A vector-valued function takes a number t (time) and returns a vector — the arrow from the origin to where the particle is at that instant:
That's it — it's the parametric equations from the last lesson, just written as one object instead of two. Everything happens component-wise: whatever you'd do to x(t) and y(t) separately, you do to the vector. Differentiation, integration — all of it splits cleanly into the two components.
Two reasons. First, it's compact — one symbol \(\vec r(t)\) instead of two equations. Second, the derivative of a position vector is itself a vector with a direction, which turns out to be exactly the direction the particle is moving. That geometric meaning is the whole point.
Differentiate the position vector component-by-component and you get the velocity vector — it points in the direction of motion (tangent to the path) and its length is the speed:
Differentiate again for the acceleration vector — it points toward the inside of the curve (the concave side) and tells you how the velocity is changing:
Watch it happen. The particle below moves along an ellipse. The cyan arrow is velocity — see how it always points along the path and stretches where the particle speeds up. The amber arrow is acceleration — see how it always leans toward the inside of the curve, bending the path.
• Direction — velocity always points the way the particle is
currently moving (tangent to the path). Acceleration points toward
the side the path is curving into — the concave side.
• Length — each arrow is scaled to its magnitude, so a longer
velocity means faster, and a longer acceleration
means the velocity is changing more sharply (a tighter bend).
• The counterintuitive part — the particle never moves in the
acceleration direction. Acceleration doesn't push the particle along; it
turns the velocity. That sideways pull is exactly what keeps the particle on a curved path
instead of flying off straight.
Notice the two arrows are independent: near the flat ends of the ellipse the particle moves slowly (short velocity) but curves sharply (long acceleration); along the long sides it's the opposite. Velocity is about how fast and which way; acceleration is about how that's changing.
Think you've got it? Hit 🎯 Predict mode on the visual above. The arrows hide, the particle jumps to a spot on the ellipse, and you call which way velocity and acceleration point — then reveal to check. Predicting before you see the answer is what makes it stick.
If you're given velocity and want position, integrate — again component-by-component. But an antiderivative always carries a constant, and here the constant is a vector \(\vec C=\langle C_1, C_2\rangle\):
To pin down that constant you need an initial condition — a known position at some time, usually \(\vec r(0)\). Plug it in, solve for \(C_1\) and \(C_2\), and you have the exact position function. The same logic takes you from acceleration back to velocity.
Integrating velocity gives position only up to a constant vector. Without an initial condition like \(\vec r(0)=\langle 1,4\rangle\) you can't find a unique position — leaving off \(+\vec C\) (or never solving for it) is a guaranteed lost point.
This is where the points are won or lost. Three different questions about motion, three different answers — and they are not interchangeable. (You met displacement vs. distance in 1-D back in Unit 8; this is the same idea in two dimensions.)
The distance integral is the one the AP exam asks for most — and it's almost always evaluated on a calculator, because \(\sqrt{(x')^2+(y')^2}\) rarely has a clean antiderivative. Total distance is always ≥ the magnitude of displacement (a straight line is the shortest path between two points), which is a handy sanity check.
Speed is \(|\vec v|=\sqrt{(x')^2+(y')^2}\) — not \(x'\), not \(y'\), and not \(x'+y'\). And total distance integrates that whole square root, \(\int|\vec v|\,dt\), not the components separately. Mixing up displacement \(\big(\int\vec v\,dt\big)\) with distance \(\big(\int|\vec v|\,dt\big)\) is the single most common motion error.
A particle's position is \(\vec r(t)=\langle t^2,\ t^3-3t\rangle\). Find its velocity, acceleration, and speed at \(t=2\).
Differentiate each component.
Evaluate at \(t=2\). \(\vec v(2)=\langle 4,\ 9\rangle\) and \(\vec a(2)=\langle 2,\ 12\rangle\). Speed is the magnitude of velocity: v=⟨4,9⟩, a=⟨2,12⟩, speed=√97≈9.85
A particle has velocity \(\vec v(t)=\langle 2t,\ \cos t\rangle\) and starts at \(\vec r(0)=\langle 1,\ 4\rangle\). Find its position \(\vec r(t)\).
Integrate each component. \(\int 2t\,dt=t^2+C_1\) and \(\int\cos t\,dt=\sin t+C_2\).
Use \(\vec r(0)=\langle 1,4\rangle\). At \(t=0\): \(x(0)=0+C_1=1\Rightarrow C_1=1\); \(y(0)=\sin 0+C_2=4\Rightarrow C_2=4\). So r(t)=⟨t²+1, sin t + 4⟩
A particle moves with velocity \(\vec v(t)=\langle 3t^2,\ 2t\rangle\) for \(0\le t\le 2\). How far does it travel?
Distance integrates speed.
Factor and substitute. \(\sqrt{9t^4+4t^2}=t\sqrt{9t^2+4}\) (since \(t\ge 0\)). With \(u=9t^2+4\):
Compare to displacement. Displacement \(=\int_0^2\vec v\,dt=\langle 8,4\rangle\), whose magnitude is \(\sqrt{80}\approx 8.94\) — less than the distance 9.07, exactly as it must be. distance ≈ 9.07
Five questions on velocity, acceleration, integrating back, and the speed/displacement/distance distinction.