Differentiate a vector trajectory and distinguish its directions

Start with quantities, not arrows of identical meaning

A scalar such as speed has a magnitude; a vector such as velocity has magnitude and direction. In a fixed Cartesian basis, î and ĵ are dimensionless unit vectors, and r=xî+yĵ locates a point relative to an origin. A third component z k̂ extends the notation, but this course’s quantitative motion cases are planar. Adding vectors means adding matching components, not adding their magnitudes.

Differentiate a trajectory

The supplied path is r(t)=Atî+Bt²ĵ, with A in m/s and B=1 m/s². Differentiate componentwise: v=Aî+2Btĵ and a=2Bĵ. Each derivative reduces one power of time and changes the units appropriately. Position points from the origin; velocity is tangent to the path; acceleration describes how velocity changes. These directions need not coincide.

Normalize only a nonzero vector

A unit vector in a vector’s direction is that vector divided by its magnitude. At A=3 and t=1, r=(3,1) m, so r̂=(3,1)/√10. Velocity is (3,2) m/s, so its direction is (3,2)/√13, not r̂. At t=0 the position is zero and r̂ is undefined. The velocity remains (3,0), so its direction still exists. Omitting an undefined result is more honest than assigning an arbitrary angle.

Differentiate speed rather than guessing

Speed is |v|=√(A²+4B²t²). Its derivative is 4B²t/√(A²+4B²t²), equivalently (v·a)/|v|. At the reference point this is 4/√13≈1.1094 m/s², whereas |a|=2 m/s². The difference reflects acceleration changing direction as well as speed. At t=0 speed has zero instantaneous rate of change even though acceleration is nonzero.

Transfer and evidence

For r=4tî+1.5t²ĵ, v(1)=(4,3) and speed is 5 m/s. Explain why the position vector (4,1.5) would give a wrong speed. In a supplied trajectory, differentiating correctly does not establish what forces produced it. The plot’s axes may scale independently; use components and units to justify angles.

Construct equal-scale arrows before reading their lengths

On axes with the same physical scale in both directions, draw displacement u=(3,0) m as a 3 cm arrow and v=(0,4) m as a 4 cm arrow using 1 cm per metre. Place v head-to-tail after u without rotating it. Predict the resultant endpoint and arrow length before checking (3,4) m and 5 cm. The vector sum is not 7 m. Reversing v gives (3,−4) m with the same 5 cm magnitude but a different direction. A diagram with independently stretched axes cannot support this ruler comparison; reconstruct an equal-scale drawing or calculate components instead.

Public numerical transfer practice

Try a new practice case

For r(t)=4t i+1.5t² j in SI units, find the speed at t=1 s.

Numerical result in the requested units Check this practice result

This public, unsaved check is not a secure assessment. Explain the derivative, integral or boundary in the retained course evidence workspace.

Sources and evidence boundary

Original explanations and cases, checked 30 October 2026 against the official AP Physics C Mechanics framework, OpenStax calculus kinematics and vector kinematics. All numerical records are supplied or generated, not actual laboratory measurements. No physical launch, drop or moving apparatus is requested. Public feedback is practice, not independent certification.