Derive a vector trajectory from recipient-specific forces

Select a recipient and an inertial frame

A free-body diagram contains external interactions exerted on the chosen object, not the reactions that object exerts on its neighbors. Label each force by agent and recipient. A component representation resolves an interaction along axes; it must not add the original vector and its components as separate forces. For a system containing several objects, internal reciprocal forces cancel in the sum but remain relevant to individual recipients.

Separate the three Newtonian claims

An inertial frame is one in which a force-free object maintains constant velocity. Zero net force therefore permits rest or nonzero constant velocity; it does not demand zero position. Newton’s second law gives acceleration from net force and inertial mass. The third law relates opposite interactions on different recipients, so it cannot cancel one recipient’s acceleration merely because the interaction has a partner.

Integrate the stipulated vector force

Here mass is 2 kg, Fx=F0+ct and Fy=−4 N, with initial velocity (1,2) m/s and initial position (0,0). These are already net components, so do not add another weight term. Acceleration is ((F0+ct)/2,−2). Integrating gives vx=1+F0t/2+ct²/4, vy=2−2t. Integrating again gives x=t+F0t²/4+ct³/12 and y=2t−t².

Work and verify the original case

For F0=2,c=2,t=2, horizontal force is 6 N and acceleration 3 m/s². Velocity is (5,−2) m/s and position (16/3,0) m. The return to vertical coordinate zero does not mean vertical velocity or acceleration is zero. Differentiate the positions to verify both velocities rather than treating agreement at one endpoint as a complete check.

Examine a force reversal

For negative F0 and positive c, horizontal force can cross zero while horizontal velocity remains nonzero. At that instant horizontal acceleration is zero, but prior integrated acceleration still determines velocity. A force snapshot cannot establish the full history. If the chosen frame accelerates, this inertial formulation cannot be carried over unchanged.

Transfer and causal limits

For mass 3 and Fx=6t, acceleration 2t integrates to vx=−1+t², giving 3 m/s at 2 s. A reciprocal force acts on the external agent, not as an additional negative term on this object. The prescribed net force establishes a conditional trajectory; it does not identify a particular mechanism or certify that a real actuator delivers it.

Public numerical transfer practice

Try a new practice case

A 3 kg object has Fx=6t N and initial vx=−1 m/s. Find vx at 2 s.

Numerical result in the requested units Check this practice result

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Original prose, models and cases, checked 30 October 2026 against the current official AP Physics C Mechanics CED and OpenStax University Physics. Values are supplied or generated, not measured apparatus records. No physical launch, suspension, moving machinery or practical safety validation is requested. Virtual captures and public answers do not certify independent mastery or laboratory participation.