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1: Rectilinear motion

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Andersson

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Andersson

2,093 pts

6 days ago

Choose your name

Andersson

Your opponent is

Andersson

2,093 pts
6 days ago
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Rectilinear Motion

Rectilinear motion describes a particle moving along a straight line. It is the simplest form of motion in kinematics and serves as the foundation for analyzing more complex trajectories. Key parameters include position (ss), velocity (vv), and acceleration (aa), all functions of time (tt).

Position, Velocity, and Acceleration Relationships
  • Position (s(t)s(t)): Defines the particle’s location on the line.
  • Velocity (vv): The rate of change of position with time. Mathematically, v=dsdtv = \frac{ds}{dt}. Velocity indicates both speed and direction (positive/negative for forward/backward motion).
  • Acceleration (aa): The rate of change of velocity with time: a=dvdt=d2sdt2a = \frac{dv}{dt} = \frac{d^2s}{dt^2}. Acceleration can be constant or time-dependent.
Constant Acceleration Equations

For constant aa, kinematic equations relate ss, vv, aa, and tt without calculus:

  1. v=v0+atv = v_0 + at
  2. s=s0+v0t+12at2s = s_0 + v_0 t + \frac{1}{2}at^2
  3. v2=v02+2a(ss0)v^2 = v_0^2 + 2a(s - s_0)
    Here, s0s_0 and v0v_0 are initial position and velocity at t=0t = 0.
Variable Acceleration: Calculus Approach

When acceleration is not constant, use differentiation and integration:

  • Given s(t)s(t), find v(t)v(t) by differentiating: v=dsdtv = \frac{ds}{dt}.
  • Given v(t)v(t), find a(t)a(t) by differentiating: a=dvdta = \frac{dv}{dt}.
  • Given a(t)a(t), integrate to find velocity: v=adt+C1v = \int a dt + C_1.
  • Given v(t)v(t), integrate to find position: s=vdt+C2s = \int v dt + C_2.
    Constants C1C_1 and C2C_2 are determined from initial conditions (e.g., v(0)=v0v(0) = v_0, s(0)=s0s(0) = s_0).
Graphical Interpretation
  • Velocity: Slope of the ss-tt graph.
  • Acceleration: Slope of the vv-tt graph.
  • Displacement: Area under the vv-tt graph.
  • Velocity change: Area under the aa-tt graph.
Key Problem Types
  1. Motion prediction: Given a(t)a(t) and initial conditions, derive v(t)v(t) and s(t)s(t) via integration.
  2. Time/distance calculations: Use constant-acceleration equations to solve for unknowns (e.g., stopping distance given initial velocity and deceleration).
  3. Graph analysis: Extract motion parameters from ss-tt, vv-tt, or aa-tt plots.

Example: A particle starts at s0=2ms_0 = 2 \text{m} with v0=5m/sv_0 = 5 \text{m/s}, undergoing constant acceleration a=3m/s2a = -3 \text{m/s}^2. Its velocity at t=2st = 2 \text{s} is:
v=v0+at=5+(3)(2)=1m/sv = v_0 + at = 5 + (-3)(2) = -1 \text{m/s}.
The position is: s=s0+v0t+12at2=2+5(2)+12(3)(2)2=6ms = s_0 + v_0 t + \frac{1}{2}at^2 = 2 + 5(2) + \frac{1}{2}(-3)(2)^2 = 6 \text{m}.