Motion in Two Dimensions: 10 Applications, Worked Step by Step

A Ferry That Must Land at the Slipway Directly Opposite

methodChoose the Heading So That the Upstream Component Cancels the Current, Leaving a Resultant Straight Across

A river 400 m wide flows due east at 3 m/s. A ferry moves at 5 m/s through the water, and it must land at the slipway directly north of the one it leaves. Take east and north as the components. (a) Find the velocity the ferry must have through the water, and the angle it must be steered from the straight-across direction. (b) How long does the crossing take?

Take east and north as the components. The current is 30 m/s, and the ferry moves at 5 m/s through the water. Each arrow drawn is 100 seconds of the velocity beside it.

Take east and north as the components. The current is 30 m/s, and the ferry's velocity through the water is xy m/s, where x² + y² = 5² because the ferry moves at 5 m/s through the water.

step 1 of 5

The ferry's velocity over the ground is its velocity through the water added to the velocity of the current. Landing directly opposite means the east component of that sum is zero, and that condition fixes the heading; the north component left over then gives the crossing time.

Take east and north as the components. The current is 30 m/s, and the ferry's velocity through the water is xy m/s, where x² + y² = 5² because the ferry moves at 5 m/s through the water.
The velocity over the ground is the sum of the two, x + 3y m/s. Landing at the slipway directly opposite means the ferry never moves east or west at all, so x + 3 = 0 and x = −3: the ferry must be steered upstream.
Put x = −3 into x² + y² = 25: 9 + y² = 25, so y² = 16 and y = 4. The positive root is the one to take, because the ferry crosses toward the north.
(a) The ferry must be steered with velocity −34 m/s through the water. The angle θ from the straight-across direction satisfies sin θ = 35 = 0.6, so θ = 36.9^° upstream of straight across.
The velocity over the ground is −34 + 30 = 04, which is 4 m/s straight north. (b) The crossing takes 4004 = 100 s. Check: in 100 s the ferry is carried 3 × 100 = 300 m east by the river and swims 3 × 100 = 300 m west through the water, so the two cancel.

answer(a) −34 m/s through the water, steered 36.9^° upstream of straight across; (b) 100 s

techniquePosition and Velocity as Vectors

Common pitfalls

Steering straight across and then adding the current on, giving a resultant speed of √3² + 5². That is the answer to a different question. A ferry pointed straight across is carried downstream, so it does not land at the slipway opposite, and the 5 m/s is the speed through the water, not the speed across the river.
Taking the crossing time as 4005= 80 s. The 5 m/s is spread between going upstream and going across; only the 4 m/s north carries the ferry toward the far bank, so the time must be found from that component alone.

Two Ships on Straight Courses: Whether They Collide, and How Close They Come

methodSubtract the Position Vectors for the Gap, Test Both Components for a Collision, and Make the Gap Perpendicular to the Relative Velocity for the Closest Approach

At noon a coaster A is at the origin and a tanker B is at 4020, in kilometers east and north of a harbor. A sails with velocity 2128 km/h and B with velocity 912 km/h, both steady. Let m be the number of hours after noon. (a) Show that the two ships do not collide. (b) Find their closest approach and the time at which it happens.

At m hours after noon, r_A = m2128 and r_B = 4020 + m912, in kilometers east and north.

The positions m hours after noon are r_A = m2128 and r_B = 4020 + m912.

step 1 of 5

Write each ship's position m hours after noon, and subtract to get the vector from one to the other. A collision needs both components of that vector to vanish at the same instant. The ships are closest when the gap is perpendicular to the relative velocity, which is a dot product set to zero.

The positions m hours after noon are r_A = m2128 and r_B = 4020 + m912.
Subtract to get the vector from A to B: AB = r_B − r_A = 40 − 12m20 − 16m. The relative velocity is −12−16 km/h, so B closes on A at 20 km/h.
(a) A collision needs both components to be zero at the same time. The east component gives 40 − 12m = 0, that is m = 103, and the north component gives 20 − 16m = 0, that is m = 1.25. The two times are different, so the ships never occupy the same point and they do not collide.
The ships are closest when the gap is perpendicular to the relative velocity, so the dot product is zero: (40 − 12m)(−12) + (20 − 16m)(−16) = 0, which is −800 + 400m = 0 and m = 2.
(b) At m = 2 the gap is 40 − 2420 − 32 = 16−12, of length √16² + 12² = 20 km, at 2 pm. Check: at m = 1 the gap is 284 and at m = 3 it is 4−28, both of length √800 ≈ 28.3 km, which is more.

answer(a) they do not collide, because the east components agree only at m = 103 hours and the north components only at m = 1.25 hours; (b) 20 km apart, 2 hours after noon

techniqueThe Closest Approach of Two Objects · Meeting Points and Travel Times

Common pitfalls

Solving one component only, finding a time, and announcing a collision. One component agreeing means the ships share a line of longitude or of latitude at that instant, not a point. Both components must give the same time.
Minimizing the gap by trying whole numbers of hours and picking the smallest. That can only land on the answer by luck: the least gap need not fall on a whole hour, so the perpendicular condition, or the minimum of the quadratic, is what settles it.

Two Aircraft on One Radar Screen, and the Separation They Must Keep

methodWrite the Square of the Separation as a Quadratic in the Time and Complete the Square, Then Confirm the Same Minimum by Differentiating

A radar screen gives the positions of two aircraft, in kilometers east and north of the station, m minutes from now, as r_P = m120 and r_Q = 500 + m68. (a) Find the least distance between the two aircraft and the time at which it happens. (b) The controller must keep them at least 25 km apart. Is the rule broken, and by what margin is it kept or broken?

The separation vector is r_Q − r_P = 50 − 6m8m kilometers, and at m = 0 the aircraft are 50 km apart.

The separation vector is r_Q − r_P = 50 − 6m8m kilometers, so at this moment, with m = 0, the aircraft are 50 km apart.

step 1 of 5

The separation vector is one position minus the other. Its length involves a square root, so work with the square of the length: that is a quadratic in the time, and completing the square reads the least value and where it happens straight off. Differentiating is the second route to the same time.

The separation vector is r_Q − r_P = 50 − 6m8m kilometers, so at this moment, with m = 0, the aircraft are 50 km apart.
Square its length: d² = (50 − 6m)² + (8m)² = 2500 − 600m + 36m² + 64m² = 100m² − 600m + 2500.
Complete the square: d² = 100(m² − 6m + 25) = 100[(m − 3)² + 16] = 100(m − 3)² + 1600. A square is never negative, so the least value of d² is 1600, and it is reached when m = 3.
(a) The least distance is √1600 = 40 km, 3 minutes from now. Differentiating is the second route to the same time: ddm(d²) = 200m − 600, which is zero at m = 3, and the second derivative 200 is positive, so that stationary value is a minimum.
(b) The least separation, 40 km, is more than 25 km, so the rule is not broken: it is kept with 40 − 25 = 15 km to spare. Check: at m = 2 the separation is 3816, of length √1700 ≈ 41.2 km, which is more than 40 km.

answer(a) 40 km, 3 minutes from now; (b) the rule is not broken, and it is kept with 15 km to spare on the 25 km

techniqueThe Closest Approach of Two Objects · Position and Velocity as Vectors

Common pitfalls

Minimizing 50 − 6m and 8m one at a time. Each component is smallest at its own time, and neither is the time when the distance itself is least; the two components must be combined into one expression before anything is minimized.
Taking the square root first and differentiating √100m² − 600m + 2500. That is correct but needs the chain rule for no gain, because a square root increases with what is under it, so the time that makes d² least is the time that makes d least.

A Coastguard Launch That Leaves Twenty Minutes After the Boat It Is Sent To

methodMeasure Both Motions from One Clock by Writing the Later Start as a Position Vector in $m - 20$, Then Match Both Components

A cabin cruiser leaves a harbor at 09:00, and m minutes later its position, in kilometers east and north of the harbor, is r_C = 21 + m0.30.4. A coastguard launch leaves a station at −25 at 09:20 with steady velocity 0.80.6 km per minute. (a) Write the launch's position vector for m ≥ 20, and find where it is at 09:30. (b) Show that the launch reaches the cruiser, and find when and where.

With m the minutes after 09:00, the cruiser is at 21 + m0.30.4, in kilometers east and north of the harbor.

Keep m as the minutes after 09:00 for both vessels. At time m the launch has been under way for m − 20 minutes, so r_L = −25 + (m − 20)0.80.6 for m ≥ 20.

step 1 of 5

One clock must serve both vessels. The launch has been under way for twenty minutes less than the cruiser, so its position is written with m − 20 in place of m. The launch reaches the cruiser only if both components agree at the same value of m.

Keep m as the minutes after 09:00 for both vessels. At time m the launch has been under way for m − 20 minutes, so r_L = −25 + (m − 20)0.80.6 for m ≥ 20.
(a) At 09:30, m = 30 and m − 20 = 10, so r_L = −25 + 100.80.6 = 611: the launch is 6 km east and 11 km north of the harbor.
The launch reaches the cruiser when both components agree. East: 2 + 0.3m = −2 + 0.8(m − 20), which is 2 + 0.3m = 0.8m − 18, so 0.5m = 20 and m = 40.
Test the north components at m = 40. The cruiser: 1 + 0.4 × 40 = 17. The launch: 5 + 0.6 × 20 = 17. They agree, so the launch really does reach the cruiser and does not merely cross its wake.
(b) They meet at 09:40, 14 km east and 17 km north of the harbor. Check: the launch covers 1612, a distance of √16² + 12² = 20 km, in 20 minutes, which agrees with its speed of √0.8² + 0.6² = 1 km per minute.

answer(a) r_L = −25 + (m − 20)0.80.6, and at 09:30 the launch is at 611; (b) they meet at 09:40, at 1417

techniqueMotion That Starts Later · Meeting Points and Travel Times

Common pitfalls

Writing the launch's position as −25 + m0.80.6. That puts the launch at its station at 09:00, twenty minutes before it sails, and the meeting it predicts is too early. The later start is what m − 20 carries.
Solving the east components for m and stopping there. Two straight paths that are not parallel always have one value of m that matches one component, so the north components have to be tested as well before a meeting can be claimed.

A Survey Drone Whose Velocity Changes as It Flies

methodRead the Instant from the Component That Vanishes, and Integrate Each Component of the Velocity for the Position

A survey drone leaves a mast and flies with velocity v = 3m4 − m m/s, where m is the number of seconds since it left and the components are east and north. (a) At what time is the drone flying due east? (b) Find its displacement from the mast at that time, and explain why the drone is then as far north as it ever gets.

The velocity is v = 3m4 − m m/s. (a) Flying due east means the north component is zero: 4 − m = 0, so m = 4 s.

Flying due east means the drone has no northward or southward velocity at that instant, so the north component is zero: 4 − m = 0. (a) The drone flies due east 4 seconds after leaving the mast.

step 1 of 5

Flying due east means the north component of the velocity is zero, which is an equation in the time alone. The position is the integral of the velocity, taken component by component, with the constant fixed by the drone starting at the mast.

Flying due east means the drone has no northward or southward velocity at that instant, so the north component is zero: 4 − m = 0. (a) The drone flies due east 4 seconds after leaving the mast.
The position is the integral of the velocity, component by component: east ∫ 3m dm = 1.5m² + c₁ and north ∫ (4 − m) dm = 4m − 0.5m² + c₂.
The drone starts at the mast, so both components are zero when m = 0, giving c₁ = 0 and c₂ = 0. The displacement from the mast is r = 1.5m²4m − 0.5m² meters.
(b) At m = 4: east 1.5 × 16 = 24 and north 16 − 8 = 8, so the displacement is 248 meters, which is 24 m east and 8 m north of the mast.
The north component of the velocity, 4 − m, is positive while m < 4 and negative while m > 4, so the drone climbs northward up to m = 4 and then turns back south. Its greatest distance north is therefore the 8 m reached at m = 4. Check: at m = 3 the drone is 12 − 4.5 = 7.5 m north, and at m = 5 it is 20 − 12.5 = 7.5 m north, both less than 8 m.

answer(a) 4 s after leaving the mast; (b) 248 meters, that is 24 m east and 8 m north, and the north velocity changes sign there, so 8 m is the furthest north the drone gets

techniqueVelocity That Varies with Time · Position and Velocity as Vectors

Common pitfalls

Multiplying the velocity by the time, as 4 × 120. Distance is speed multiplied by time only while the velocity is constant. Here the velocity changes every instant, so the position comes from integrating.
Leaving the constants of integration out. They carry where the drone started, and dropping them silently assumes the start was the origin; here it happens to be, but the step that says so is what makes the answer true rather than lucky.

A Walker and a Cyclist on Two Paths That Cross in a Park

methodGive the Two Paths Their Own Time Letters to Find Where the Lines Cross, Then Compare the Two Times to See Whether Anyone Is There Together

At four o'clock a walker leaves the west gate of a park, at the origin, with velocity 6080 meters per minute, and a cyclist leaves the east gate at 6000 with velocity −150200 meters per minute. The components are meters east and north of the west gate. (a) Find the point where the two paths cross. (b) Do the walker and the cyclist meet there, and if not, by how long do they miss each other?

Give each traveler a time letter: the walker p minutes out is at 60p80p and the cyclist q minutes out at 600 − 150q200q.

Let the walker be p minutes out and the cyclist q minutes out, each measured from four o'clock. The walker is at 60p80p and the cyclist at 600 − 150q200q.

step 1 of 5

Where two paths cross is a question about the lines, not about the clock, so each traveler is given a time letter of their own. Whether anyone meets is then decided by comparing the two times that the crossing point is reached.

Let the walker be p minutes out and the cyclist q minutes out, each measured from four o'clock. The walker is at 60p80p and the cyclist at 600 − 150q200q.
The paths cross where the two positions are the same point, whatever the times: 60p = 600 − 150q and 80p = 200q. The second equation gives p = 2.5q.
Put p = 2.5q into the first: 150q = 600 − 150q, so 300q = 600 and q = 2, and then p = 5.
(a) The crossing point is 60 × 580 × 5 = 300400, which is 300 m east and 400 m north of the west gate. The cyclist's position agrees: 600 − 150 × 2 = 300 and 200 × 2 = 400.
(b) They do not meet. The cyclist is at the crossing at 16:02 and the walker at 16:05, so they miss each other by 5 − 2 = 3 minutes. Check: at 16:02 the walker is at 120160, still 300 m short of the crossing, which is the 100 m per minute the walker manages for the 3 minutes.

answer(a) 300400, that is 300 m east and 400 m north of the west gate; (b) they do not meet, and they miss each other by 3 minutes

techniqueMeeting Points and Travel Times · Position and Velocity as Vectors

Common pitfalls

Using one letter for both travelers and concluding that the paths never cross when no solution appears. One letter asks a stronger question, whether they are at the same place at the same time; the paths can cross perfectly well with nobody meeting.
Reading the crossing of two lines on a map as a collision. The map shows where, and says nothing about when. Only after the two times are compared can a meeting be claimed or ruled out.

A Rescue Boat Steered to Intercept a Raft Carried by the Tide

methodAim the Velocity Relative to the Drifting Target Along the Line of Sight, Which Fixes One Component and Leaves the Other to Pythagoras

A raft is 24 km due north of a lifeboat station and is carried due east by the tide at 5 km/h. A rescue boat leaves the station and travels at 13 km/h over the ground. (a) On what course must it be steered to reach the raft, and what is its velocity in components east and north? (b) How long does the interception take, and where does it happen?

With the station at the origin, the raft is at 024 + m50 and the boat at mxy, where x² + y² = 13².

Take east and north as the components, with the station at the origin. The raft starts at 024 and drifts with velocity 50 km/h. Let the boat's velocity be xy km/h, where x² + y² = 13².

step 1 of 5

Seen from the raft, the rescue boat must come straight at it, so the velocity of the boat relative to the raft points along the line of sight, which here is due north. That kills the east component of the relative velocity and fixes the boat's own east component; Pythagoras' theorem supplies the north one.

Take east and north as the components, with the station at the origin. The raft starts at 024 and drifts with velocity 50 km/h. Let the boat's velocity be xy km/h, where x² + y² = 13².
The velocity of the boat relative to the raft is x − 5y. For the boat to reach the raft this must point along the line of sight, which is due north, so its east component is zero: x − 5 = 0 and x = 5.
Then y² = 169 − 25 = 144 and y = 12, the positive root because the boat heads north. (a) The boat's velocity is 512 km/h, and the course satisfies sin θ = 513, so it is steered 22.6^° east of north.
Relative to the raft the boat closes the line of sight at 12 km/h, and the gap along it is 24 km. (b) The interception takes 2412 = 2 hours.
In 2 hours the raft drifts 5 × 2 = 10 km east, so the interception is at 1024: 10 km east and 24 km north of the station. Check: the boat covers √10² + 24² = 26 km, which is 13 × 2, as it must be.

answer(a) steered 22.6^° east of north, with velocity 512 km/h; (b) after 2 hours, at 1024, the boat having covered 26 km

techniqueMeeting Points and Travel Times · Position and Velocity as Vectors

Common pitfalls

Steering straight at the raft's present position, due north. The raft does not wait there; a boat aimed due north at 13 km/h arrives 10 km west of the raft, because nothing in its velocity answers the tide.
Dividing the 24 km by the boat's own speed of 13 km/h. Only the part of the boat's velocity that closes the line of sight shortens the gap, and that is the 12 km/h north, not the full 13 km/h.

A Sprayer and the Water Bowser It Must Meet in a Field

methodSolve One Component for the Time, Substitute It into Both Position Vectors, and Accept the Meeting Only If the Two Agree

A field is 220 m from west to east and 400 m from south to north. A tractor towing a water bowser sets off from the south-west corner, the origin, with velocity 3040 meters per minute, and at the same moment a sprayer sets off from the north-east corner with velocity −25−60 meters per minute. The sprayer must meet the bowser to refill. (a) Show that they do meet, and find when and where. (b) How far has each driven by then?

With m the minutes after they set off, the bowser is at 30m40m and the sprayer at 220 − 25m400 − 60m.

Let m be the number of minutes after they set off. The bowser is at 30m40m, and the sprayer starts at the north-east corner 220400, so it is at 220 − 25m400 − 60m.

step 1 of 5

Both vehicles are timed from the same moment, so one letter serves both. Solving the east components gives a candidate time; substituting it into both position vectors and comparing the north components is what turns the candidate into a meeting.

Let m be the number of minutes after they set off. The bowser is at 30m40m, and the sprayer starts at the north-east corner 220400, so it is at 220 − 25m400 − 60m.
Set the east components equal: 30m = 220 − 25m, so 55m = 220 and m = 4.
Substitute m = 4 into both north components. The bowser: 40 × 4 = 160. The sprayer: 400 − 60 × 4 = 160. They agree, so at m = 4 the two vehicles are at the same point.
(a) They meet 4 minutes after setting off, at 120160: 120 m east and 160 m north of the south-west corner, which is inside the field.
(b) The bowser has driven √120² + 160² = 200 m and the sprayer √100² + 240² = 260 m. Check: the bowser's speed is √30² + 40² = 50 m per minute and 50 × 4 = 200; the sprayer's is √25² + 60² = 65 m per minute and 65 × 4 = 260.

answer(a) they meet 4 minutes after setting off, at 120160; (b) the bowser has driven 200 m and the sprayer 260 m

techniqueMeeting Points and Travel Times · Position and Velocity as Vectors

Common pitfalls

Adding the two speeds, 50 + 65 = 115, and dividing the distance between the corners by it. That closing-speed shortcut is for two vehicles driving straight at each other along one line; these two start from opposite corners on courses that are not along the line joining them, so it gives the wrong time.
Reporting the meeting from the east components alone. If the north components had disagreed, the vehicles would have been level with each other and still hundreds of meters apart, so the substitution back is the step that proves the meeting.

An Aircraft Blown off Its Intended Track by a Crosswind

methodAdd the Wind Velocity to the Velocity Through the Air for the Track Made Good, and Read the Drift Angle from the Two Components

An aircraft is flown on a heading of due north at 240 km/h through the air. The wind blows from the west at 70 km/h. Take east and north as the components. (a) Find the aircraft's velocity over the ground and its ground speed. (b) Find the angle between the track made good and due north, and how far east of its intended track the aircraft is after half an hour.

The heading is due north at 240 km/h, so the velocity through the air is 0240 km/h. Each arrow drawn is half an hour of the velocity beside it.

The heading is due north at 240 km/h, so the velocity through the air is 0240 km/h. A wind from the west blows toward the east, so the wind velocity is 700 km/h.

step 1 of 5

The aircraft is carried by the air it flies in, so its velocity over the ground is its velocity through the air added to the wind's. The ground speed is the magnitude of that sum, and the drift angle comes from the ratio of its two components.

The heading is due north at 240 km/h, so the velocity through the air is 0240 km/h. A wind from the west blows toward the east, so the wind velocity is 700 km/h.
The velocity over the ground is the sum: 0240 + 700 = 70240 km/h.
(a) The ground speed is its magnitude: √70² + 240² = √4900 + 57600 = √62500 = 250 km/h. The aircraft covers ground faster than it flies through the air, because part of the wind is behind it.
(b) The drift angle θ from due north satisfies tan θ = 70240, so θ = 16.3^°: the track made good is 16.3^° east of due north.
In half an hour the wind alone carries the aircraft 70 × 0.5 = 35 km east, and the intended track is due north, so the aircraft is 35 km east of it. Check: it is also 240 × 0.5 = 120 km north, and √35² + 120² = √15625 = 125 km, which is 250 × 0.5.

answer(a) 70240 km/h, a ground speed of 250 km/h; (b) the track made good is 16.3^° east of north, and after half an hour the aircraft is 35 km east of its intended track

techniquePosition and Velocity as Vectors

Common pitfalls

Subtracting the wind, as though a crosswind held the aircraft back. A wind at right angles to the heading adds a sideways velocity and leaves the northward one untouched, so the ground speed rises from 240 to 250 km/h rather than falling.
Giving the drift angle as tan θ = 24070. The angle asked for is measured from due north, so the side opposite it is the 70 east and the side adjacent is the 240 north; the reversed ratio answers for the angle from due east instead.

A Robot Vacuum That Changes Course Partway Through Its Sweep

methodTake Each Leg from Its Own Start by Counting the Time Since That Leg Began, and Carry the End of One Leg into the Next

A robot vacuum leaves its dock at the origin and moves with velocity 0.30.4 m/s for the first 20 seconds. It then turns and moves with velocity 0.2−0.1 m/s for the next 30 seconds. The components are meters east and north. (a) Where is the robot 40 seconds after it leaves the dock? (b) It stops after the second leg and then returns straight to the dock at 0.5 m/s. How far does it have to travel, and how long after setting out is it back?

On the first leg the displacement is the velocity multiplied by the time: 200.30.4 = 68, so after 20 s the robot is at 68.

First leg: the displacement is the velocity multiplied by the time, 200.30.4 = 68. After 20 s the robot is at 68.

step 1 of 5

On each leg the velocity is constant, so the displacement is the velocity multiplied by the time spent on that leg. The second leg is counted from its own start, m − 20 seconds, and begins where the first leg ended.

First leg: the displacement is the velocity multiplied by the time, 200.30.4 = 68. After 20 s the robot is at 68.
The second leg starts there, and at time m it has been running for m − 20 seconds, so r = 68 + (m − 20)0.2−0.1 for 20 ≤ m ≤ 50.
(a) At m = 40, m − 20 = 20, so r = 68 + 200.2−0.1 = 106: the robot is 10 m east and 6 m north of the dock.
At the end of the second leg, m = 50 and m − 20 = 30, so the robot is at 68 + 300.2−0.1 = 125.
(b) The way home is the straight line back to the dock, of length √12² + 5² = √169 = 13 m. At 0.5 m/s that takes 130.5 = 26 s, so the robot is back 50 + 26 = 76 s after setting out.

answer(a) at 106, that is 10 m east and 6 m north of the dock; (b) it ends the second leg at 125, so it travels 13 m home, taking 26 s, and it is back 76 s after setting out

techniqueMotion That Starts Later · Velocity That Varies with Time

Common pitfalls

Using the whole time in the second leg, as 68 + 400.2−0.1 at m = 40. The second velocity only applies from 20 s onward, so the time to multiply by is the 20 s the robot has spent on that leg, not the 40 s since it left the dock.
Adding the lengths of the two legs to get the distance home. The robot does not retrace its route; it goes straight back, so the distance is the magnitude of the position vector at the end of the second leg.