Crossing paths are not a collision
Draw the paths of two moving objects on a map. If the two lines cross, the routes share a point. That says nothing about when each object gets there.
Two ships whose courses cross may pass that point an hour apart and never come near each other. A collision, or a meeting, needs both objects at the same place at the same time.
Both coordinates, one time
Object A is at rA = (1 + 3t, 2 + 2t) and object B at rB = (10 − t, 2 + 3t), with t in seconds. They collide only if there is one value of t at which rA = rB. That is two equations, one for each coordinate: 1 + 3t = 10 − t and 2 + 2t = 2 + 3t, and the same t must solve both.
Solve one coordinate, test the other
Solve the x-coordinates first: 1 + 3t = 10 − t gives 4t = 9, so t = 2.25. At that moment both objects are at x = 7.75.
Now test the y-coordinates at t = 2.25. A is at 2 + 2 × 2.25 = 6.5 and B is at 2 + 3 × 2.25 = 8.75. They disagree, so at the one moment the objects are level across, B is 2.25 above A. They never collide.
The y-coordinates on their own agree only when 2 + 2t = 2 + 3t, that is at t = 0, and then A is at x = 1 while B is at x = 10. Each coordinate agrees at its own time, and the two times are different.
The gold path of A and the plain path of B. The two solid dots are A and B at t = 2.25, level across at x = 7.75 but 2.25 apart up the page. The hollow dot is where the paths cross: B passes it first, and A arrives later.
Where the paths cross, and when
To find where the paths cross, give each object its own time: A at time p and B at time q. The paths share a point when 1 + 3p = 10 − q and 2 + 2p = 2 + 3q. The second equation gives p = 1.5q, and then the first gives 1 + 4.5q = 10 − q, so 5.5q = 9 and . So .
The crossing point is , about (8.36, 6.91). B is there at , about 1.64, and A at , about 2.45. B passes the crossing first, of a second, about 0.82, before A.
Change B to rB = (10 − t, 6 − t). The x-coordinates still agree at t = 2.25, and now the y-coordinates are 6.5 for A and 6 − 2.25 = 3.75 for B, so again there is no collision. These two paths cross at (16, 12). A gets there at t = 5, because 1 + 3 × 5 = 16, but B's path passes through it at t = −6, six seconds before B set off. The objects pass the same point 11 seconds apart.
A pair that does meet
Object C is at rC = (11 − 2t, 12 − 3t). Solve the x-coordinates of A and C: 1 + 3t = 11 − 2t gives 5t = 10, so t = 2. Test the y-coordinates at t = 2: A is at 2 + 4 = 6 and C at 12 − 6 = 6. They agree, so A and C meet at (7, 6) after 2 seconds.
Solving the y-coordinates first gives the same time, 2 + 2t = 12 − 3t, so 5t = 10 and t = 2. When the objects meet, both coordinates lead to one time.
A on the gold path and C on the plain one, at t = 0, 1 and 2. Each second A moves (3, 2) and C moves (−2, −3), and at t = 2 both are at (7, 6).
Two objects on one line
On a single axis there is only one coordinate. A is at x = 2 + 3t and B at x = 17 − 2t, moving toward each other. They are level when 2 + 3t = 17 − 2t, so 5t = 15 and t = 3, when both are at 11. The gap of 15 closes at 3 + 2 = 5 per second, which is another way to see the 3 seconds.
The usual mistakes
Taking a crossing on the map as a collision. The map shows where the paths meet, not when each object is there.
Solving one coordinate and stopping. One coordinate agreeing means the objects are level in one direction at that moment, not that they are at the same point. Substitute the time into the other coordinate as well.
Using one time letter to find where the paths cross. One letter asks whether the objects are at the same place at the same time; when the answer is no, the paths can still cross. Give each object its own letter for the crossing point.
Giving the gap as the time. The gap of 15 still has to be divided by the rate at which it closes.
Two applications
In the first application below, a sprayer and a water bowser set off from opposite corners of a field, and the test of both coordinates shows that they meet. In the second, a walker and a cyclist cross a park on paths that cross, and they miss each other by 3 minutes.
Worked example: A Sprayer and the Water Bowser It Must Meet in a Field
Question 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?
1.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.
With m the minutes after they set off, the bowser is at 30m40m and the sprayer at 220 − 25m400 − 60m. 2.Set the east components equal: 30m = 220 − 25m, so 55m = 220 and m = 4.
Set the east components equal: 30m = 220 − 25m, so 55m = 220 and m = 4. 3.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.
Substitute m = 4 into both north components: the bowser gives 160 and the sprayer 400 − 240 = 160, so they agree. 4.(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.
(a) They meet 4 minutes after setting off, at 120160, which is inside the field. 5.(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.
(b) The bowser has driven √1202 + 1602 = 200 m and the sprayer √1002 + 2402 = 260 m, which agree with their speeds of 50 and 65 meters per minute.
Answer: (a) they meet 4 minutes after setting off, at 120160; (b) the bowser has driven 200 m and the sprayer 260 m
Common mistakes
- 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.
More motion in two dimensions problems, worked step by step →
Worked example: A Walker and a Cyclist on Two Paths That Cross in a Park
Question 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?
1.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.
Give each traveler a time letter: the walker p minutes out is at 60p80p and the cyclist q minutes out at 600 − 150q200q. 2.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.
The paths cross where the two positions are the same point, whatever the times: 60p = 600 − 150q and 80p = 200q, so p = 2.5q. 3.Put p = 2.5q into the first: 150q = 600 − 150q, so 300q = 600 and q = 2, and then p = 5.
Substituting, 150q = 600 − 150q, so 300q = 600 and q = 2, and then p = 5. 4.(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.
(a) The crossing is at 300400: 300 m east and 400 m north of the west gate. The cyclist's position agrees at q = 2. 5.(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.
(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 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
Common mistakes
- 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.
More motion in two dimensions problems, worked step by step →