come to a stop.
- Objects in freefall dropping vertically under gravity.
- Examples of Nonuniform Motion Non-Confined to a Straight Line:
- Motion along a curved or circular path.
- The parabolic path of thrown objects, such as baseballs.
Motion Diagrams for Nonuniform Motion:
- Motionless: Represented by a single image, indicating zero velocity (e.g., a runner waiting at the starting line for a race signal).
- Uniform Motion: Consecutive positions of the object are separated by equal distances, indicating constant velocity.
- Speeding Up: Distances between successive positions increase with each time interval.
- Slowing Down: Distances between successive positions decrease with each time interval.

The distance the jogger moves in each time interval indicates the type of motion.

Particle Model Motion Diagrams and Acceleration Vectors

Indicators of Velocity Change in Particle Models:
1. Dot Spacing: Increasing distance between consecutive dots indicates speeding up; decreasing distance between consecutive dots indicates slowing down.
2. Vector Length: The lengths of velocity vectors directly correspond to speed. Longer subsequent vectors indicate speeding up, whereas shorter subsequent vectors indicate slowing down.
Acceleration ( $\mathbf{a}$ ):
- Defined as the rate at which an object's velocity changes over time.
- Including acceleration vectors on a motion diagram provides a complete description of an object's changing motion.
Determining and Drawing Acceleration Vectors:
- For constant acceleration, the length and direction of an acceleration vector are calculated by finding the change in velocity ( $\Delta v$ ) over a specific time interval ( $\Delta t$ ): $\Delta v = v_f - v_i = v_f + (-v_i)$
- Here, $v_i$ represents the initial velocity vector at the beginning of the time interval, and $v_f$ represents the final velocity vector at the end of the interval.
- Procedure to Draw Acceleration Vector:
1. Draw the final velocity vector $v_f$ .
2. Draw the negative initial velocity vector $-v_i$ with its tail placed directly at the tip of $v_f$
3. Draw the velocity change vector $\Delta v$ starting from the tail of $v_f$ to the tip of $-v_i$
4. The acceleration vector $\mathbf{a}$ points in the exact same direction as $\Delta v$ and has a magnitude equal to $\frac{\Delta v}{\Delta t}$ .
- Color Conventions in Diagrams:
- Velocity vectors ( $v$ ) are drawn in red.
- Acceleration vectors ( $a$ ) are drawn in violet.
Directional Relationship Between Velocity and Acceleration:
- Speeding Up: The velocity vector ( $v$ ) and acceleration vector ( $a$ ) point in the same direction.
- Slowing Down: The velocity vector ( $v$ ) and acceleration vector ( $a$ ) point in opposite directions.
- Four Motion Scenarios (assuming time interval $\Delta t = 1.0\,\text{s}$ ):
1. Moving in the positive direction and speeding up: Positive velocity ( $+v$ ), Positive acceleration ( $+a$ ).
2. Moving in the positive direction and slowing down: Positive velocity ( $+v$ ), Negative acceleration ( $-a$ ).
3. Moving in the negative direction and speeding up: Negative velocity ( $-v$ ), Negative acceleration ( $-a$ ).
4. Moving in the negative direction and slowing down: Negative velocity ( $-v$ ), Positive acceleration ( $+a$ ).
Mathematical Sign Caveat:
- An object has positive acceleration when its acceleration vector points in the positive direction and negative acceleration when it points in the negative direction.
- The algebraic sign of acceleration alone does not indicate whether an object is speeding up or slowing down. Both the direction of velocity and the direction of acceleration are required to determine whether speed is increasing or decreasing.

Velocity-Time Graphs

Plotting Velocity-Time Graphs ( $v-t$ Graphs):
- Velocity ( $v$ ) is plotted on the vertical axis ( $y$ -axis).
- Time ( $t$ ) is plotted on the horizontal axis ( $x$ -axis).
Slope of a Velocity-Time Graph:
- The slope of a line on a velocity-time graph equals the acceleration of the object: $\text{Slope} = \frac{\text{rise}}{\text{run}} = \frac{\Delta v}{\Delta t} = a$
- Constant Acceleration: A straight line on a $v-t$ graph indicates that velocity changes at a constant rate over time.
- Example Analysis: A car starts from rest and accelerates linearly. Between $t = 4.00\,\text{s}$ ( $v = 20.0\,\text{m/s}$ ) and $t = 5.00\,\text{s}$ ( $v = 25.0\,\text{m/s}$ ): $a = \frac{25.0\,\text{m/s} - 20.0\,\text{m/s}}{5.00\,\text{s} - 4.00\,\text{s}} = \frac{5.0\,\text{m/s}}{1.00\,\text{s}} = 5.00\,\text{m/s}^2$
Interpreting Five Runner Motions on a $v-t$ Graph (East Defined as Positive):
- Graph A: Horizontal line above the x-axis. Slope is zero ( $a = 0\,\text{m/s}^2$ ). Motion at constant velocity toward the East.
- Graph B: Slanted upward above the x-axis. Positive slope ( $+a$ ) and positive velocity ( $+v$ ). Runner moves East while speeding up.
- Graph C: Slanted downward starting above the x-axis. Negative slope ( $-a$ ) with positive velocity ( $+v$ ). Runner moves East while slowing down to a stop.
- Intersection Point of Graphs B and C: Indicates the exact time at which both runners have identical velocities. It does not provide position information.
- Graph D: Starts below the x-axis (moving West) slanting upward with a positive slope ( $+a$ ). Runner initially moves West while slowing down, comes to a complete stop for an instant where the line crosses the x-axis ( $v = 0\,\text{m/s}$ ), and then moves East with increasing speed ( $+v$ , $+a$ ).
- Graph E: Horizontal line below the x-axis. Slope is zero ( $a = 0\,\text{m/s}^2$ ). Motion at constant velocity toward the West.
Meaning of Crossing the x-axis:
- When a line on a $v-t$ graph crosses the horizontal x-axis, the object changes its direction of motion.
Nonlinear Velocity-Time Graphs:
- A curved line on a velocity-time graph indicates changing (non-constant) acceleration.

A curved line on a velocity-time graph shows that acceleration is changing.

Average vs. Instantaneous Acceleration

Average Acceleration ( $\bar{a}$ ):
- Defined as the change in velocity during a measurable time interval divided by that time interval: $\bar{a} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i}$
- Measured in units of meters per second per second ( $\text{m/s/s}$ ) or meters per second squared ( $\text{m/s}^2$ ).
- Depends strictly on the initial and final velocities during the time interval, regardless of speed variations within that interval.
Instantaneous Acceleration ( $a$ ):
- Defined as the change in an object's velocity at a specific instant of time.
- Determined by drawing a line tangent to the curve on a $v-t$ graph at the specified time and measuring the tangent line's slope.
- In ideal cases where acceleration is constant throughout an interval, the average acceleration equals the instantaneous acceleration at every point.
Three Modes of Acceleration:
1. Increasing speed.
2. Decreasing speed.
3. Changing direction of motion (even if speed remains constant).

Quantitative Worked Examples

Example Problem 1: Sprinter Motion and Tangent Lines
- Context: A sprinter accelerates rapidly from rest, reaching near maximum speed around $10.0\,\text{m/s}$ .
- Instantaneous Acceleration at $t = 1.00\,\text{s}$ :
- Tangent line points selected: $(1.00\,\text{s}, 6.00\,\text{m/s})$ and $(2.4\,\text{s}, 10.0\,\text{m/s})$ .
- $a = \frac{\text{rise}}{\text{run}} = \frac{10.0\,\text{m/s} - 6.00\,\text{m/s}}{2.4\,\text{s} - 1.00\,\text{s}} = \frac{4.00\,\text{m/s}}{1.4\,\text{s}} = 2.9\,\text{m/s}^2$
- Instantaneous Acceleration at $t = 5.00\,\text{s}$ :
- Tangent line points selected: $(0.00\,\text{s}, 10.0\,\text{m/s})$ and $(10.0\,\text{s}, 10.3\,\text{m/s})$ .
- $a = \frac{\text{rise}}{\text{run}} = \frac{10.3\,\text{m/s} - 10.0\,\text{m/s}}{10.0\,\text{s} - 0.00\,\text{s}} = \frac{0.30\,\text{m/s}}{10.0\,\text{s}} = 0.030\,\text{m/s}^2$
- Conclusion: Acceleration is variable because its magnitude decreases from $2.9\,\text{m/s}^2$ at $1.00\,\text{s}$ to $0.030\,\text{m/s}^2$ at $5.00\,\text{s}$ .
Example Problem 2: Ball Rolling Up a Slanted Driveway
- Context: A ball rolls up a slanted driveway starting at $v_i = +2.50\,\text{m/s}$ . Uphill is designated as the positive direction, with origin at the starting point. The ball slows down for $5.00\,\text{s}$ , stops instantaneously ( $v_f = 0.00\,\text{m/s}$ ), and rolls back down.
- Calculation:
- $\Delta v = v_f - v_i = 0.00\,\text{m/s} - 2.50\,\text{m/s} = -2.50\,\text{m/s}$
- $\Delta t = t_f - t_i = 5.00\,\text{s} - 0.00\,\text{s} = 5.00\,\text{s}$
- $\bar{a} = \frac{\Delta v}{\Delta t} = \frac{-2.50\,\text{m/s}}{5.00\,\text{s}} = -0.500\,\text{m/s}^2$
- Interpretation: The magnitude of acceleration is $0.500\,\text{m/s}^2$ , directed down the driveway.
Example Problem 3: Wind Sprints and Directional Change
- Context: A runner sprints toward a wall at $4.0\,\text{m/s}$ . $10.0\,\text{s}$ later, the runner is sprinting away from the wall at $4.0\,\text{m/s}$ . Direction toward the wall is positive.
- Calculation:
- Initial velocity $v_i = +4.0\,\text{m/s}$
- Final velocity $v_f = -4.0\,\text{m/s}$ (negative due to reversed direction)
- $\bar{a} = \frac{v_f - v_i}{\Delta t} = \frac{-4.0\,\text{m/s} - (+4.0\,\text{m/s})}{10.0\,\text{s}} = \frac{-8.0\,\text{m/s}}{10.0\,\text{s}} = -0.80\,\text{m/s}^2$
- Explanation: Although speed remains constant at $4.0\,\text{m/s}$ , velocity changes because of the direction change, giving an acceleration directed away from the wall.

Practice Problems Solutions and Progress Assessment

Velocity-time graphs for Steven's motion and a toy train

Practice Problem 1 (Steven's Motion):
- Analysis of Figure 8 shows velocity changing over time. Corresponding motion diagram requires velocity vectors proportional to graph values for each second.
Practice Problem 2 (Toy Train Graph Analysis - Figure 9):
- a. Constant Speed: Occurs during horizontal graph segments where velocity does not change ( $t = 5.0\,\text{s}$ to $15.0\,\text{s}$ and $t = 20.0\,\text{s}$ to $25.0\,\text{s}$ ).
- b. Positive Acceleration: Occurs where the slope is positive and line slants upward ( $t = 0.0\,\text{s}$ to $5.0\,\text{s}$ ).
- c. Most Negative Acceleration: Occurs where line has the steepest downward slope ( $t = 15.0\,\text{s}$ to $20.0\,\text{s}$ ).
Practice Problem 3 (Toy Train Average Acceleration Calculations):
- a. Interval $0.0\,\text{s}$ to $5.0\,\text{s}$ : $\bar{a} = \frac{10.0\,\text{m/s} - 0.0\,\text{m/s}}{5.0\,\text{s} - 0.0\,\text{s}} = 2.0\,\text{m/s}^2$
- b. Interval $15.0\,\text{s}$ to $20.0\,\text{s}$ : $\bar{a} = \frac{4.0\,\text{m/s} - 10.0\,\text{m/s}}{20.0\,\text{s} - 15.0\,\text{s}} = \frac{-6.0\,\text{m/s}}{5.0\,\text{s}} = -1.2\,\text{m/s}^2$
- c. Interval $0.0\,\text{s}$ to $40.0\,\text{s}$ : $\bar{a} = \frac{0.0\,\text{m/s} - 0.0\,\text{m/s}}{40.0\,\text{s} - 0.0\,\text{s}} = 0.0\,\text{m/s}^2$
Practice Problem 4 (Elevator Motion Challenge Graph):
- Interval 1: $t = 0.0\,\text{s}$ to $2.0\,\text{s}$ , accelerates at $0.5\,\text{m/s}^2$ to reach $v = 1.0\,\text{m/s}$ .
- Interval 2: $t = 2.0\,\text{s}$ to $14.0\,\text{s}$ , travels at constant velocity $v = 1.0\,\text{m/s}$ for $12.0\,\text{s}$ .
- Interval 3: $t = 14.0\,\text{s}$ to $18.0\,\text{s}$ , decelerates at $-0.25\,\text{m/s}^2$ for $4.0\,\text{s}$ back to $0.0\,\text{m/s}$ .
Practice Problem 5 (Race Car Acceleration):
- $\bar{a} = \frac{36\,\text{m/s} - 4.0\,\text{m/s}}{4.0\,\text{s}} = \frac{32\,\text{m/s}}{4.0\,\text{s}} = 8.0\,\text{m/s}^2$
Practice Problem 6 (Race Car Deceleration):
- $\bar{a} = \frac{15\,\text{m/s} - 36\,\text{m/s}}{3.0\,\text{s}} = \frac{-21\,\text{m/s}}{3.0\,\text{s}} = -7.0\,\text{m/s}^2$
Practice Problem 7 (Braking Bus):
- a. Average Acceleration (assuming West is positive): $\bar{a} = \frac{0\,\text{m/s} - 25\,\text{m/s}}{3.0\,\text{s}} = -8.3\,\text{m/s}^2$ (Directed East at $8.3\,\text{m/s}^2$ ).
- b. Comparison when time doubles to $6.0\,\text{s}$ : $\bar{a} = \frac{-25\,\text{m/s}}{6.0\,\text{s}} = -4.2\,\text{m/s}^2$ Doubling time halves the required average acceleration magnitude.
Practice Problem 8 (Coasting Car Hill Turnaround):
- Uphill is positive. Initial velocity $v_i = -3.0\,\text{m/s}$ (downhill), final velocity $v_f = +4.5\,\text{m/s}$ (uphill), time $\Delta t = 2.5\,\text{s}$ .
- $\bar{a} = \frac{+4.5\,\text{m/s} - (-3.0\,\text{m/s})}{2.5\,\text{s}} = \frac{7.5\,\text{m/s}}{2.5\,\text{s}} = 3.0\,\text{m/s}^2$
Practice Problem 9 (Jogger Slowing Down):
- $v_i = 3.5\,\text{m/s}$ East, $v_f = 0.75\,\text{m/s}$ East, time $\Delta t = 10.0\,\text{s}$ .
- $\bar{a} = \frac{0.75\,\text{m/s} - 3.5\,\text{m/s}}{10.0\,\text{s}} = \frac{-2.75\,\text{m/s}}{10.0\,\text{s}} = -0.28\,\text{m/s}^2$
Practice Problem 10 (Continental Drift Challenge):
- Rate changes from $1.0\,\text{cm/y}$ to $0.5\,\text{cm/y}$ over $\Delta t = 1.0\,\text{y}$ .
- $\bar{a} = \frac{0.5\,\text{cm/y} - 1.0\,\text{cm/y}}{1.0\,\text{y}} = -0.5\,\text{cm/y}^2$
Check Your Progress Solutions:
- 11. Three Ways to Accelerate:
1. Speeding up (increasing velocity magnitude).
2. Slowing down (decreasing velocity magnitude).
3. Changing direction of motion.
- 12. Two Joggers Motion Comparison:
- Context: Both joggers run at a constant velocity of $7.5\,\text{m/s}$ East. At $t = 0\,\text{s}$ , Jogger 1 is $15\,\text{m}$ West of origin ( $-15\,\text{m}$ ), Jogger 2 is $15\,\text{m}$ East of origin ( $+15\,\text{m}$ ).
- a. Position-Time Graphs: Two parallel straight lines with equal positive slopes ( $7.5\,\text{m/s}$ ), but different y-intercepts ( $-15\,\text{m}$ vs. $+15\,\text{m}$ ).
- b. Velocity-Time Graphs: Identical horizontal lines plotted at $v = 7.5\,\text{m/s}$ with zero slope.
- 13. East/West Car Velocity-Time Graph Sketch:
- Segment 1 ( $0\,\text{s}$ to $100\,\text{s}$ ): Horizontal line at $v = +25\,\text{m/s}$ .
- Segment 2 ( $100\,\text{s}$ to $200\,\text{s}$ ): Horizontal line at $v = -25\,\text{m/s}$ .
- 14. Canoeist Motion Analysis:
- Context: Paddles upstream at $+2.0\,\text{m/s}$ for $4.0\,\text{s}$ , then floats downstream at $-4.0\,\text{m/s}$ for $4.0\,\text{s}$ .
- Upstream displacement: $d_1 = (+2.0\,\text{m/s}) \times (4.0\,\text{s}) = +8.0\,\text{m}$
- Downstream displacement: $d_2 = (-4.0\,\text{m/s}) \times (4.0\,\text{s}) = -16.0\,\text{m}$
- Total displacement: $\Delta d = +8.0\,\text{m} + (-16.0\,\text{m}) = -8.0\,\text{m}$
- a. Average Velocity: $\bar{v} = \frac{\Delta d}{\Delta t} = \frac{-8.0\,\text{m}}{8.0\,\text{s}} = -1.0\,\text{m/s}$
- b. Average Acceleration: $\bar{a} = \frac{v_f - v_i}{\Delta t} = \frac{-4.0\,\text{m/s} - (+2.0\,\text{m/s})}{8.0\,\text{s}} = \frac{-6.0\,\text{m/s}}{8.0\,\text{s}} = -0.75\,\text{m/s}^2$
- 15. Speeding Ticket Critical Thinking:
- Argument: A driver clocked at $32\,\text{km/h}$ over the speed limit claims the adjacent vehicle must have the same speed because they were side-by-side when observed.
- Analysis: The driver is incorrect. Occupying the same position at a single instant ( $d_1 = d_2$ ) means their locations are equal, not their velocities. A faster car passing a slower car occupies the exact same spatial coordinate at the precise moment of passing, but has a significantly higher rate of position change (steeper slope on a position-time graph).

Real-World Career Applications

Delivery Truck Drivers:
- Logistical distribution centers rely on precise monitoring of acceleration, velocity, and transit time.
- Delivery schedules depend on predictable vehicle motion parameters to ensure safe, on-time arrivals at residential and commercial locations.