Motion as Derivatives

You’ll be able to

Define velocity and acceleration as the first and second derivatives of position
Differentiate a polynomial position function x(t) to find v(t) and a(t)
Interpret the signs of v and a to decide whether an object speeds up or slows down

Velocity is the rate of change of position

Average velocity is just displacement over a time interval, Δx / Δt. But motion rarely happens at one steady rate — you want the velocity right now. Shrink the interval toward zero and the average becomes the instantaneous velocity, which is exactly the derivative of position with respect to time. Geometrically, v(t) is the slope of the tangent line to the x-versus-t graph at that instant.

Velocity as a derivative

v(t) = dx/dt = lim (Δt→0) Δx / Δt

The slope of the position–time graph at an instant. Units: meters per second.

Acceleration is the derivative of velocity

Apply the same idea one level up. Acceleration is how fast velocity itself is changing, so it is the derivative of v — and therefore the second derivative of position. A large acceleration means velocity is changing quickly, whether the object is speeding up, slowing down, or turning.

Acceleration as a derivative

a(t) = dv/dt = d²x/dt²

First derivative of velocity; second derivative of position. Units: meters per second².

Worked example

A particle moves along a line with x(t) = 3t³ − 2t² + 5t (meters, seconds). Find v(t) and a(t), and evaluate both at t = 2 s.

1.Differentiate x term by term using the power rule d/dt(tⁿ) = n·tⁿ⁻¹: v(t) = dx/dt = 9t² − 4t + 5.
2.Differentiate again to get acceleration: a(t) = dv/dt = 18t − 4.
3.Evaluate the velocity at t = 2: v(2) = 9(2²) − 4(2) + 5 = 36 − 8 + 5 = 33 m/s.
4.Evaluate the acceleration at t = 2: a(2) = 18(2) − 4 = 36 − 4 = 32 m/s².

Answer: v(t) = 9t² − 4t + 5, a(t) = 18t − 4; at t = 2 s, v = 33 m/s and a = 32 m/s²

Checkpoint

A particle has position x(t) = 4t² − 3t (meters, seconds). What is its velocity at t = 3 s?

Speeding up versus slowing down

A common trap: a negative acceleration does not always mean slowing down. What matters is whether a and v share a sign. If a and v point the same way, the object speeds up; if they point opposite ways, the object slows down. Acceleration only sets how velocity is changing, not the direction of travel.

Checkpoint

At one instant a particle has velocity v = −4 m/s and acceleration a = +2 m/s². Which best describes its motion right then?

On the exam

On the AP exam, given x(t) you should be able to produce v(t) and a(t) instantly by differentiating, and read the reverse too: where the x–t slope is zero the object is momentarily at rest, and where the v–t slope is zero the acceleration is zero.

Answer the 2 checkpoints as you read.