Relative motion across a river
You will be able to: Add perpendicular relative velocities and distinguish heading from ground-track direction.
Why does aiming straight across not land you directly opposite?
A boat moves north at 3 m/s relative to the water while the river flows east at 4 m/s. Relative to the bank, the boat moves northeast with speed 5 m/s.
A useful starting point: Reference frames and relative velocity →
Words and symbols before equations
- Heading
- The direction the boat points and moves relative to still water in this model.
- Ground track
- The path measured relative to the river bank.
- Perpendicular components
- Independent parts of the vector along x and y.
- Upstream
- Opposite the river flow direction.
What this picture assumes
Boat heads straight north relative to uniform water flow. Current is eastward and constant. Screen axis scales differ; use coordinates for distance and direction.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Crossing time 10 s; eastward drift 40 m. Ground velocity (4, 3) m/s; speed 5 m/s.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Use +x east along the river and +y north across it. Add v_boat/bank = v_boat/water + v_water/bank = (0, 3) + (4, 0) = (4, 3) m/s. The magnitude is 5 m/s, but that is not the cross-river component.
For a 30 m wide river, crossing time is width/vᵧ = 30/3 = 10 s. Eastward drift is vₓt = 4(10) = 40 m. The two dimensions share time; one speed magnitude cannot replace both components.
To land directly opposite, the westward boat-relative-water component must cancel the river flow. This requires boat speed greater than river speed for a positive remaining cross-river component. Pointing straight across minimizes crossing time for a fixed boat speed in a uniform current, but it does not remove drift.
A worked example, step by step
A boat can travel at 5 m/s relative to water flowing east at 3 m/s. Find a heading that gives zero drift and the crossing time for 40 m.
- Require the boat’s water-relative x component to be −3 m/s.
- With magnitude 5, its positive y component is √(25 − 9) = 4 m/s.
- Head 36.9° west of north; adding current gives ground velocity (0, 4) m/s.
- Crossing time = 40/4 = 10 s, with zero east–west drift.
The magnitude of the ground velocity is not automatically the cross-river speed.
Does a stronger current change crossing time when the northward boat speed stays fixed?
Compare with an explanation
No in this uniform-current model; it changes eastward drift and ground speed.
Predict. Change one thing. Explain.
Keep the boat pointed north. Change only river speed. Predict which of crossing time, drift and ground speed will change.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Crossing time 10 s; eastward drift 40 m. Ground velocity (4, 3) m/s; speed 5 m/s.
Boat heads straight north relative to uniform water flow. Current is eastward and constant. Screen axis scales differ; use coordinates for distance and direction.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant position, velocity, acceleration or integral relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA boat heads north at 4 m/s relative to water, with current 3 m/s east and river width 24 m. Find ground velocity, speed, crossing time and drift.
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Compare with the answer and four-point rubric
- 1 point: Ground velocity is (3, 4) m/s.
- 1 point: Ground speed is 5 m/s.
- 1 point: Time is 24/4 = 6 s.
- 1 point: Drift is 3(6) = 18 m east.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Which component determines crossing time?
The component perpendicular to the river banks.
RECALL 2How do you cancel drift?
Aim upstream so the boat’s current-opposing component cancels the current.
RECALL 3What links the two component motions?
The same elapsed time.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Relative motion across a river
- v_boat/bank = v_boat/water + v_water/bank.
- Crossing time = width/vᵧ for constant positive vᵧ.
- Drift = vₓ × crossing time.
Remember: The magnitude of the ground velocity is not automatically the cross-river speed.
Conditions: Boat heads straight north relative to uniform water flow. Current is eastward and constant. Screen axis scales differ; use coordinates for distance and direction.
Refresh Kid · AP Physics C: Mechanics Unit 1 (official Unit 1) · Objectives 1.4.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 1.4, objectives 1.4.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is Mechanics Unit 1: Kinematics. Quantitative motion problems stay in one or two dimensions; the optional spatial view clarifies vector notation. Students should study calculus alongside this course. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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