Comprehensive Study Guide: Force, Acceleration, and Motion

Chapter Overview and Learning Objectives

Course Unit: Chapter 1 - Force and Motion
Lessons Covered:
- Lesson 1: Unbalanced Forces and Acceleration
- Lesson 2: Uniform Circular Motion
- Lesson 3: Graphing Acceleration
Academic Context: Vanie Yales Benben, College of Education, Central Mindanao University
Learning Competencies:
- Identify that forces cause objects to accelerate, and that acceleration of an object is its rate of change of velocity.
- Observe and describe examples of accelerating objects at school and in the local community, including objects that show uniform circular motion.
- Construct and annotate distance-time graphs and velocity-time graphs to represent uniform and non-uniform acceleration.

Review of Forces

Contact Forces:
- Tension ( $T$ ): Pulling force transmitted through a string, rope, or cable.
- Normal Force ( $F_N$ ): Support force exerted upon an object that is in contact with another stable object, perpendicular to the surface.
- Friction ( $f$ ): Force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other.
Noncontact Forces:
- Weight ( $W$ or $F_g$ ): Force exerted on a body by gravity.
- Magnetic Force: Attraction or repulsion that arises between electrically charged particles because of their motion.
- Electrical Force: Attractive or repulsive force between two charged objects.
Free-Body Diagrams and Force Interactions:
- Hanging Mass: A mass $m$ suspended by a rope experiences downward weight ( $W$ ) and upward tension ( $T$ ).
- Block on Surface: A bag of dog food ("BOW WOW CHOW") held by a hand or resting on a table experiences downward gravitational force ( $F_g$ ) balanced by upward hand force ( $F_{\text{hand}}$ ) or normal force ( $F_N$ ).
- Frictional Oppositions: A crate pushed along a surface experiences a friction force ( $f$ ) pointing in the direction opposite to motion or attempted motion due to microscopic surface irregularities.
- Charge/Magnet Interactions: Opposite charges ( $+\text{ and }-$ ) or magnetic poles ( $N\text{ and }S$ ) attract each other; like charges ( $+\text{ and }+$ or $-\text{ and }-$ ) or magnetic poles repel each other.

Balanced and Unbalanced Forces

Balanced Forces:
- When balanced forces act on a body, the net force is zero.
- A body under balanced forces can either be completely at rest or moving with uniform motion.
- Uniform Motion: Movement at a constant velocity, meaning the body does not change its speed or its direction.
Unbalanced Forces:
- When unbalanced forces act on a body, a non-zero net force is present.
- Unbalanced forces cause the body to accelerate (change its velocity over time).

Acceleration and Linear Motion

Velocity Definition and Equation:
- Velocity ( $v$ ) is the change in displacement ( $d$ ) over time ( $t$ ): $v = \frac{d}{t}$
Acceleration Definition and Equation:
- Acceleration ( $a$ ) is defined as the rate at which velocity changes ( $\Delta v$ ) over time ( $\Delta t$ ): $a = \frac{\Delta v}{\Delta t}$ $a = \frac{v_f - v_i}{t_f - t_i}$ $a = \frac{v_f - v_i}{t}$
- Units of Acceleration: Meters per second squared ( $\text{m/s}^2$ ).
Example 1: Positive Acceleration (Speeding Up):
- Problem: A car starting from rest inside a garage reaches a speed of $2.5\,\text{m/s}$ in $5\,\text{s}$ . What is the acceleration of the car?
- Given:
- Initial velocity $v_i = 0\,\text{m/s}$
- Final velocity $v_f = 2.5\,\text{m/s}$
- Time $t = 5\,\text{s}$
- Calculation: $a = \frac{v_f - v_i}{t}$ $a = \frac{2.5\,\text{m/s} - 0\,\text{m/s}}{5\,\text{s}}$ $a = 0.5\,\text{m/s}^2$
- Conclusion: Positive acceleration ( $+0.5\,\text{m/s}^2$ ) indicates that the car is speeding up.
Example 2: Negative Acceleration / Deceleration (Slowing Down):
- Problem: A marathon runner passes the finish line running at $6\,\text{m/s}$ and comes to a full stop after $10\,\text{s}$ . What was her acceleration?
- Given:
- Initial velocity $v_i = 6\,\text{m/s}$
- Final velocity $v_f = 0\,\text{m/s}$
- Time $t = 10\,\text{s}$
- Calculation: $a = \frac{v_f - v_i}{t}$ $a = \frac{0\,\text{m/s} - 6\,\text{m/s}}{10\,\text{s}}$ $a = -0.6\,\text{m/s}^2$
- Conclusion: Negative acceleration ( $-0.6\,\text{m/s}^2$ ) indicates that the runner was slowing down (decelerating).

Uniformly Accelerated Motion

Definition: Uniformly accelerated motion occurs when an object moves with constant acceleration, meaning its velocity changes at a steady, uniform rate over equal time intervals.
Derivation of the Final Velocity Formula:
- Starting from the base acceleration formula: $a = \frac{v_f - v_i}{t_f - t_i}$ $a = \frac{v_f - v_i}{t}$
- Multiply both sides by time $t$ : $a t = v_f - v_i$
- Rearrange to solve for final velocity $v_f$ : $a t + v_i = v_f$ $v_f = a t + v_i$
- When starting from rest ( $v_i = 0\,\text{m/s}$ ): $v_f = a t$
Example 3: Step-by-Step Velocity Progression:
- Scenario: A car starts from rest ( $v_i = 0\,\text{m/s}$ ) and accelerates uniformly at $a = 0.5\,\text{m/s}^2$ for $5\,\text{s}$ :
- At $t_1 = 1\,\text{s} \rightarrow v_{f1} = a t_1 = (0.5\,\text{m/s}^2)(1\,\text{s}) = 0.5\,\text{m/s}$
- At $t_2 = 2\,\text{s} \rightarrow v_{f2} = a t_2 = (0.5\,\text{m/s}^2)(2\,\text{s}) = 1.0\,\text{m/s}$
- At $t_3 = 3\,\text{s} \rightarrow v_{f3} = a t_3 = (0.5\,\text{m/s}^2)(3\,\text{s}) = 1.5\,\text{m/s}$
- At $t_4 = 4\,\text{s} \rightarrow v_{f4} = a t_4 = (0.5\,\text{m/s}^2)(4\,\text{s}) = 2.0\,\text{m/s}$
- At $t_5 = 5\,\text{s} \rightarrow v_{f5} = a t_5 = (0.5\,\text{m/s}^2)(5\,\text{s}) = 2.5\,\text{m/s}$

Practice Problems and Solutions: Linear Acceleration

Problem 1:
- Question: A car initially at rest accelerates uniformly to a speed of $20\,\text{m/s}$ in $5\,\text{seconds}$ . What is the car's acceleration?
- Given: $v_i = 0\,\text{m/s}$ , $v_f = 20\,\text{m/s}$ , $t = 5\,\text{s}$
- Solution: $a = \frac{v_f - v_i}{t}$ $a = \frac{20\,\text{m/s} - 0\,\text{m/s}}{5\,\text{s}}$ $a = 4\,\text{m/s}^2$
Problem 2:
- Question: A bicycle moving at $8\,\text{m/s}$ slows down uniformly and comes to a stop in $4\,\text{seconds}$ . What is the acceleration of the bicycle?
- Given: $v_i = 8\,\text{m/s}$ , $v_f = 0\,\text{m/s}$ , $t = 4\,\text{s}$
- Solution: $a = \frac{v_f - v_i}{t}$ $a = \frac{0\,\text{m/s} - 8\,\text{m/s}}{4\,\text{s}}$ $a = -2\,\text{m/s}^2$
Problem 3:
- Question: A runner crosses the finish line at $6\,\text{m/s}$ and comes to a stop $10\,\text{seconds}$ later with uniform acceleration. What was the acceleration, and how far did she travel after crossing the finish line?
- Given: $v_i = 6\,\text{m/s}$ , $v_f = 0\,\text{m/s}$ , $t = 10\,\text{s}$
- Acceleration Solution: $a = \frac{v_f - v_i}{t}$ $a = \frac{0\,\text{m/s} - 6\,\text{m/s}}{10\,\text{s}}$ $a = -0.6\,\text{m/s}^2$
- Distance Solution: $d = v_{\text{avg}} t$ $d = \left(\frac{v_i + v_f}{2}\right) t$ $d = \left(\frac{6\,\text{m/s} + 0\,\text{m/s}}{2}\right)(10\,\text{s})$ $d = (3\,\text{m/s})(10\,\text{s})$ $d = 30\,\text{m}$

Uniform Circular Motion (UCM)

Core Principles:
- Tangential Speed ( $v_t$ ): The direction of tangential speed continuously changes as an object moves in a circle, but the magnitude (speed) remains constant.
- Source of Acceleration: Because speed magnitude is constant, acceleration in uniform circular motion arises solely from the continuous change in direction.
- Centripetal Acceleration ( $a_c$ ): Acceleration is directed inward toward the center of the circular path because an inward force is required to maintain rotation.
- Constant Angular Speed: Uniform circular motion means the object moves around the center at a constant angular speed ( $\omega$ ).

Uniform circular motion vectors showing velocity and centripetal acceleration

Mathematical Definitions and Formulas:
- Frequency ( $f$ ): The number of turns or complete cycles per unit of time. $f = \frac{\text{cycles}}{\text{time}} = \frac{1\,\text{cycle}}{\text{s}} = \text{Hz}$ $f = \frac{\text{number of revolutions}}{\text{time}}$ $f = \frac{1}{T}$
- Period ( $T$ ): The amount of time required for one complete cycle. $T = \frac{\text{time}}{\text{cycles}}$ $T = \frac{\text{time}}{\text{number of revolutions}}$ $T = \frac{1}{f}$
- Angular Speed ( $\omega$ ): Describes how fast an object turns or goes around the path. $\omega = \frac{2\pi}{T} = 2\pi f$
- Tangential Speed ( $v_t$ ): $v_t = r \omega$ $v_t = \frac{2\pi r}{T} = 2\pi r f$ Note: Tangential speed depends on both radius ( $r$ ) and angular speed ( $\omega$ ). For objects on the same rotating platform, a larger radius requires a higher tangential speed to maintain the same angular speed.
- Centripetal Acceleration ( $a_c$ ) Equations: $a_c = \frac{v_t^2}{r}$ $a_c = \frac{(r \omega)^2}{r} = \frac{r^2 \omega^2}{r} = r \omega^2$ $a_c = r(2\pi f)^2 = r \left(\frac{2\pi}{T}\right)^2$
Example 4: Lazy Susan with Bug and Spoon:
- Scenario: A bug is positioned at the outer edge of a lazy Susan with a diameter of $0.75\,\text{m}$ . A spoon is situated $0.1\,\text{m}$ away from the edge. The lazy Susan rotates uniformly at a rate of $1\,\text{turn per } 5\,\text{seconds}$ .

Lazy Susan with bug on edge and spoon near edge

Part A: Tangential Speed of the Bug:
- Radius of lazy Susan ( $r$ ): $r = \frac{d}{2} = \frac{0.75\,\text{m}}{2} = 0.38\,\text{m}$
- Frequency ( $f$ ): $f = \frac{\text{cycle}}{\text{time}} = \frac{1\,\text{turn}}{5\,\text{s}} = 0.2\,\text{Hz}$
- Tangential speed calculation: $v_t = 2\pi r f$ $v_t = 2\pi (0.38\,\text{m})(0.2\,\text{Hz})$ $v_t = 0.48\,\text{m/s}$
Part B: Tangential Speed of the Spoon:
- Spoon frequency ( $f$ ): Same as lazy Susan ( $0.2\,\text{Hz}$ ).
- Spoon radius ( $r_s$ ): $r_s = r - 0.1\,\text{m} = 0.38\,\text{m} - 0.1\,\text{m} = 0.28\,\text{m}$
- Tangential speed calculation: $v_{ts} = 2\pi r_s f$ $v_{ts} = 2\pi (0.28\,\text{m})(0.2\,\text{Hz})$ $v_{ts} = 0.35\,\text{m/s}$
Part C: Centripetal Acceleration of the Bug and Spoon:
- Centripetal acceleration of bug ( $a_{cb}$ ): $a_{cb} = \frac{v_t^2}{r}$ $a_{cb} = \frac{(0.48\,\text{m/s})^2}{0.38\,\text{m}}$ $a_{cb} = 0.61\,\text{m/s}^2$
- Centripetal acceleration of spoon ( $a_{cs}$ ): $a_{cs} = \frac{v_{ts}^2}{r_s}$ $a_{cs} = \frac{(0.35\,\text{m/s})^2}{0.28\,\text{m}}$ $a_{cs} = 0.44\,\text{m/s}^2$
- Example 5: Safe Radius for a Hairpin Curve:
Problem: The speed limit at a hairpin curve is $14\,\text{m/s}$ , and the safest centripetal acceleration is $5.2\,\text{m/s}^2$ . What is a safe radius for the hairpin curve?

Hairpin curve with speed limit and acceleration sign

Given:
- Tangential speed limit $v_t = 14\,\text{m/s}$
- Safest acceleration $a_c = 5.2\,\text{m/s}^2$
Formula Rearrangement: $a_c = \frac{v_t^2}{r}$ $a_c r = v_t^2$ $\frac{a_c r}{a_c} = \frac{v_t^2}{a_c}$ $r = \frac{v_t^2}{a_c}$
Calculation: $r = \frac{(14\,\text{m/s})^2}{5.2\,\text{m/s}^2}$ $r = \frac{196\,\text{m}^2/\text{s}^2}{5.2\,\text{m/s}^2}$ $r = 37.7\,\text{m}$

Graphing Acceleration

Primary Data Table:
- Time ( $\text{s}$ ), Distance ( $\text{m}$ ), and Velocity ( $\text{m/s}$ ) values:
- At $t = 0\,\text{s}$ : $d = 0\,\text{m}$ , $v = 0\,\text{m/s}$
- At $t = 10\,\text{s}$ : $d = 800\,\text{m}$ , $v = 160\,\text{m/s}$
- At $t = 20\,\text{s}$ : $d = 3200\,\text{m}$ , $v = 320\,\text{m/s}$
- At $t = 30\,\text{s}$ : $d = 7200\,\text{m}$ , $v = 480\,\text{m/s}$
- At $t = 40\,\text{s}$ : $d = 12800\,\text{m}$ , $v = 640\,\text{m/s}$
- At $t = 50\,\text{s}$ : $d = 20000\,\text{m}$ , $v = 800\,\text{m/s}$
Distance-Time Graphs (Determining Velocity):
- The slope of a distance-time graph represents velocity ( $v = \frac{\Delta d}{\Delta t}$ ).
- Interval Calculations:
- $0 - 10\,\text{s}$ : $v = \frac{800\,\text{m} - 0\,\text{m}}{10\,\text{s} - 0\,\text{s}} = 80\,\text{m/s}$
- $10 - 20\,\text{s}$ : $v = \frac{3200\,\text{m} - 800\,\text{m}}{20\,\text{s} - 10\,\text{s}} = \frac{2400\,\text{m}}{10\,\text{s}} = 240\,\text{m/s}$
- $20 - 30\,\text{s}$ : $v = \frac{7200\,\text{m} - 3200\,\text{m}}{30\,\text{s} - 20\,\text{s}} = \frac{4000\,\text{m}}{10\,\text{s}} = 400\,\text{m/s}$
- $30 - 40\,\text{s}$ : $v = \frac{12800\,\text{m} - 7200\,\text{m}}{40\,\text{s} - 30\,\text{s}} = \frac{5600\,\text{m}}{10\,\text{s}} = 560\,\text{m/s}$
- $40 - 50\,\text{s}$ : $v = \frac{20000\,\text{m} - 12800\,\text{m}}{50\,\text{s} - 40\,\text{s}} = \frac{7200\,\text{m}}{10\,\text{s}} = 720\,\text{m/s}$
- Distance-Time Graph Line Interpretations:
- Object at rest: Horizontal flat line ( $d$ stays constant).
- Uniform motion (constant speed): Straight sloping upward line.
- Accelerated motion (speeding up): Curved upward line getting progressively steeper.
- Decelerated motion (slowing down): Curved line flattening out toward horizontal.
- Annotated Distance-Time Example (Maria's Journey):

Annotated distance-time graph showing motion segments A through E

- Segment A–B: Flat horizontal line  $\rightarrow$  Maria is at rest, waiting at the gate.
- Segment B–C: Straight sloping line  $\rightarrow$  Maria walks steadily with constant speed.
- Segment C–D: Steeper curved line  $\rightarrow$  Maria runs faster, accelerating.
- Segment D–E: Curve flattens  $\rightarrow$  Maria slows down and finally comes to a stop.

Velocity-Time Graphs (Determining Acceleration):
- The slope of a velocity-time graph represents acceleration ( $a = \frac{\Delta v}{\Delta t}$ ).
- Interval Calculations:
- $0 - 10\,\text{s}$ : $a = \frac{160\,\text{m/s} - 0\,\text{m/s}}{10\,\text{s} - 0\,\text{s}} = 16\,\text{m/s}^2$
- $10 - 20\,\text{s}$ : $a = \frac{320\,\text{m/s} - 160\,\text{m/s}}{20\,\text{s} - 10\,\text{s}} = 16\,\text{m/s}^2$
- $20 - 30\,\text{s}$ : $a = \frac{480\,\text{m/s} - 320\,\text{m/s}}{30\,\text{s} - 20\,\text{s}} = 16\,\text{m/s}^2$
- $30 - 40\,\text{s}$ : $a = \frac{640\,\text{m/s} - 480\,\text{m/s}}{40\,\text{s} - 30\,\text{s}} = 16\,\text{m/s}^2$
- $40 - 50\,\text{s}$ : $a = \frac{800\,\text{m/s} - 640\,\text{m/s}}{50\,\text{s} - 40\,\text{s}} = 16\,\text{m/s}^2$
- Velocity-Time Graph Line Interpretations:
- Object at rest: Line along the horizontal axis ( $v = 0$ ).
- Uniform motion (constant velocity): Horizontal line above zero ( $v$ is constant, $a = 0$ ).
- Accelerated motion (speeding up): Straight line sloping upward (positive slope).
- Decelerated motion (slowing down): Straight line sloping downward (negative slope).
- Annotated Velocity-Time Example (Maria's Journey):

Annotated velocity-time graph showing motion segments A through E

- Segment A–B: Velocity stays at 0  $\rightarrow$  Maria is at rest, still waiting at the gate.
- Segment B–C: Velocity increases slowly  $\rightarrow$  She starts walking at a steady pace.
- Segment C–D: Velocity rises more steeply  $\rightarrow$  She runs faster, accelerating.
- Segment D–E: Velocity decreases  $\rightarrow$  She slows down until she stops at the classroom.