Difficulty: Intermediate | Prerequisites: Basic vector operations, parametric equations, derivatives of vector functions
In Calculus III, you move from studying motion along a line to motion along curves in 2D and 3D. Velocity is the derivative of position, acceleration is the derivative of velocity, and the angle between them tells you whether an object is speeding up, slowing down or just changing direction. This is foundational for everything from projectile motion to planetary orbits. You should be comfortable differentiating vector-valued functions before diving in.
The angle between the acceleration vector and the velocity vector determines what is happening to the object's speed and direction. Acute angle means speeding up (with some turning), right angle means pure turning at constant speed, and obtuse angle means slowing down.
Velocity vector (v)
The derivative of the position vector r(t) with respect to time. It points tangent to the curve in the direction of motion. In simple terms, it tells you which way the object is going and how fast.
Speed
The magnitude (length) of the velocity vector, |v|. Speed is a scalar, not a vector. It tells you how fast without telling you which direction.
Acceleration vector (a)
The derivative of the velocity vector with respect to time. It describes the rate of change of velocity, capturing both changes in speed and changes in direction.
Tangential component of acceleration
The part of acceleration along the direction of motion. This component is responsible for changes in speed (speeding up or slowing down).
Normal component of acceleration
The part of acceleration perpendicular to the direction of motion. This component is responsible for changes in direction (turning).
Parametric curve / parametrisation
A curve described by giving x, y (and possibly z) as functions of a parameter t. In simple terms, you describe a path by saying where the object is at each moment in time.
The relationship between the acceleration vector and the velocity vector is the central idea. There are four cases to know.
Acceleration in the same direction as velocity (acute angle, less than 90°): the object is speeding up. The acceleration vector is "mostly along" the direction of travel. There is also a small turning component, so the object speeds up and turns a little.
Acceleration at exactly 90° to velocity: no speeding up and no slowing down. The object is turning at constant speed. This is what happens in uniform circular motion.
Acceleration in the opposite direction to velocity (obtuse angle, greater than 90°): the object is slowing down. The velocity vector's length is decreasing. There may also be a small turning component.
Zero acceleration (the zero vector): the object moves in a straight line at constant speed. No change in speed, no change in direction. This is what happens when you parametrise a line with a linear function of t.
If you draw the velocity vector at time t and then draw the velocity vector at a slightly later time, the difference between them points in the direction of acceleration. When that difference vector shrinks the velocity vector's length, the object is slowing down. When it stretches it, the object is speeding up.
When a curve is parametrised as a straight line (r(t) = r₀ + tv for constant vectors r₀ and v), the velocity is constant (v' = v, a constant) and the acceleration is the zero vector. This means constant speed, constant direction, and no turning.
Students often think acceleration always means speeding up. It does not. Acceleration can cause slowing down (when it opposes velocity) or pure turning (when it is perpendicular to velocity).
Students confuse speed with velocity. Speed is a scalar (just the magnitude). Velocity is a vector (magnitude and direction). An object can have constant speed but changing velocity if it is turning.
Students sometimes think zero acceleration means the object is stationary. Zero acceleration means constant velocity, which could be any constant speed in a straight line, including zero.
Students forget that acceleration perpendicular to velocity changes direction without changing speed. This is exactly why circular motion at constant speed still requires acceleration (centripetal acceleration).
⚠️ Expect a question asking you to determine whether an object is speeding up, slowing down or maintaining speed from a diagram of v and a vectors.
⚠️ The dot product v · a is positive when the object speeds up, zero when it turns at constant speed, and negative when it slows down. This is a quick computational check.
⚠️ Decomposing acceleration into tangential and normal components is a common exam calculation. The tangential component is (v · a)/|v| and the normal component is |v × a|/|v|.
True or false: if v · a = 0, the object is stationary.
Fill in the blank: when the angle between a and v is obtuse, the object is ______.
True or false: a straight-line parametrisation r(t) = (1 + 2t, 3 - t) has zero acceleration.
Fill in the blank: the tangential component of acceleration is responsible for changes in ______.
True or false: an object moving in a circle at constant speed has zero acceleration.
Q: An object has velocity v = ⟨3, 4⟩ and acceleration a = ⟨3, 4⟩. Is it speeding up, slowing down, or turning at constant speed?
A: Speeding up. The acceleration points in the same direction as velocity (the angle between them is 0°, which is acute). There is no normal component, so the object accelerates along a straight line.
Q: If v · a < 0, what is happening to the object's speed?
A: The object is slowing down. A negative dot product means the angle between v and a is obtuse (greater than 90°), so acceleration has a component opposing the direction of motion.
Q: Why does an object moving in a circle at constant speed still have acceleration?
A: Because the direction of velocity is constantly changing. The acceleration is entirely perpendicular to velocity (centripetal), pointing toward the centre of the circle. Speed does not change, but velocity (a vector) does.
Q: Given r(t) = ⟨t², t³⟩, find the velocity and acceleration at t = 1.
A: v(t) = r'(t) = ⟨2t, 3t²⟩, so v(1) = ⟨2, 3⟩. a(t) = v'(t) = ⟨2, 6t⟩, so a(1) = ⟨2, 6⟩. Since v · a = 4 + 18 = 22 > 0, the object is speeding up at t = 1.
Velocity and acceleration vectors connect to arc length and curvature (the normal component of acceleration involves curvature directly). They also feed into Newton's second law (F = ma) when you study motion in physics. The tangential/normal decomposition reappears in the study of TNB frames (tangent, normal, binormal vectors).
Velocity vector, acceleration vector, speed vs velocity, tangential acceleration, normal acceleration, centripetal acceleration, parametric curves, vector-valued functions, dot product and acceleration, speeding up slowing down, angle between v and a, uniform circular motion, TNB frame, Calculus III kinematics, MATH 126