Vectors & kinematics (calculus-based)
The idea
Calculus turns the kinematics you learned in high school from a formula sheet into two operations. Velocity is the time derivative of position, acceleration is the time derivative of velocity, and you recover position from acceleration by integrating twice and fixing the constants with initial conditions. The familiar constant-acceleration equations are not separate laws — they are what you get when you integrate a constant a, so they silently fail the moment a depends on time.
Vectors make this machinery work in two or three dimensions with almost no extra cost: differentiate and integrate each component independently, then reassemble magnitude and direction at the end with the Pythagorean theorem and an arctangent. The x motion never talks to the y motion; only the clock is shared. That decoupling is exactly why projectile problems split into uniform horizontal motion and free vertical fall.
A frequent trap is differentiating the speed to get acceleration. Acceleration is the derivative of the velocity vector, so an object turning at constant speed still accelerates — the direction of the vector is changing even though its magnitude is not. Whenever motion curves, expect a component of acceleration perpendicular to the velocity.
Worked example
A particle moves in the xy plane with position components x(t) = 2t³ − 3t² and y(t) = 8t, in meters with t in seconds. Find its velocity, speed, direction of travel, and acceleration at t = 2.0 s.
- Differentiate each position component once to get velocity: vx = dx/dt = 6t² − 6t and vy = dy/dt = 8, so the y velocity is constant while the x velocity changes with time.
- Evaluate at t = 2.0 s: vx = 6(4) − 6(2) = 24 − 12 = 12 m/s and vy = 8.0 m/s.
- Reassemble the vector: speed = √(12² + 8²) = √208 ≈ 14.4 m/s, pointing at arctan(8/12) ≈ 33.7° above the x axis.
- Differentiate again for acceleration: ax = dvx/dt = 12t − 6 = 12(2) − 6 = 18 m/s², and ay = 0 because vy is constant.
- Interpret the result: the acceleration is purely along x while the velocity has both components, so the velocity direction is changing and the path curves — exactly what nonparallel velocity and acceleration vectors imply.
Answer. At t = 2.0 s the velocity is (12, 8.0) m/s — speed about 14.4 m/s at 33.7° above the x axis — and the acceleration is 18 m/s² in the +x direction.
Check your understanding
- Why do the constant-acceleration equations fail when a depends on t, and what replaces them?
- How can an object have zero velocity at an instant but nonzero acceleration — what does its position graph look like there?
- What information do the initial conditions supply when you integrate acceleration back to position, and why can the math not supply it?
- If velocity and acceleration vectors are perpendicular at some instant, what is happening to the speed at that moment?
Build the foundations first
Vectors & kinematics (calculus-based) builds on these concepts. If any feel shaky, start there.