Linear acceleration is the rate of change of an object's linear (translational) velocity with time, a = dv/dt, measured in m/s². In AP Physics C Mechanics, it connects to rotational motion through a = αr, which ties a point's tangential acceleration to the object's angular acceleration.
Linear acceleration tells you how fast an object's velocity is changing as it moves through space. In calculus terms, it's the first derivative of velocity and the second derivative of position, so a = dv/dt = d²x/dt². If acceleration points the same direction as velocity, the object speeds up. If it points opposite, the object slows down. That's the whole idea, and it's the same acceleration you've used since Unit 1.
The reason this term lives in Topic 5.2 (Rotational Kinematics) is the bridge between straight-line motion and spinning motion. Every point on a rotating object traces a circle, and its tangential (linear) acceleration relates to the object's angular acceleration by a = αr, where r is the distance from the axis. So linear acceleration is the translational twin of angular acceleration. Same physics, different coordinates. When a wheel rolls without slipping, this bridge becomes an equation you'll actually use, because the acceleration of the wheel's center equals αR.
Linear acceleration shows up in Topic 5.2, Rotational Kinematics, where the CED expects you to translate between linear and angular quantities. Every rotational kinematics equation (ω = ω₀ + αt, θ = ω₀t + ½αt², and friends) is a direct copy of the linear kinematics equations with x → θ, v → ω, and a → α. If you know the linear versions cold, you already know the rotational ones. Beyond kinematics, the a = αr relationship is the hinge that connects translational dynamics (F = ma) to rotational dynamics (τ = Iα), which is exactly what rolling, pulley, and yo-yo problems test in Units 5 and 6 of Physics C Mechanics.
Keep studying AP Physics C: Mechanics Unit 5
Angular Acceleration (Unit 5)
Angular acceleration α is the rotational version of linear acceleration, measured in rad/s² instead of m/s². The two are locked together by a = αr, so a point farther from the axis has a larger linear acceleration even though the whole object shares one α.
Rolling Without Slipping (Unit 5)
Rolling without slipping is the constraint that makes a = αR an equation you must use, not just a fact. It lets you combine Newton's second law for translation (F = ma) with the rotational version (τ = Iα) and solve for both at once.
Kinematics (Unit 1)
Linear acceleration is the same a = dv/dt from your very first unit. Topic 5.2 just recycles those kinematics equations with new symbols, so a constant-α rotation problem solves exactly like a constant-a free-fall problem.
Velocity (Unit 1)
Acceleration is the derivative of velocity, so any time you're given v(t), differentiate to get a(t). The same calculus move works in rotation, where α is the derivative of ω.
You won't see "define linear acceleration" as a question; instead, the exam tests whether you can move between linear and angular descriptions of the same motion. Common MCQ setups give you ω(t) or α and ask for the tangential acceleration of a point at radius r, or give a rolling object and ask how the center's acceleration relates to α. On FRQs, the classic move is writing F = ma for translation, τ = Iα for rotation, then using a = αR (the rolling constraint) to solve the system. Watch for one trap in particular. A point on a rotating object can have linear acceleration in two perpendicular pieces, tangential (αr, from spinning faster or slower) and centripetal (ω²r, from changing direction). Constant angular velocity means zero tangential acceleration but nonzero centripetal acceleration.
Linear acceleration (a, in m/s²) measures how fast a point's velocity changes along its path. Angular acceleration (α, in rad/s²) measures how fast the whole object's spin rate changes. A rotating object has exactly one α, but different points on it have different linear accelerations because a = αr depends on distance from the axis. A spinning merry-go-round speeding up has one α, yet a kid at the edge feels more linear acceleration than a kid near the center.
Linear acceleration is the rate of change of velocity, a = dv/dt, and it's the second derivative of position with respect to time.
The bridge between linear and angular motion is a = αr, which gives the tangential acceleration of a point at distance r from the rotation axis.
Every rotational kinematics equation is the linear kinematics equation with x, v, a swapped for θ, ω, α, so you only need to memorize one set.
For rolling without slipping, the acceleration of the object's center of mass equals αR, and this constraint lets you solve F = ma and τ = Iα together.
A point moving in a circle at constant angular velocity has zero tangential (linear) acceleration but still has centripetal acceleration ω²r pointing toward the center.
Linear acceleration is the rate of change of an object's velocity, a = dv/dt, measured in m/s². In Topic 5.2 it connects to rotation through a = αr, which links a point's tangential acceleration to the object's angular acceleration.
Linear acceleration (m/s²) describes changing velocity along a path, while angular acceleration (rad/s²) describes a changing spin rate. They're related by a = αr, so points farther from the axis have larger linear acceleration even though the whole object has the same α.
No, not necessarily. Constant ω means the tangential acceleration (αr) is zero, but any point moving in a circle still has centripetal acceleration ω²r pointing toward the axis, because its velocity direction is constantly changing.
Tangential linear acceleration equals angular acceleration times the radius, a = αr. For an object rolling without slipping, the center of mass acceleration equals αR, where R is the object's radius.
For a point on a rotating object, tangential acceleration is one component of its linear acceleration; the full linear acceleration also includes the centripetal piece ω²r. For an object moving in a straight line, linear acceleration is just the ordinary a = dv/dt with no centripetal component.
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