Two-Dimensional Motion: Quick Notes
Two-Dimensional Motion: Independence of x and y
- Base equations and relationships in 2D/3D are unchanged; x and y components act independently.
- Treat as two separate 1D problems (one for x, one for y) and combine results vectorially when needed.
Decomposing Motion into Components
- Break any 2D problem into its x and y parts.
- Solve each part with its own initial velocity and acceleration, then recombine.
- Final velocity components: vx,\; vy
- Magnitude: v = \sqrt{vx^2 + vy^2}
- Direction: \theta = \tan^{-1}\left(\frac{vy}{vx}\right)
Projectile Motion Basics (gravity only)
- ax = 0,\quad vx(t) = v_{0x}
- ay = -g,\quad vy(t) = v_{0y} - g t
- y(t) = v_{0y} t - \tfrac{1}{2} g t^2
- g \approx 9.8\ \mathrm{m/s^2}
Example Problem: Separate x and y (7 s)
- Given: v{0x}=22\ \mathrm{m/s},\; ax=24\ \mathrm{m/s^2};\quad v{0y}=14\ \mathrm{m/s},\; ay=12\ \mathrm{m/s^2};\; t=7\ \mathrm{s}
- Displacements:
- x = v{0x} t + \tfrac{1}{2} ax t^2
- y = v{0y} t + \tfrac{1}{2} ay t^2
- Final components:
- vx = v{0x} + a_x t = 190\ \mathrm{m/s}
- vy = v{0y} + a_y t = 98\ \mathrm{m/s}
- Final speed: v = \sqrt{vx^2 + vy^2} \approx 214\ \mathrm{m/s}
- Direction: \theta = \tan^{-1}\left(\frac{vy}{vx}\right) \approx 27.5^{\circ}
Final Velocity Vector (components)
- Velocity components: vx = 190\ \mathrm{m/s},\quad vy = 98\ \mathrm{m/s}
- Vector form can give magnitude and angle as above
Time of Flight and Range: Horizontal drop example
- Scenario: plane speed v_{0x}=115\ \mathrm{m/s}, height H=1050\ \mathrm{m}
- Time in air: t = \sqrt{\dfrac{2H}{g}} \approx 14.6\ \mathrm{s}
- Final velocities: vx = 115\ \mathrm{m/s},\quad vy = -g t \approx -143\ \mathrm{m/s}
- Final speed: v = \sqrt{115^2 + 143^2} \approx 184\ \mathrm{m/s}
- Flight angle: \theta = \tan^{-1}\left(\dfrac{vy}{vx}\right) \approx -51.2^{\circ}
- Horizontal range: R = v_x t \approx 1.68 \times 10^3\ \mathrm{m}
Strategy and Tips for 2D Motion Problems
- Draw a diagram; set positive directions clearly.
- Treat x and y as separate problems; verify at least three kinetic values per component.
- Break any initial velocity given as magnitude + angle into components: v{0x} = v0\cos\theta,\; v{0y} = v0\sin\theta
- If motion occurs in segments (varying accelerations), solve per segment and chain results.
- Combine x and y results to obtain the final vector (magnitude and direction).
- Note: air resistance is neglected in these baseline problems; real motion may differ.
Quick takeaway
- Any 2D motion can be analyzed by independently solving x(t) and y(t) using their own initial conditions and accelerations, then recombining to get the overall motion.