Physics Kinematics Review for Test

Physics Notes for Test

Important Note: The information provided here is for study purposes. Always double-check critical details and calculations.

1. Fundamental Kinematic Concepts

• Displacement:

  • Defined as the difference between an object's final position and its initial position.

  • It is a vector quantity, meaning it has both magnitude and direction.

  • Formula: $\Delta d = d<em>{final} - d</em>{initial}$

• Distance Travelled:

  • The total path length an object covers, irrespective of direction.

  • It is a scalar quantity, only having magnitude.

• Speed:

  • Describes the rate of motion of an object.

  • It is a scalar quantity.

  • Average Speed Formula: $\text{Average Speed} = \frac{\text{total distance travelled}}{\text{total time of travel}}$

• Velocity:

  • Describes an object's speed and its direction.

  • It is a vector quantity.

  • The velocity of an object can change even if its speed remains constant, which occurs if the direction of motion changes.

  • Instantaneous Velocity: The velocity of an object at some instant or at a specific point in its path.

    • Measured by the slope (rise over run) on a position-time graph.

  • Average Velocity Formula: $V_{avg} = \frac{\Delta d}{t} = \frac{\text{total displacement}}{\text{total time}}$

• Acceleration:

  • Defined as the rate at which velocity changes over time.

  • It is a vector quantity.

  • Acceleration indicates speeding up, slowing down, or changing direction.

  • Formula: $a = \frac{\Delta V}{t}$

2. Graphical Analysis of Motion

Velocity-Time Graphs

• Area Under the Graph:

  • Used to find distance (sum of absolute values of areas, without considering negatives).

  • Used to find displacement (sum of areas, considering positives and negatives).

• Slope of the Graph: Represents acceleration.

• Interpreting Velocity-Time Graphs:

  • Positive Velocity (above the x-axis): Object is moving forward.

  • Negative Velocity (below the x-axis): Object is moving backward or in the opposite direction.

  • Straight horizontal line: Velocity is constant (not changing with time).

    • The object is moving at the same speed the entire time.

  • Object at rest: Velocity equals $0$ and would appear as a line segment on the x-axis.

  • Lowest Acceleration: Occurs with a flat slope (zero acceleration), not necessarily a negative slope (which indicates slowing down or accelerating in the negative direction, but still a change in velocity).

Position-Time Graphs

• Slope of the Graph: Represents velocity.

  • The slope (rise over run) at any point gives the instantaneous velocity.

Relationship Between Velocity and Acceleration

• If velocity and acceleration have the same sign, the object is speeding up in that direction.

  • Example: If velocity is negative and acceleration is negative, the car is gaining speed in reverse.

• If velocity and acceleration have opposite signs, the object is slowing down.

  • Example: If velocity is positive and acceleration is negative, the car is slowing down while moving forwards.

3. Units and Conversions

• SI Unit of Velocity: meters per second ($m/s$)

• SI Unit of Acceleration: meters per second squared ($m/s^2$ or $m/s/s$)

• Metric Prefixes:

  • Kilo ($K$): $1000$

  • Hecto ($H$): $100$

  • Deka ($Da$): $10$

  • Base (e.g., meter, gram, liter): $1$

  • Deci ($d$): $0.1$

  • Centi ($c$): $0.01$

  • Milli ($m$): $0.001$

• Conversion Example:

  • $1 \text{ km/hr} = 1000 \text{ m} / 3600 \text{ s}$ (approx. $0.2778 \text{ m/s}$)

4. Key Definitions (Summary)

• Instantaneous Velocity: Velocity of an object at a specific instant.

• Speed: Rate of motion.

• Frame of Reference: Helps define and describe motion.

• Displacement: Difference between final and initial position.

• Acceleration: Ratio of the change in an object's velocity to the time required for the change to occur.

• Velocity: Ratio of an object's displacement to the time interval during which the displacement occurred.

5. Kinematic Formulas (Equations of Motion)

These equations are used for motion with constant acceleration:

1. Average Velocity: $V_{avg} = \frac{\Delta d}{t}$

2. Acceleration Definition: $a = \frac{\Delta V}{t} = \frac{V<em>f - V</em>i}{t}$

3. Displacement with initial velocity and acceleration: $\Delta d = V_i t + \frac{1}{2} a t^2$

4. Final Velocity with initial velocity and acceleration: $V<em>f = V</em>i + at$

5. Displacement with average velocity: $\Delta d = \left( \frac{V<em>i + V</em>f}{2} \right) t$

6. Final Velocity Squared: $V<em>f^2 = V</em>i^2 + 2a \Delta d$

6. Practice Problems & Solutions

Problem 1: Average Speed and Velocity

A car travels $23 \text{m}$ north in $8 \text{s}$ and then travels $15 \text{m}$ south in $3 \text{s}$.

• Average Speed:

  • Total Distance = $23 \text{m} + 15 \text{m} = 38 \text{m}$

  • Total Time = $8 \text{s} + 3 \text{s} = 11 \text{s}$

  • Average Speed = $38 \text{m} / 11 \text{s} \approx 3.45 \text{ m/s}$

• Average Velocity:

  • Total Displacement (assigning North as positive) = $+23 \text{m} - 15 \text{m} = +8 \text{m}$

  • Total Time = $11 \text{s}$

  • Average Velocity = $+8 \text{m} / 11 \text{s} \approx 0.73 \text{ m/s}$ (North)

Problem 2: Car Applying Brakes

A car is traveling at $20 \text{ m/s}$ when the driver applies the brakes, taking $2 \text{s}$ to come to a complete stop.

• Given: $V<em>i = 20 \text{ m/s}$, $V</em>f = 0 \text{ m/s}$, $t = 2 \text{ s}$

• Average Acceleration:

  • $a = \frac{V<em>f - V</em>i}{t} = \frac{0 \text{ m/s} - 20 \text{ m/s}}{2 \text{ s}} = -10 \text{ m/s}^2$

Problem 3: Sports Car Displacement

A sports car traveling at $27.8 \text{ m/s}$ slows at a constant rate to a stop in $8.00 \text{ s}$. What is the displacement?

• Given: $V<em>i = 27.8 \text{ m/s}$, $V</em>f = 0 \text{ m/s}$, $t = 8.00 \text{ s}$

• Displacement:

  • $\Delta d = \left( \frac{V<em>i + V</em>f}{2} \right) t = \left( \frac{27.8 \text{ m/s} + 0 \text{ m/s}}{2} \right) \times 8.00 \text{ s} = 13.9 \text{ m/s} \times 8.00 \text{ s} = 111.2 \text{ m}$

Problem 4: Racehorse Running

A racehorse is running with a uniform speed of $69 \text{ km/hr}$ along a straightaway. What is the time it takes for the horse to cover $400 \text{ meters}$?

• Given: $V = 69 \text{ km/hr}$, $d = 400 \text{ m}$

• Convert Velocity:

  • $V = 69 \text{ km/hr} = 69 \times \frac{1000 \text{ m}}{3600 \text{ s}} \approx 19.167 \text{ m/s}$

• Time:

  • $t = \frac{d}{V} = \frac{400 \text{ m}}{19.167 \text{ m/s}} \approx 20.87 \text{ s}$

Problem 5: Skater Slowing Down

A skater glides at $1.8 \text{ m/s}$ onto ground and is slowed at a constant rate of $-3.00 \text{ m/s}^2$. How fast is the skater moving when she has slid $0.37 \text{ m}$?

• Given: $V_i = 1.8 \text{ m/s}$, $a = -3.00 \text{ m/s}^2$, $\Delta d = 0.37 \text{ m}$

• Final Velocity:

  • $V<em>f^2 = V</em>i^2 + 2a \Delta d$

  • $V_f^2 = (1.8 \text{ m/s})^2 + 2(-3.00 \text{ m/s}^2)(0.37 \text{ m})$

  • $V_f^2 = 3.24 \text{ m}^2/\text{s}^2 - 2.22 \text{ m}^2/\text{s}^2 = 1.02 \text{ m}^2/\text{s}^2$

  • $V_f = \sqrt{1.02} \text{ m/s} \approx 1.01 \text{ m/s}$

Problem 6: Toy Car on Position Graph

A toy car rolls from $+5 \text{ m}$ to $+3 \text{ m}$ and then to $+8 \text{ m}$. (Assuming a 1D position axis)

• Car's Displacement:

  • Final position ($+8 \text{ m}$) - Initial position ($+5 \text{ m}$) = $+3 \text{ m}$

• Car's Distance Traveled:

  • $|3 \text{ m} - 5 \text{ m}| + |8 \text{ m} - 3 \text{ m}| = |-2 \text{ m}| + |5 \text{ m}| = 2 \text{ m} + 5 \text{ m} = 7 \text{ m}$

Problem 7: Velocity-Time Graph Analysis (Cyclist)

• Displacement after 6 seconds: (Area under the curve from $0$ to $6 \text{ s}$)

  • Looks like a triangle from $0-2 \text{s}$ and a rectangle from $2-6 \text{s}$ assuming the velocity increases from $0$ to $2 \text{ m/s}$ at $2 \text{s}$ and then stays at $2 \text{ m/s}$ until $6 \text{s}$. The graph in the document indicates a different path. If we assume the annotated answer of $15 \text{m}$ is correct, this would require a specific interpretation. From the graph on page 2, the velocity increases linearly from $0$ to $4 \text{ m/s}$ in $4 \text{s}$ and then stays constant at $4 \text{ m/s}$ up to $10 \text{s}$. Let's use the visible graph for calculation.

  • From $0-4 \text{s}$ (triangle): $0.5 \times \text{base} \times \text{height} = 0.5 \times 4 \text{ s} \times 4 \text{ m/s} = 8 \text{ m}$

  • From $4-6 \text{s}$ (rectangle): $\text{base} \times \text{height} = 2 \text{ s} \times 4 \text{ m/s} = 8 \text{ m}$

  • Total displacement after $6 \text{s}$ = $8 \text{ m} + 8 \text{ m} = 16 \text{ m}$ (Discrepancy with handwritten $15 \text{m}$, calculations based on visible graph.)

• Velocity after 4 seconds: (Read directly from the graph)

  • Velocity = $4 \text{ m/s}$ (Matches graph).

Problem 8: Velocity-Time Graph Analysis (A,B,C,D,E Sections)

• Acceleration between A and B: (Slope of the line segment from $(0, 100)$ to $(20, 400))$

  • $a = \frac{400 - 100}{20 - 0} = \frac{300}{20} = 15 \text{ m/s}^2$

• Acceleration between B and C: (Slope of the line segment from $(20, 400)$ to $(40, 400))$

  • $a = \frac{400 - 400}{40 - 20} = \frac{0}{20} = 0 \text{ m/s}^2$

• Acceleration between D and E: (Slope of the line segment from roughly $(50, 250)$ to $(70, 0))$ - using values that fit handwritten answer.

  • $a = \frac{0 - 250}{70 - 50} = \frac{-250}{20} = -12.5 \text{ m/s}^2$

• Total distance from B to C: (Area under the curve from $20 \text{s}$ to $40 \text{s}$)

  • Velocity is constant at $400 \text{ m/s}$ for $20 \text{s}$ duration ($40 \text{s} - 20 \text{s}$).

  • Distance = $400 \text{ m/s} \times 20 \text{ s} = 8000 \text{ m} = 8 \text{ km}$. (Note: This calculation differs from the handwritten answer of $3000 \text{m}$ or $3 \text{k}$. The calculation is based on the visual representation of the graph on page 3.)

Problem 9: Cat's Motion (Displacement-Time Graph)

• Cat's displacement during $10.0-15.0 \text{s}$:

  • At $10.0 \text{s}$, displacement is $3.0 \text{ m}$. At $15.0 \text{s}$, displacement is $1.5 \text{ m}$.

  • Displacement = $1.5 \text{ m} - 3.0 \text{ m} = -1.5 \text{ m}$. (Based on graph reading, differs from handwritten 1m.)

• Cat's average velocity during $15.0-20.0 \text{s}$:

  • At $15.0 \text{s}$, displacement is $1.5 \text{ m}$. At $20.0 \text{s}$, displacement is approx $1.0 \text{ m}$.

  • Average Velocity = $\frac{\Delta d}{\Delta t} = \frac{1.0 \text{ m} - 1.5 \text{ m}}{20.0 \text{ s} - 15.0 \text{ s}} = \frac{-0.5 \text{ m}}{5.0 \text{ s}} = -0.1 \text{ m/s}$.

• Cat's displacement during $0.0-20.0 \text{s}$:

  • At $0.0 \text{s}$, displacement is $3.0 \text{ m}$. At $20.0 \text{s}$, displacement is approx $1.0 \text{ m}$.

  • Displacement = $1.0 \text{ m} - 3.0 \text{ m} = -2.0 \text{ m}$. (Based on graph, differs from handwritten -1.5m.)

• Cat's average velocity during $0.0-5.0 \text{s}$:

  • At $0.0 \text{s}$, displacement is $3.0 \text{ m}$. At $5.0 \text{s}$, displacement is $4.0 \text{ m}$.

  • Average Velocity = $\frac{\Delta d}{\Delta t} = \frac{4.0 \text{ m} - 3.0 \text{ m}}{5.0 \text{ s} - 0.0 \text{ s}} = \frac{1.0 \text{ m}}{5.0 \text{ s}} = 0.2 \text{ m/s}$.

7. Useful Formulas

• $V_{avg} = \frac{\Delta d}{t}$

• $a = \frac{\Delta V}{t}$

• $\Delta d = V_i t + \frac{1}{2} a t^2$

• $V<em>f = V</em>i + at$

• $\Delta d = \left( \frac{V<em>i + V</em>f}{2} \right) t$

• $V<em>f^2 = V</em>i^2 + 2a \Delta d$

• $\text{Speed} = \frac{\text{total distance travelled}}{\text{total time of travel}}$

• $\text{Velocity} = \frac{\text{total displacement}}{\text{total time}}$