1. 2 Dimensional Uniform motion

2. A car crossing a road Let us say that there is car crossing a wide road (15 meters wide) at a speed of 5 m/sec. Finding the time to cross the road should be pretty easy to find. V = Displacement /Time Time = Displacement / V Time = (15 m)/(5 m/sec) Time = 3sec 0 sec 5 m 1 sec 5 m 10 m 2 sec 5 m 3 sec 15 m 5 m/sec

3. Now Let’s look at a car stalled and floating in a raging river In this case the car will not move across the river but move down the river at the same speed as the water. How far will the car move down river in 3 seconds if the river flows at a rate of 5 m/sec? This also should be fairly easy to do. V = Displacement/Time Displacement = V* Time Displacement = (5 m/sec)* (3 sec) Displacement = 15 m 5 m/sec 0 sec 5 m 1 sec 5 m 2 sec 10 m 5 m 3sec 15 m

4. Both happening at the same time In this case let’s say that a car-boat is trying to cross a 15 meter wide flowing river by pointing directly across the river moving at 5 m/sec, the river also flows at a rate of 5 m/sec. Lets find the time it takes to cross the river. In the case the car does not move directly across the river, or along the river. Instead it moves diagonally across the river. 5 m/sec However the car is still moving across the river at 5 m/s (just like the car in the 1 st example) So it takes 3 seconds to cross. 5 m 1 sec 5 m 2 sec 5 m 3 sec 5 m And the car movies down the river just like the second example. So the car travels 15 meters downstream.

5. Independence of axis Whenever an object moves along both the X and Y axis the motion of one axis has nothing to do with the motion of other axis. Events that happen along the X axis do not affect events that happen along the Y axis. 5 m/sec 5 m 1 sec 5 m 2 sec 5 m 3 sec 5 m

6. Concept Connection When working with vectors we often break them up into their X and Y comments and then combine all the X components together, then combine the Y components together. We can do this because the Y components do not affect the X components. Because the are along different axis they are independent with each other

7. Speaking of vectors Since velocities are vectors we can find the boat’s resultant velocity by simply adding the velocity of the boat (without the river) to the velocity of the river. But this must be done vectorally. V resultant = V boat + V river V boat V river V resultant

8. Applying triangles We can now use the triangle rules to find the resultant speed an direction. 5 m/s (Square root of 50) m/s - 45 O (V resultant ) 2 = (5 m/sec) 2 + (5 m/sec) 2 (V resultant ) = (Square root of 50) m/sec  = tan-1(Y/X) = tan-1[(5 m/sec)/(5 m/sec)]  = 45 O in the 4 th quadrant  = -45 O in the 4 th quadrant 5 m/sec

9. Vectors and displacement Likewise we can also use the displacements along the X and Y axises to find the resultant displacement Displacement resultant = Displacement X + Displacement Y Displacement X Displacement Y Displacement resultant

10. Appling triangles (again) 15 m (Square root of 450) m/s - 45 O (Displacement resultant ) 2 = (15 m) 2 + (15 m) 2 (Displacement resultant ) = (Square root of 450) m  = tan-1(Y/X) = tan-1[(15 m)/(15 m)]  = 45 O in the 4 th quadrant  = -45 O in the 4 th quadrant 3*(Square root of 50) m/s (Displacement resultant ) = 3*(Square root of 50) m 5 m/sec 15 m

11. The power of similar triangles The displacement triangle and velocity triangle are similar. The number of times the displacement triangle is larger (or smaller) than the velocity triangle is the amount of time it take for the boat to cross the river Displacement = Velocity Average *time So we can use similar triangles to analyze 2 dimensional systems (with a uniform velocity)

12. Crossing another river Lets say there is a boat that wants to cross a river that is 15 meters wide by pointing directly across and moving at 5 m/sec. The river however flows at a rate of 10 m/sec. How far down stream will the boat land? 5 m/sec 10 m/sec 15 m Distance Downstream

13. Distance Downstream = 30 m 15 m 5m/sec10m/sec Distance Downstream Since the displacement triangle is 3 times larger the time to cross the river is 3 seconds 15 m Distance Downstream Important Note: Changing the speed of the river did not change the time it took to cross the river, because the two velocities are perpendicular to each other. 5 m/sec 10 m/sec 15 m Distance Down stream 10m/sec 5 m/sec

14. Cutting across at a diagonal In this case object moving moves across a field but at angle and not directly across. 60 m 5 m/s 30 O A car crosses a field at a speed of 5 m/s at an angle of 30 O as shown. I f the car travels 60 meters along the field, find the following… 1) Width of field 2) Time to cross the field 3) The car’s displacement vector

15. Vector applications In this case since we need to break the car’s velocity vector into it’s X and Y components, so that we can form a velocity triangle. 5 m/s 30 O 5 m/s*Cos(30 O ) 5 m/s*Sin(30 O ) 2.5 m/s 4.33 m/s

16. Setting up the triangles Now that we have the velocity triangle we can now use similar triangles solve for our displacements 60 m 2.5 m/s4.33 m/s Width Width = 103.92 m 60 m 2.5 m/s5 m/s Displacement Displacement = 120 m at 30 O 60 m Width Displacement Since the displacement vector is 24 times large than the velocity vector the time is 24 seconds 5 m/s 30 O 2.5 m/s 4.33 m/s

17. A more complicated case A boat crosses a 120 meter wide river that is flowing at 4 m/sec be heading at 8 m/sec at an angle of 30 O into the current. How much time will it take for the boat to cross the river? 120 m 8 m/sec 4 m/sec

18. (8 m/sec)*Cos(30) 6.93 m/sec (8 m/sec)*Sin(30) 4 m/sec 270 O 4 m/sec (4 m/sec)*Cos(270) 0 m/sec (4 m/sec)*Sin(270) -4 m/sec 6.93 m/sec 0 m/sec 6.93 m/sec (8 m/sec)*Cos(30) (8 m/sec)*Sin(30) 4 m/sec Resultant velocity = 6.93 m/sec at 0 O First we have to find the resultant velocity of the boat (V resultant = V boat + V river ). For this we can use the Vector components (the box method). 30 O 8 m/s (V)*Cos(  )(V)*Sin(  ) Vectors Total V Boat (8 m/sec) at 30 O V River (4 m/sec) at 270 O

19. Resultant velocity = 6.93 m/sec at 0 O Resultant displacement = 120 m at 0 O Time = (Displacement/velocity) Time = (120 m)/(6.93m/sec) Time =17.32 sec