1. d1d1 d2d2 Mr. Bean travels from position 1 (d 1 ) to position 2 (d 2 )

2. d1d1 d2d2 d3d3 Mr. Bean then travels from position 2 (d 2 ) to position 3 (d 3 )

3. d1d1 d2d2 d3d3 d4d4 Mr. Bean then travels from position 3 (d 3 ) to position 4 (d 4 )

4. d1d1 d2d2 d3d3 d4d4  d 1-2  d 2-3  d 3-4 Each change in position is a displacement (  d)

5.  d 1-2  d 2-3  d 3-4 d initial d final dRdR The overall change in position is the resultant displacement (  d R ) initial position (d initial ) tofinal position (d final )

6. d initial d final dRdR  d 1-2  d 2-3  d 3-4

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

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

9.  d1y  d2y  d3y  d1x  d2x  d3x   dy   dx 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.   dy   dx The next step involves redrawing the   dx and   dy vectors tail to tip.

11.   dy   dx The next step involves redrawing the   dx and   dy vectors tail to tip.

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

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

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

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

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