d 1

1. d2 d1 Mr. Bean travels from position 1 (d1) to position 2 (d2)

2. d3 d2 d1 Mr. Bean then travels from position 2 (d2) to position 3 (d3)

3. d3 d2 d4 d1 Mr. Bean then travels from position 3 (d3) to position 4 (d4)

4. d3 d2-3 d3-4 d2 d4 d1-2 d1 Each change in position is a displacement (d)

5. d2-3 d3-4 d1-2 dfinal dR dinitial The overallchange in position is the resultantdisplacement (dR) initial position (dinitial) to final position (dfinal)

6. d2-3 d3-4 d1-2 dfinal dR dinitial

7. d2x d2y d3 d3y d2 dfinal d1x d3x d1y d1 dinitial To find the resultant displacement algebraically, we need to find the x and y components of each individual vector.

8. d2x d2y d3y d1x d3x dx dfinal d1y dy dinitial To find the resultant displacement algebraically, we need to find the x and y components of each individual vector.

9. d2y dx d3y d2x d3x d1x d1y dy • The next step involves finding the vector sum in the x and y. • dx = vector sum of x components • dy = vector sum of y components

10. dx dy The next step involves redrawing the dxanddyvectors tail to tip.

11. dx dy The next step involves redrawing the dxanddyvectors tail to tip.

12. dx dfinal dy dR dinitial To find the resultant displacement dR draw a new vector from the initial to final position.

13. dR= dx2 + dy2 dx dy dR Use the Pythagorean Theorem to find the magnitude (size) of the resultant.

14. dx dy  dR dR= dx2 + dy2 dx tan = dy Use the Pythagorean Theorem to find the magnitude (size) of the resultant and the tangent function to determine the direction of the resultant.

15. dx dy dy dR dR   dx or Notice that there are two possible ways of drawing the resultant vector diagram. Each is correct!

16. dx dy dy dR dR   dx or • Notice that there are two possible ways of drawing the resultant vector diagram. Each is correct! • Both the magnitude (size) and directionof the dRremain the same.