Mathfolis

Motion with Vectors

Unit 9 · Parametric, Polar & Vector

Solving Motion Problems with Vectors

For a particle moving in the plane, velocity v(t) = r′(t), speed = |v(t)| = √((x′)² + (y′)²), and acceleration a(t) = r″(t). The particle speeds up when v and a point in the same direction (v·a > 0) and slows down when they point in opposite directions (v·a < 0).

Key Motion Quantities

Velocity vector
v⃗(t)=⟨x′(t),  y′(t)⟩\vec{v}(t) = \langle x'(t),\; y'(t) \rangle
Speed (scalar)
∣v⃗(t)∣=[x′(t)]2+[y′(t)]2|\vec{v}(t)| = \sqrt{[x'(t)]^2+[y'(t)]^2}
Total distance
∫t1t2∣v⃗(t)∣ dt\int_{t_1}^{t_2}|\vec{v}(t)|\,dt
Displacement vector
∫t1t2v⃗(t) dt=r⃗(t2)−r⃗(t1)\int_{t_1}^{t_2}\vec{v}(t)\,dt = \vec{r}(t_2)-\vec{r}(t_1)
Caution: Displacement (magnitude of net position change) ≠ total distance traveled. They are equal only if the particle never reverses direction.
AP Tip: To determine speeding up/slowing down: compute v·a. Positive → speeding up; negative → slowing down.
Type 1

Velocity Vector and Speed

Differentiate position to get the velocity vector, then compute its magnitude for speed.

Example 1

r(t) = ⟨t² − 1, 2t³⟩. Find velocity and speed at t = 1.

Example 2

x(t) = t² − 4t, y(t) = t³ − 3t. Find velocity and speed at t = 2.

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Type 2

Displacement vs. Total Distance

Displacement is ∫v dt (a vector). Total distance is ∫|v| dt (a scalar, always ≥ 0).

Example 3

v(t) = ⟨2t − 4, 1⟩ for t ∈ [0, 4]. Find (a) displacement, (b) total distance integral.

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Type 3

Speeding Up or Slowing Down

Compute the dot product v·a. If v·a > 0, the particle is speeding up; if v·a < 0, it is slowing down.

Example 4

At t = 1: v(1) = ⟨−1, 3⟩ and a(1) = ⟨1, 0⟩. Is the particle speeding up or slowing down?

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