Physics Summative #1

1D Motion

Key Concepts:

  • Distance: Total length of the path traveled (scalar).

  • Displacement: The straight-line distance between the starting and ending points, including direction (vector).

Example: If you walk 3 meters east and then 4 meters west, your distance is 7 meters, but your displacement is 1 meter west.

Speed and Velocity

  • Speed: How fast an object moves, calculated as speed=distance/time (scalar)

  • Velocity: Speed with direction, calculated as velocity=displacement/time (vector)

Example: A car moving 60 km/h east has a velocity of 60 km/h east.

Acceleration

  • Definition: The rate of change of velocity over time, Acceleration = deltaV/deltaT

Example: A car accelerates from 0 to 20 m/s in 5 seconds. The acceleration is:

a=20−05=4 m/s2a = \frac{20 - 0}{5} = 4 \, \text{m/s}^2a=520−0​=4m/s2

Position and Velocity Graphs

  1. Position vs. Time (xxx-ttt):

    • Slope = velocity.

    • A straight line = constant velocity.

    • A curved line = changing velocity (acceleration).

  2. Velocity vs. Time (vvv-ttt):

    • Slope = acceleration.

    • Area under the curve = displacement.

Kinematics Equations

These equations describe 1D motion under constant acceleration:

  1. v=v0+atv = v_0 + atv=v0​+at

  2. x=x0+v0t+12at2x = x_0 + v_0t + \frac{1}{2}at^2x=x0​+v0​t+21​at2

  3. v2=v02+2a(x−x0)v^2 = v_0^2 + 2a(x - x_0)v2=v02​+2a(x−x0​)

Where:

  • vvv: final velocity

  • v0v_0v0​: initial velocity

  • x,x0x, x_0x,x0​: final and initial position

  • aaa: acceleration

  • ttt: time

Freefall

  • Objects near Earth’s surface accelerate downward at g≈9.8 m/s2g \approx 9.8 \, \text{m/s}^2g≈9.8m/s2.

  • Equations of motion apply, with a=−ga = -ga=−g.

Example: Dropping a ball from rest at a height of 20 m:

x=x0+v0t−12gt2x = x_0 + v_0t - \frac{1}{2}gt^2x=x0​+v0​t−21​gt2

Solve for ttt to find how long it takes to hit the ground.

Vectors

  • Vectors have magnitude and direction (e.g., velocity, displacement).

  • Represented graphically by arrows.

Basic Operations:

  1. Use the Pythagorean Theorem to find the magnitude: ∣A⃗∣=Ax2+Ay2|\vec{A}| = \sqrt{A_x^2 + A_y^2}∣A∣=Ax2​+Ay2​​

  2. Use trigonometry to find direction: θ=tan⁡−1(AyAx)\theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)θ=tan−1(Ax​Ay​​)

Vector Addition

  • Add components: Rx=Ax+Bx,Ry=Ay+ByR_x = A_x + B_x, \quad R_y = A_y + B_yRx​=Ax​+Bx​,Ry​=Ay​+By​

  • Combine into a resultant vector using Pythagoras and trigonometry.

2D Motion

  • Break motion into x and y components.

  • Horizontal (xxx) motion:

    • Constant velocity (vx=v0cos⁡θv_x = v_0 \cos \thetavx​=v0​cosθ).

  • Vertical (yyy) motion:

    • Accelerated motion (a=−ga = -ga=−g).

Projectile Motion:

  • Combine x and y motions to describe parabolic trajectories.

Key Equations:

  1. Horizontal displacement: x=v0cos⁡θ⋅tx = v_0 \cos\theta \cdot tx=v0​cosθ⋅t

  2. Vertical displacement: y=v0sin⁡θ⋅t−12gt2y = v_0 \sin\theta \cdot t - \frac{1}{2} g t^2y=v0​sinθ⋅t−21​gt2

Relative Motion

  • Combine velocities of objects relative to each other.

  • Use vector addition to solve problems.

Example: A boat crossing a river with a current:

  • Boat’s velocity: v⃗boat\vec{v}_{\text{boat}}vboat​.

  • River’s velocity: v⃗river\vec{v}_{\text{river}}vriver​.

  • Resultant velocity: v⃗boat,ground=v⃗boat+v⃗river\vec{v}_{\text{boat,ground}} = \vec{v}_{\text{boat}} + \vec{v}_{\text{river}}vboat,ground​=vboat​+vriver​.