Motion in a Plane · Lesson 7 of 13

Relative Motion in Two Dimensions

“Combine velocities across reference frames and solve apparent-rain and river-crossing problems.”

Learning Objectives

• Distinguish adding velocities across frames from subtracting object velocities. • Calculate a relative-velocity vector using components. • Apply the result to rain and moving observers. • Separate shortest-time and no-drift river crossings.

Attach an observer to every velocity

A boat’s velocity relative to water and the water’s velocity relative to land are different descriptions. To obtain the boat’s land velocity, add those two vectors. To obtain the velocity of one moving object as seen from another, subtract their velocities measured in the same frame. Labels make these operations understandable and prevent choosing a sign from memory.

Velocity addition across framesLaTeX

v_{B / G} = v_{B / W} + v_{W / G}

B is the boat, W the water and G the ground. The shared W labels connect the two frame descriptions.

Velocity relative to another objectLaTeX

v_{A / B} = v_{A / G} - v_{B / G}

Both velocities on the right are measured relative to the same ground frame. Reversing A and B changes the sign.

These rules follow by differentiating relative position: r_A/B = r_A/G − r_B/G. For ordinary speeds, time is common to these descriptions. In two dimensions, subtract x-components and y-components separately, then find magnitude and direction. The source’s rain and boat examples illustrate addition, while a moving observer requires the relative-velocity subtraction as well.

Why rain looks slanted to a moving person

Running through vertically falling rain

Problem
Rain falls vertically at 10 m/s relative to the ground. A person runs east at 6 m/s. Find the rain velocity relative to the person.

1.Choose east as +x and upward as +y. Rain ground velocity is −10 j-hat m/s; person ground velocity is +6 i-hat m/s.
2.Rain relative velocity = −6 i-hat − 10 j-hat m/s. Its speed is √136 = 11.66 m/s.
3.The rain moves down and west relative to the person and arrives from above and east. The angle with vertical is arctan(6/10) = 30.96°.
4.Tilt an umbrella toward the incoming side, east. This slant exists even though the rain has no ground-frame horizontal component.

Crossing a river: aim and outcome differ

Let x point downstream and y directly across a river. A boat aimed straight across with speed u relative to the water has velocity u j-hat. If water flows at speed w downstream, ground velocity is w i-hat + u j-hat. The crossing time depends on the across-river component, while downstream drift depends on the current acting for that same time.

Straight-across headingLaTeX

t = \frac{d}{u}, x_{drift} = w \frac{d}{u}, v_{ground} = u^{2} + w^{2}

d is river width. A straight-across heading gives the shortest crossing time for a fixed speed relative to water, if landing position is unrestricted.

A crossing with drift

Problem
A river is 120 m wide and flows at 2 m/s. A boat can move at 5 m/s relative to water and points straight across. Find crossing time and downstream displacement.

1.The across component is 5 m/s, so t = 120/5 = 24 s.
2.Downstream displacement is 2 × 24 = 48 m.
3.The ground speed is √29 = 5.39 m/s. It is not the speed to divide into river width, because part of it is along the bank.

Landing directly opposite requires upstream aiming

To cancel downstream motion, give the boat an upstream x-component of −w relative to water. If α is its angle upstream from straight across, u sinα = w. The remaining across component is u cosα = √(u² − w²). This crossing takes longer than a straight-across heading because some of the boat’s speed is spent countering current.

No-drift crossingLaTeX

sin α = \frac{w}{u}, t = \frac{d}{u ^{2} - w ^{2}}

A finite directly-opposite crossing requires u > w. At u = w, exact current cancellation leaves zero across speed; for u < w, cancellation is impossible.

Aim to cancel the current

Problem
Use the same 120 m river, current 2 m/s and boat speed 5 m/s. Find the heading and time to land opposite the start.

1.Heading angle upstream from straight across is arcsin(2/5) = 23.58°.
2.The across speed is √(25 − 4) = √21 = 4.583 m/s.
3.The time is 120/√21 = 26.19 s. Ground velocity points directly across, even though the boat’s heading points partly upstream.
4.The shortest-time and no-drift conditions answer different goals; first decide which landing condition the question requires.

Relative acceleration and encounters

Differentiating the relative-velocity relation gives a_A/B = a_A/G − a_B/G. If both objects have the same acceleration vector, their relative acceleration is zero, so their relative velocity stays constant. For an encounter, both relative position components must become zero at one common time. Two paths crossing geometrically do not guarantee a collision: the objects must reach the crossing at the same instant.

Quick check

A boat has velocity 4 j-hat m/s relative to water, while water moves at 3 i-hat m/s relative to land. What is its ground speed?

Key Takeaways

• Label the object and observer for each velocity. • Add velocities across connected frames; subtract common-frame velocities for relative motion. • Treat both components with the same time. • An object’s heading can differ from its ground-track direction. • Shortest-time and no-drift crossings impose different conditions.

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Motion in a Plane with Constant Acceleration

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Projectile Motion and the Parabolic Path