Physics 1005: Kinematics and Free Fall Notes

Physics 1005 - Lecture 3: Kinematics and Free Fall

Kinematic Equations for Constant Acceleration

These equations are fundamental tools for analyzing motion where acceleration remains constant. It is crucial to select the correct equation based on the information provided in a problem and the desired unknown.

  • Displacement:
    \Delta x = x - x_0

    • Where x is the final position and x_0 is the initial position.

  • Average Velocity:
    v{avg} = \frac{s}{\Delta t} = \frac{d}{t - t0}
    v{avg} = \frac{v0 + v}{2}

    • Where s or d is displacement/distance, \Delta t or (t - t0) is the time interval, v0 is the initial velocity, and v is the final velocity.

  • Average Acceleration:
    a = \frac{\Delta v}{\Delta t} = \frac{v - v0}{t - t0}

    • Where \Delta v is the change in velocity.

  • Full Set of Kinematic Equations (Constant Acceleration):

    1. v = v_0 + at

    2. x = x0 + \frac{1}{2}(v + v0)t

    3. x = x0 + v0t + \frac{1}{2}at^2

    4. v^2 = v0^2 + 2a(x - x0)

Steps for Solving Kinematics Problems

  1. Identify Variables: Each equation typically involves 4 or 5 variables. To solve a problem, only one variable can be unknown.

  2. Unit Conversion: Convert all known values to proper, consistent units (e.g., meters, seconds, m/s, m/s^2).

  3. Equation Selection: Choose the appropriate kinematic equation that relates the known values to the single unknown variable.

  4. Rearrange and Solve: Rearrange the chosen equation if necessary to isolate the unknown variable, then substitute the known values and solve.

Examples: Constant Acceleration

Example 1: Car Acceleration

  • Problem Statement: A car accelerates from rest to a final velocity of 80m/s in 6.5s. What is the car's displacement?

  • Given:

    • Initial velocity (v_0) = 0m/s (from rest)

    • Final velocity (v) = 80m/s

    • Time (t) = 6.5s

    • Initial position (x_0) = 0m (assumed)

  • Find: Displacement (x)

  • Equation to use: x = x0 + \frac{1}{2}(v + v0)t

Example 2: Soccer Ball

  • Problem Statement: A soccer ball is kicked along the ground at an initial velocity of 25m/s. If it comes to a stop 4s later, what is its average acceleration?

  • Given:

    • Initial velocity (v_0) = 25m/s

    • Final velocity (v) = 0m/s (comes to a stop)

    • Time (t) = 4s

  • Find: Average acceleration (a)

  • Equation to use: a = \frac{v - v0}{t - t0} (assuming t0 = 0) or v = v0 + at

Example 3: Plane Takeoff

  • Problem Statement: A plane accelerates from rest at a rate of 4.0m/s^2. How long does it take the plane to cover a 100m distance on the runway?

  • Given:

    • Initial velocity (v_0) = 0m/s (from rest)

    • Acceleration (a) = 4.0m/s^2

    • Displacement (x) = 100m

    • Initial position (x_0) = 0m (assumed)

  • Find: Time (t)

  • Equation to use: x = x0 + v0t + \frac{1}{2}at^2

Example 4: Sprinter

  • Problem Statement: A sprinter comes out of the starting blocks and runs along a 60m long track. If the sprinter accelerated at a uniform rate and achieves a final velocity of 10m/s, what is their average acceleration?

  • Given:

    • Initial velocity (v_0) = 0m/s (comes out of starting blocks, implies rest)

    • Displacement (x - x_0) = 60m

    • Final velocity (v) = 10m/s

  • Find: Average acceleration (a)

  • Equation to use: v^2 = v0^2 + 2a(x - x0)

Free Fall

Galileo's Discovery

  • Galileo challenged the long-held belief that heavier objects fall with greater acceleration.

  • He discovered that, in the absence of air resistance, all objects experience the same acceleration due to gravity.

Definition of Free Fall

  • Free fall refers to the motion of an object solely under the influence of gravity.

Acceleration Due to Gravity (g)

  • An object in free fall has a constant acceleration due to Earth's gravity, provided air resistance is negligible.

  • Value: g = -9.81m/s^2

    • The negative sign indicates that the acceleration due to gravity is directed downwards when the positive y-direction is defined as upwards.

    • This value can also be expressed as g = -9.81N/kg.

  • Variability: There is some slight variability in the value of g depending on the specific location on Earth.

Role of Air Resistance

  • Differences observed in the falling patterns of distinct objects (e.g., a rock vs. a leaf) are primarily due to air resistance.

  • In a vacuum, without any air resistance, all objects fall with the same acceleration due to gravity.

Free Fall Kinematic Equations

These are the same constant-acceleration kinematic equations, specialized for vertical motion under gravity.

  • Convention: The positive y-direction is typically chosen to be upwards. If chosen downwards, the sign of g would need to be positive (+9.81m/s^2).

  • In these equations, acceleration (a) is replaced by g, and position (x, x0) is replaced by (y, y0).

  1. v = v_0 + gt

  2. y = y0 + \frac{1}{2}(v + v0)t

  3. y = y0 + v0t + \frac{1}{2}gt^2

  4. v^2 = v0^2 + 2g(y - y0)

When solving problems:

  • For horizontal one-dimensional motion, use x for position and a for acceleration.

  • For vertical one-dimensional motion, use y for position and g = -9.81m/s^2 for acceleration.

Cartesian Coordinate System Conventions

  • X-dimension:

    • + generally describes motion to the right.

    • - generally describes motion to the left.

  • Y-dimension:

    • + generally describes motion upwards.

    • - generally describes motion downwards.

Concept Check: Vacuum Experiment

  • Question: What happens if you drop a rock and a leaf in a vacuum?

  • Answer: Both the rock and the leaf fall with the same acceleration. This is because air resistance, which normally causes the leaf to fall slower, is absent in a vacuum.

Examples: Free Fall

Example 5: Ball Thrown Up - Displacement

  • Problem Statement: If a ball is thrown straight up with an initial velocity of 11.2m/s, what is its displacement after 3.5s?

  • Given:

    • Initial velocity (v_0) = 11.2m/s

    • Time (t) = 3.5s

    • Acceleration (g) = -9.81m/s^2

    • Initial position (y_0) = 0m (assumed)

  • Find: Displacement (y)

  • Equation to use: y = y0 + v0t + \frac{1}{2}gt^2

Example 6: Ball Thrown Up - Final Velocity

  • Problem Statement: If a ball is thrown upwards from a window with an initial velocity of 15m/s, what will its velocity be after 2.5s?

  • Given:

    • Initial velocity (v_0) = 15m/s

    • Time (t) = 2.5s

    • Acceleration (g) = -9.81m/s^2

  • Find: Final velocity (v)

  • Equation to use: v = v_0 + gt

Example 7: Object Thrown Up - Time in Air

  • Problem Statement: An object is thrown straight up with an initial velocity of 2.5m/s. It later hits the ground with a velocity -8.0m/s. For how long a period of time was the object in the air? (Assume no air resistance).

  • Given:

    • Initial velocity (v_0) = 2.5m/s

    • Final velocity (v) = -8.0m/s

    • Acceleration (g) = -9.81m/s^2

  • Find: Time (t)

  • Equation to use: v = v_0 + gt

  • Calculation Note: Using the given values, -8.0 = 2.5 + (-9.81)t, which yields t \approx 1.07s.

Example 8: Skydiver - Falling Speed

  • Problem Statement: A skydiver jumps out of a plane. How fast is he falling after falling 100m?

  • Given:

    • Initial velocity (v_0) = 0m/s (jumps implies starting from rest)

    • Displacement (y - y_0) = -100m (falling 100m, so negative displacement)

    • Acceleration (g) = -9.81m/s^2

  • Find: Final speed (|v|)

  • Equation to use: v^2 = v0^2 + 2g(y - y0)

Example 9: Ball from Window - Velocity at different points

  • Problem Statement: A ball is thrown from a window with an upwards velocity of 8.5m/s. What is the approximate velocity of the ball at points a, b, and c?

  • Points Description (based on typical projectile motion):

    • a: Initial launch point (or just after launch, going up).

    • b: The peak of its trajectory (maximum height).

    • c: The point where the ball returns to the initial launch height but is now moving downwards.

  • Expected Velocities (due to symmetry and gravity):

    • At point a (initial velocity): 8.5m/s

    • At point b (peak height): 0m/s

    • At point c (same height as launch, moving down): -8.5m/s

  • Correct Option: 8.5m/s; 0m/s; -8.5m/s (This corresponds to initial velocity, velocity at max height, and velocity upon return to initial height, respectively).

Example 10: Ball from Window - Maximum Height

  • Problem Statement: A ball is thrown from a window with an upwards velocity of 8.5m/s. At what height above the window will the ball turn and start coming back down?

  • This refers to the maximum height (y_{max}) where the vertical velocity (v) momentarily becomes zero.

  • Given:

    • Initial velocity (v_0) = 8.5m/s

    • Final velocity at max height (v) = 0m/s

    • Acceleration (g) = -9.81m/s^2

  • Find: Displacement (y - y_0) or height above the window.

  • Equation to use: v^2 = v0^2 + 2g(y - y0)

  • Calculation: 0^2 = (8.5)^2 + 2(-9.81)(y - y0) \implies y - y0 = \frac{72.25}{19.62} \approx 3.68m

Example 11: Vertical Jump

  • Problem Statement: Haneef Munir (5'10") jumps so high that his head reaches a height of approximately 10'10''. Find his upwards velocity at take-off.

  • Given:

    • Initial head height (y_{0, \text{head}}) = 5'10''

    • Maximum head height (y_{\text{max, head}}) = 10'10''

    • Displacement (\Delta y) = y{\text{max, head}} - y{0, \text{head}} = 10'10'' - 5'10'' = 5' = 1.524m

      • (1 \text{ foot} = 12 \text{ inches}, 1 \text{ inch} = 2.54 \text{ cm})

    • Velocity at maximum height (v) = 0m/s

    • Acceleration (g) = -9.81m/s^2

  • Find: Initial upwards velocity (v_0) at take-off.

  • Equation to use: v^2 = v0^2 + 2g(y - y0)

  • Calculation: 0^2 = v0^2 + 2(-9.81)(1.524) \implies v0^2 = 29.897 \implies v_0 \approx 5.47m/s

Example 12: World Record Dive

  • Problem Statement: Dana Kunze's 172 ft World Record Dive. Neglecting air resistance, what would be Dana's downward velocity when he hits the water?

  • Given:

    • Height of the dive (|y - y_0|) = 172ft

      • Convert to meters: 172ft \times 0.3048m/ft = 52.4256m

    • Displacement (y - y_0) = -52.4256m (downwards, if ext{up} is positive)

    • Initial velocity (v_0) = 0m/s (assumed from rest before the dive)

    • Acceleration (g) = -9.81m/s^2

  • Find: Final velocity (v) when hitting the water.

  • Equation to use: v^2 = v0^2 + 2g(y - y0)

  • Calculation: v^2 = 0^2 + 2(-9.81)(-52.4256) \implies v^2 = 1028.5 \implies v \approx -32.07m/s

    • The negative sign indicates the downward direction of the velocity.

General Problem-Solving Strategy in Physics

  1. Read and Understand: Carefully read the problem statement to fully grasp what is being asked.

  2. Identify Given and Unknowns: Write down all the provided data and clearly state what you are supposed to find.

  3. Sketch: Draw a quick, clear sketch of the physical situation. This helps visualize the motion and identify relevant directions and positions.

  4. Determine Principles: Identify the underlying physics principles or equations that are applicable to the problem (e.g., kinematics, conservation of energy).

  5. Perform Calculations: Execute the calculations on paper. This methodical approach allows for easy double-checking of your work.

  6. Evaluate Reasonableness: After obtaining a result, consider whether the answer is reasonable in the context of the problem. Does the magnitude and sign of your answer make physical sense?