Introduction
One-dimensional motion along a straight line provides the simplest introduction to kinematics. By restricting motion to a single axis, vector equations simplify to scalar equations with sign indicating direction. Understanding 1D motion thoroughly is essential before progressing to higher dimensions.
Uniform Motion
Uniform motion occurs when velocity is constant (acceleration = 0). The position equation is x(t) = x0 + vt where x0 is initial position and v is constant velocity. The x-t graph is a straight line with slope v. Distance traveled equals |v|t. Area under v-t graph gives displacement. This simplest case applies to objects moving at steady speed along straight paths.
Uniformly Accelerated Motion
When acceleration a is constant, we derive the standard kinematic equations: (1) v = v0 + at (velocity as function of time), (2) x = x0 + v0t + 1/2at2 (position as function of time), (3) v2 = v02 + 2a(x - x0) (velocity as function of position). These apply to freely falling bodies (a = -g = -9.8 m/s2 near Earth) and many other physical situations.
\nIntroduction
One-dimensional motion along a straight line provides the simplest introduction to kinematics. By restricting motion to a single axis, vector equations simplify to scalar equations with sign indicating direction. Understanding 1D motion thoroughly is essential before progressing to higher dimensions.
Uniform Motion
Uniform motion occurs when velocity is constant (acceleration = 0). The position equation is x(t) = x0 + vt where x0 is initial position and v is constant velocity. The x-t graph is a straight line with slope v. Distance traveled equals |v|t. Area under v-t graph gives displacement. This simplest case applies to objects moving at steady speed along straight paths.
Uniformly Accelerated Motion
When acceleration a is constant, we derive the standard kinematic equations: (1) v = v0 + at (velocity as function of time), (2) x = x0 + v0t + 1/2at2 (position as function of time), (3) v2 = v02 + 2a(x - x0) (velocity as function of position). These apply to freely falling bodies (a = -g = -9.8 m/s2 near Earth) and many other physical situations.
\nFree Fall Motion
Free fall is motion under gravity alone with acceleration g ≈ 9.8 m/s2 downward. Taking upward as positive: a = -g. An object thrown upward has decreasing velocity, reaches zero velocity at maximum height, then accelerates downward. Time to reach max height: t = v0/g. Maximum height: h = v02/(2g). Time of flight for object returning to launch level: T = 2v0/g.
Variable Acceleration
When acceleration varies with time a(t), position and velocity require integration: v(t) = v0 + ∫0ᵗ a(t')dt' and x(t) = x0 + ∫0ᵗ v(t')dt'. If acceleration depends on position a(x), use chain rule: a = dv/dt = (dv/dx)(dx/dt) = v(dv/dx). Then ∫v dv = ∫a(x)dx giving 1/2v2 - 1/2v02 = ∫a(x)dx. This technique solves problems with position-dependent forces.
\nMotion Diagrams
Motion diagrams visualize particle motion by showing position at equal time intervals. Dots represent particle position; arrows show velocity vectors. Spacing between dots indicates speed (closer = slower). Direction of arrows shows motion direction. Acceleration shown by change in arrow length and direction between successive positions. Essential tool for qualitative analysis before quantitative calculations.
Solved Example: Car Braking
A car moving at 20 m/s brakes with constant acceleration of -5 m/s2. Find stopping distance and time. Solution: Given v0 = 20 m/s, v = 0, a = -5 m/s2. Using v = v0 + at: 0 = 20 - 5t, so t = 4 s. Using v2 = v02 + 2a(x-x0): 0 = 400 + 2(-5)(x-x0), so x-x0 = 400/10 = 40 m. The car stops in 4 seconds after traveling 40 m. Verify using x = x0 + v0t + 1/2at2: x-x0 = 20(4) + 1/2(-5)(16) = 80 - 40 = 40 m.
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