Giải Các Bài Toán Chuyển Động Sử Dụng Phương Trình Tham Số và Hàm Giá Trị Vector

Full planar motion, start to finish

• Now combine everything: a particle moving in the plane, described by $\mathbf{r}(t)=\langle x(t),y(t)\rangle$.
• Differentiate for velocity and acceleration; integrate to go the other way.
• Compute speed and distance traveled from the velocity components.
• This lesson is the toolbox for any BC plane-motion problem.

Position, velocity, acceleration

• Position: $\mathbf{r}(t)=\langle x(t),\,y(t)\rangle$.
• Velocity: $\mathbf{v}(t)=\langle x'(t),\,y'(t)\rangle$ — direction of motion, tangent to the path.
• Acceleration: $\mathbf{a}(t)=\langle x''(t),\,y''(t)\rangle$.
• Move up the chain by differentiating, down by integrating with initial conditions.

The position vector points to the particle; the velocity vector is tangent to its path

Speed and distance traveled

• Speed at time $t$: $|\mathbf{v}(t)|=\sqrt{\big(x'(t)\big)^2+\big(y'(t)\big)^2}$ — a scalar.
• Distance traveled over $[a,b]$: integrate the speed, $\displaystyle\int_a^b |\mathbf{v}(t)|\,dt$.
• That's the same arc-length integral — total path length, never negative.
• Speed answers "how fast right now"; the integral answers "how far overall."

Position at a later time

• To find where the particle is at time $b$: start from a known position and add the accumulated change.
• $x(b)=x(a)+\displaystyle\int_a^b x'(t)\,dt$, and likewise for $y$ — each component separately.
• This is "final = initial + accumulated change," done per coordinate.
• Combine the two results into the position vector at time $b$.