sin! =! d y =! d T ! d y = 15 m = m = 8.6 m cos! =! d x ! d x ! d T 2 =! d x 2 +! d y =! d x 2 +! d y = 27.2 m = 30.0 m tan! =!

Transcription

1 Section. Motion in Two Dimensions An Algebraic Approach Tutorial 1 Practice, page Gien d 1 = 7 m [W]; d = 35 m [S] Required d T Analysis d T = d 1 + d Solution Let φ represent the angle d T with the d T = d 1 + d d T = d 1 + d d T = d 1 + d ( ) + ( 35 m) = 7 m d T = 44 m tan = d d 1 tan = 35 m 7 m tan = 1.96 = tan " ( ) = 5 Statement The sum of the two ectors is 44 m [W 5 S].. Gien d 1 = 13. m [S]; d = 17.8 m [E] Required d T Analysis d T = d 1 + d Solution Let φ represent the angle d T with the d T = d 1 + d d T = d 1 + d d T = d 1 + d ( ) + ( 17.8 m) = 13. m d T =. m tan = d d 1 tan = 17.8 m 13. m tan = = tan " ( ) = 53 The sum of the two ectors is. m [S 53 E] or [E 37 S]. Statement The sum of the two ectors is. m [E 37 S]. Tutorial Practice, page Gien d T = 15 m [W 35 N] Required d x ; d y Analysis d T = d x + d y Solution Since the direction of d T west and north, the direction of d x is [W] and the direction of d y is [N]. sin = d y d T d y = d T sin ( )( sin35 ) = 15 m = m d y = 8.6 m cos = d x d T d x = d T cos ( )( cos35 ) = 15 m = 1.9 m d x = 1 m Statement The ector has a horizontal or x-component of 1 m [W] and a ertical or y-component of 8.6 m [N].. Gien d x = 7. m [E]; d y = 1.7 m [N] Required d T Analysis d T = d 1 + d Solution Let φ represent the angle d T with the d T = d x + d y d T = d x + d y d T = d x + d y ( ) + ( 1.7 m) = 7. m d T = 30.0 m tan = d y d x tan = 1.7 m 7. m tan = = tan " ( ) = 5 Copyright 011 Nelson Education Ltd. Chapter Motion in Two Dimensions.-1

2 Statement The sum of the two ectors is 30.0 m [E 5 N], which is the original ector from Sample Problem 1. Tutorial 3 Practice, page Gien d 1 =.78 cm [W]; d = 6.5 cm [S 40 E] Required d T Analysis d T = d 1 + d Solution Determine the total x-component and y-component of d T = d 1x + d x =.78 cm [W] cm ( )( sin40 ) [E] =.78 cm [W] cm [E] =.78 cm [E] cm [E] = cm [E] = 1.4 cm [E] = d 1y + d y = 0 cm cm = cm [S] = 4.79 cm [S] ( )( cos40 ) [S] Determine the magnitude of d T d T = + d T = + ( ) + ( cm) = cm d T = 4.95 cm Let φ represent the angle d T tan = cm tan = cm tan = = tan " ( ) = 76 Statement The ant s total displacement is 4.95 cm [E 76 S].. Gien d 1 =.64 m [W 6 N]; d = 3.1 m [S 1 E] Required d T Analysis d T = d 1 + d Solution Determine the total x-component and y-component of d T = d 1x + d x =.64 m ( )( cos6 ) [W] + ( 3.1 m) ( sin1 ) [E] =.378 m [W] m [E] =.378 m [W] m [W] = m [W] = 1.71 m [W] = d 1y + d y =.64 m ( )( sin6 ) [N] + ( 3.1 m) ( cos1 ) [S] = m [N] m [S] = m [S] m [S] = m [S] = 1.98 m [S] Determine the magnitude of d T d T = + d T = + ( ) + ( m) = m d T =.6 m Let φ represent the angle d T tan = tan = m m tan = = tan " ( ) = 49 Statement The total displacement of the paper airplane is.6 m [W 49 S]. Tutorial 4 Practice, page Answers may ary. Sample answer Imagine the rier current is flowing south and the canoe is pointed east. As long as the boat is pointed perpendicular to the current, the current has no effect on the time it takes to cross the rier. Think about the component ectors. Een though the canoe will be traelling in a direction between south and east, all its eastbound elocity is the same, no matter how fast the current is, or if there s any current at all. Copyright 011 Nelson Education Ltd. Chapter Motion in Two Dimensions.-

3 . (a) Gien d = 0.0 m; = 1.3 m/s Required t Analysis = d t t = d Solution t = d = 0.0 m 1.3 m s = s t = 15 s Statement It will take the swimmer 15 s to cross the rier. (b) Gien x =.7 m/s [W] Required d x Analysis x = d x t d x = x t Solution d x = x t =.7 m " # s [W] $ % & s d x = 4 m [W] ( ) Statement The swimmer lands 4 m downstream from his intended location. Section. Questions, page (a) Gien d T = 0 km [W 50 N] Required d x ; d y Analysis d T = d x + d y Solution Since the direction of d T west and north, the direction of d x is [W] and the direction of d y is [N]. sin = d y d T d y = d T sin ( )( sin50 ) = 0 km d y cos = d x d T d x = d T cos ( )( cos50 ) = 0 km d x = 13 km Statement The ector has a ertical or y-component of 15 km [N] and a horizontal or x-component of 13 km [W]. (b) Gien d T [W 80 S] Required d x ; d y Analysis d T = d x + d y Solution Since the direction of d T west and south, the direction of d x is [W] and the direction of d y is [S]. sin = d y d T d y = d T sin ( )( sin80 ) = km d y cos = d x d T d x = d T cos ( )( cos80 ) d x =.6 km Statement The ector has a ertical or y-component of 15 km [S] and a horizontal or x-component of.6 km [W]. (c) Gien d T = 40 km [N 65 E] Required d x ; d y Analysis d T = d x + d y Solution Since the direction of d T north and east, the direction of d x is [E] and the direction of d y is [N]. cos = d y d T d y = d T cos ( )( cos65 ) = 40 km d y = 17 km sin = d x d T d x = d T sin ( )( sin65 ) = 40 km d x = 36 km Statement The ector has a ertical or y-component of 17 km [N] and a horizontal or x-component of 36 km [E]. Copyright 011 Nelson Education Ltd. Chapter Motion in Two Dimensions.-3

4 . Gien d 1 = 5.1 km [E]; d = 14 km [N] Required d T Analysis d T = d 1 + d Solution Let φ represent the angle d T with the d T = d 1 + d d T = d 1 + d d T = d 1 + d ( ) + ( 14 km) = 5.1 km d T tan = d d 1 14 km tan = 5.1 km tan =.745 = tan "1.745 ( ) = 70 Statement The sum of the two ectors is 15 km [E 70 N]. 3. Gien d 1 = 11 m [N 0 E]; d = 9.0 m [E] Required d T Analysis d T = d 1 + d Solution Determine the total x-component and y-component of d T = d 1x + d x = 11 m ( )( sin0 ) [E] m [E] = 3.76 m [E] m [E] = 1.76 m [E] = 13 m [E] = d 1y + d y = 11 m ( )( cos0 ) [N] + 0 m = m [N] = 10 m [N] Determine the magnitude of d T d T = + d T = + ( ) + ( m) = 1.76 m d T = 16 m Let φ represent the angle d T tan = tan = 1.76 m m tan = 1.34 = tan " ( ) = 51 Statement The total displacement of the football player is 16 m [N 51 E]. 4. Gien d 1 = 00.0 m [S 5 W]; d = m [N 30 E] Required d T Analysis d T = d 1 + d Solution Determine the total x-component and y-component of d T = d 1x + d x = 00.0 m ( )( sin5 ) [W] + ( m) ( sin30 ) [E] = m [W] + 75 m [E] = m [W] 75 m [W] = m [W] = 9.54 m [W] = d 1y + d y = 00.0 m ( )( cos5 ) [S] + ( m) ( cos30 ) [N] = m [S] m [N] = m [S] m [S] = m [S] = m [S] Determine the magnitude of d T d T = + d T = + ( ) + ( m) (one extra digit carried) = m d T = 5.3 m Let φ represent the angle d T tan = tan = m m (one extra digit carried) Copyright 011 Nelson Education Ltd. Chapter Motion in Two Dimensions.-4

5 tan = = tan " ( ) = 11 Statement The total displacement of the boat is 5.3 m [S 11 W]. 5. Gien d 1 = 5 m [N 0 W]; d = 35 m [S 15 E] Required d T Analysis d T = d 1 + d Solution Determine the total x-component and y-component of d T = d 1x + d x = 5 m ( )( sin0 ) [W] + ( 35 m) ( sin15 ) [E] = m [W] m [E] = m [E] m [E] = m [E] = 0.51 m [E] = d 1y + d y = 5 m ( )( cos0 ) [N] + ( 35 m) ( cos15 ) [S] = 3.49 m [N] m [S] = 3.49 m [S] m [S] = 10.3 m [S] = 1.0 " 10 m [S] Determine the magnitude of d T d T = + d T = + ( ) + ( 10.3 m) = m d T = m Let φ represent the angle d T tan = tan = m 10.3 m (one extra digit carried) tan = = tan " ( ) = 3 Statement The total displacement of the object is m [S 3 E]. 6. (a) Gien d 1 = 4.3 km [W]; d = 8.0 km [W 54 N] Analysis d T = d 1 + d Solution Determine the total x-component and y-component of d T = d 1x + d x = 4.3 km [W] km ( )( cos54 ) [W] = 4.3 km [W] km [W] = 9.00 km [W] = 9.0 km [W] = d 1y + d y = 0 km km = 6.47 km [N] = 6.5 km [N] ( )( sin54 ) [N] Determine the magnitude of d T d T = + d T = + ( ) + ( 6.47 km) = 9.00 km d T = 11 km Let φ represent the angle d T tan = 6.47 km tan = 9.00 km tan = = tan " ( ) = 36 Statement The total displacement gien by the two ectors is 11 km [W 36 N]. (b) Gien d 1 = 35 m [E 65 N]; d = m [E 37 S] Required d T Analysis d T = d 1 + d Solution Determine the total x-component and y-component of d T = d 1x + d x = 35 m ( )( cos65 ) [E] + ( m) ( cos37 ) [E] = m [E] m [E] = 3.36 m [E] = 3 m [E] Required d T Copyright 011 Nelson Education Ltd. Chapter Motion in Two Dimensions.-5

6 = d 1y + d y = 35 m ( )( sin65 ) [N] + ( m) ( sin37 ) [S] = 31.7 m [N] m [S] = 31.7 m [N] 13.4 m [N] = m [N] = 18 m [N] Determine the magnitude of d T d T = + d T = + ( ) + ( m) = 3.36 m d T = 37 m Let φ represent the angle d T tan = tan = m 3.36 m (one extra digit carried) tan = = tan " ( ) = 30 Statement The total displacement of the boat is 37 m [E 30 N]. 7. Gien d 1 = 5 m [N 30 W]; d = 30.0 m [N 40 E]; d 3 = 35 m [S 5 W] Required d T Analysis d T = d 1 + d + d 3 Solution Determine the total x-component and y-component of d T = d 1x + d x + d 3x = 5 m ( )( sin30 ) [W] + ( 30.0 m) ( sin40 ) [E] ( )( sin5 ) [W] + 35 m = 1.5 m [W] m [E] m [W] = 1.5 m [W] 19.8 m [W] m [W] = 8.01 m [W] = 8.0 m [W] = d 1y + d y + d 3y ( )( cos30 ) [N] + ( 30.0 m) ( cos40 ) [N] ( )( cos5 ) [S] = 5 m + 35 m = 1.65 m [N] +.98 m [N] m [S] = 1.65 m [N] +.98 m [N] 31.7 m [N] = 1.91 m [N] = 13 m [N] Determine the magnitude of d T d T = + d T = + ( ) + ( 1.91 m) = 8.01 m d T = 15 m Let φ represent the angle d T tan = tan = 8.01 m 1.91 m tan = = tan " ( ) = 3 Statement The total displacement of the ectors is 15 m [N 3 W]. 8. (a) Gien d = 5.1 km; = 0.87 km/h Required t Analysis = d t t = d Solution t = d = 5.1 km 0.87 km h = 5.86 h t = 5.9 h Statement It will take the swimmer 5.9 h to cross the rier. (b) Gien x =.0 km/h [W] Required d x Analysis x = d x t d x = x t Solution d x = x t =.0 km " # h [W] $ % & 5.86 h d x = 1 km [W] ( ) Statement The current has moed the swimmer 1 km downstream by the time she reaches the other side. Copyright 011 Nelson Education Ltd. Chapter Motion in Two Dimensions.-6

7 9. Time for the conductor to reach the other side Gien d = 4.0 m; = 1. m/s Required t Analysis = d t t = d Solution t = d = 4.0 m 1. m s t = 3.3 s Statement It will take the conductor 3.3 s to cross the railcar. The conductor s elocity relatie to the ground Gien 1 = 4.0 m/s [N]; = 1. m/s [E] Required T Analysis T = 1 + Solution Let φ represent the angle T with the T = 1 + T = 1 + T = 1 + ( ) + ( 1. m/s) = 4.0 m/s T = 4. m/s tan = 1 1. m/s tan = 4.0 m/s = 17 Statement The elocity of the conductor relatie to the ground is 4. m/s [N 17 E]. 10. Answers may ary. Sample answer (a) I prefer the algebraic method of adding ectors. It takes more time, but I think the answers are more accurate because there is no chance of making errors measuring. I prefer to use a diagram only to double check my answer by sketching the ectors. (b) If the naigator on a boat were working on a large map, it would probably be more useful to plot the ectors directly on the map. There would be no need for calculations and the naigator could use other map features to double check the resultant (like comparing distances and directions to landmarks on the map). Copyright 011 Nelson Education Ltd. Chapter Motion in Two Dimensions.-7