End Ch2 + Beginning Ch3

Position Equations

• $x = bt$ (most common form, assumes initial time and order gain are zero)
  • x: position
  • b: constant
  • t: time

Alternative Expression for Position

• $a = \frac{\Delta v}{\Delta t}$ (acceleration equals change in velocity over change in time)
• Constant acceleration means it does not change over time.
• $x = v_0 t + \frac{1}{2} a t^2$ (position with initial velocity and constant acceleration)
  • $v_0$ (v naught): initial velocity
  • $x_0$ (x naught) initial position; unless stated otherwise, assumed to be zero

Example Problem: Car at a Red Light

• Problem: Where is your car at five seconds after the light goes green?
• Solving for $x$, assuming $x<em>0 = 0$ and $v</em>0 = 0$.
• Given: acceleration
• Result: After calculation, the car is 18.75 meters away from the red light.

Importance of Problem Formulation

• Real-world problems require careful reading to extract necessary information.
• First, read the problem and identify the relevant equation.
• Then, assess if you have all the necessary values or can derive them.

Situation Where $v_0$ is Not Zero

• Example: Starting the vehicle after the green light, implying an initial velocity.

Additional Useful Equation

• $v^2 = v_0^2 + 2 a \Delta x$ (velocity squared equals initial velocity squared plus two times acceleration times change in position)
  • Advantage: Does not involve time ($t$).
  • Derived: By combining previous equations to eliminate time.

Airplane Landing Example

• Scenario: Airplane landing and coming to a complete stop.
  • Landing speed: Approximately 300 miles per hour.
  • Time to stop: 20 seconds.
• Problem: Calculate the stopping acceleration.

Conversion Factors:

• 1 mile = 1.609 kilometers = 1609 meters
• 1 hour = 60 minutes = 3600 seconds

Calculation

• Convert speed to meters per second.$300 \text{ miles/hour} \approx 134.1 \text{ m/s}$
• Final velocity ($v$) = 0 m/s (complete stop)
• Using $v^2 = v_0^2 + 2 a \Delta x$ to solve for $a$
• $0 = (134.1)^2 + 2 a (15)$
• Solving for $a$ will yield the stopping acceleration.

Acceleration of Gravity

• $g = -9.8 \text{ m/s}^2$ (negative indicates downward direction)
  • $y$ axis typically represents vertical motion.
• One-dimensional motion scenarios (straight up or straight down).

Meaning of Negative Sign

• Negative sign indicates direction relative to the coordinate axis.
• It does not necessarily mean the object is decelerating or stopping.
• Example: Dropping a ball accelerates downwards, despite the negative sign.

Interpretation of Acceleration Value

• For every second, the speed increases by 9.8 m/s.

Free Fall Problem

• Problem: How long does it take a ball to fall if dropped from a certain height?
• Given: Height of the building.
• Right before touching the ground, speed is 20 m/s.

Solving for Time

• Identify the appropriate equation to solve for time.
• $a = \frac{\Delta v}{\Delta t}$
• $\Delta t = \frac{\Delta v}{a} = \frac{v<em>{\text{final}} - v</em>{\text{initial}}}{g}$
  • $v_{\text{final}} = -20 \text{ m/s}$ (negative because it's going down)
  • $v_{\text{initial}} = 0 \text{ m/s}$
  • $g = -9.8 \text{ m/s}^2$

Supplement Information

• Practice relaunch at an angle initial velocity or the initial velocity in x and y directions.
• $v_x = v \cos(\theta)$
• $v_y = v \sin(\theta)$
• If the angle is 45 degrees, sine and cosine have the same value ($7.071$). Angle is rarely $45$.