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 (s), velocity (v), and acceleration (a), all functions of time (t).
Position, Velocity, and Acceleration Relationships
- Position (s(t)): Defines the particle’s location on the line.
- Velocity (v): The rate of change of position with time. Mathematically, v=dtds. Velocity indicates both speed and direction (positive/negative for forward/backward motion).
- Acceleration (a): The rate of change of velocity with time: a=dtdv=dt2d2s. Acceleration can be constant or time-dependent.
Constant Acceleration Equations
For constant a, kinematic equations relate s, v, a, and t without calculus:
- v=v0+at
- s=s0+v0t+21at2
- v2=v02+2a(s−s0)
Here, s0 and v0 are initial position and velocity at t=0.
Variable Acceleration: Calculus Approach
When acceleration is not constant, use differentiation and integration:
- Given s(t), find v(t) by differentiating: v=dtds.
- Given v(t), find a(t) by differentiating: a=dtdv.
- Given a(t), integrate to find velocity: v=∫adt+C1.
- Given v(t), integrate to find position: s=∫vdt+C2.
Constants C1 and C2 are determined from initial conditions (e.g., v(0)=v0, s(0)=s0).
Graphical Interpretation
- Velocity: Slope of the s-t graph.
- Acceleration: Slope of the v-t graph.
- Displacement: Area under the v-t graph.
- Velocity change: Area under the a-t graph.
Key Problem Types
- Motion prediction: Given a(t) and initial conditions, derive v(t) and s(t) via integration.
- Time/distance calculations: Use constant-acceleration equations to solve for unknowns (e.g., stopping distance given initial velocity and deceleration).
- Graph analysis: Extract motion parameters from s-t, v-t, or a-t plots.
Example: A particle starts at s0=2m with v0=5m/s, undergoing constant acceleration a=−3m/s2. Its velocity at t=2s is:
v=v0+at=5+(−3)(2)=−1m/s.
The position is: s=s0+v0t+21at2=2+5(2)+21(−3)(2)2=6m.