Chapter 3 Motion in a Plane Practice Proble Solution Student Textbook page 80 1. Frae the Proble - Sketch and label a diagra of the otion. 40 v(/) 30 0 10 0 4 t () - The equation of otion apply to the proble, ince the car i aued to be oving with contant acceleration. The car average acceleration after 4.0 v i = +6.0 v i = +38 = 4.0 a Ue the equation for acceleration. The Indy race car acceleration i +8.0 /. a = v f v i = +38 =+8.0 ( +6.0 ) 4.0 The unit in the anwer were /, which i correct for acceleration. The average acceleration i poitive, ince the velocity of the Indy car wa increaing while travelling in the poitive direction.. Frae the Proble - Sketch and label a diagra of the otion. 16
a V f = +3.5 V i = 4.0 = 3.0 = 3.0 (/) v y 5 4 3 1 t() 0 1 1 3 4 5 3 4 5 - The equation of otion apply to the proble, ince the acceleration wa contant. The car acceleration once the driver got it tarted v i = 4.0 a v f =+3.5 = 3.0 - Ue the equation for acceleration. - Let uphill be the poitive direction. a = v f v i = +3.5 ( 4.0 ) 3.0 = 7.5 3.0 =+.5 The car acceleration once the driver got it tarted wa.5 / [uphill]. The unit in the anwer were /, which i correct for acceleration. The acceleration i poitive, ince the velocity of the Indy car wa increaing uphill. 3. Frae the Proble - Sketch and label a diagra of the otion. V i = 3.0 V f = 0 a = 8.0 - The equation of otion apply to the proble, ince the acceleration wa contant. - The final velocity of the bu will be zero. The velocity the bu wa travelling when the brake were applied 17
a = 8.0 v f = 0.0 = 3.0 v i Select the equation that relate the initial v i = v f a velocity to the final velocity, acceleration, and tie interval. All of the needed quantitie are known, o ubtitute the into the equation. v i = 0.0 ( 8.0 )(3.0 () Siplify. v i =+4 The bu wa travelling at +4 / when the brake were applied. The unit in the anwer were /, which i correct for velocity. The initial velocity wa poitive, which i correct. Practice Proble Solution Student Textbook page 89 4. Frae the Proble - Make a diagra of the otion of the field hockey player that include the known variable. (/) v y 5 4 3 1 t() 0 1 3 4 5 - The equation of otion apply to the proble, ince the acceleration wa contant. (a) The ditance he travelled (b) Her acceleration v i = 0.0 v f = 4.0 =.5 d a Ue the equation of otion that relate d = ( v i + v f ) tie, initial velocity, and final velocity ( 0.0 d = + 4.0 to diplaceent..0 All of the needed quantitie are known, o ubtitute the into the equation. d =.0 (.5 ) Siplify. d = 5.0 ) (.5 ) 18
(a) The ditance wa 5.0. Ue the inforation to find the acceleration. (b) Her acceleration i 1.6 /. a = v a = 4.0 a = 4.0.5 a = 1.6 = v v 1 0.0.5 The unit for diplaceent are etre and etre per econd for acceleration, which are correct. Both diplaceent and acceleration are poitive, a they hould be. 5. Frae the Proble - Let tie zero be the oent that Michael begin to accelerate. - At tie zero, Michael i 75 behind Robert and will thu ut run 75 further than Robert in order to catch up with hi. - When Michael catche up to Robert, they will have run for the ae aount of tie. - Michael i travelling with unifor acceleration. Thu, the equation of otion that relate diplaceent, initial velocity, acceleration, and tie interval decribe Michael otion. - Robert travel with contant velocity or unifor otion. Robert otion can therefore be decribed by uing the equation that define velocity. Length of tie it will take Michael to catch up with Robert in the race a M = 0.15 v M(i) = 3.8 v R = 4. d M d R + 75 Write a atheatical equation that d M = d R + 75 tate that the ditance Michael run i v M + 1 a M = 75 + v R equal to the ditance Robert run during the tie interval, plu the 75 Michael ha to ake up. Subtitute the equation that define the 3.8 + 1 (0.15 ) velocity for Robert for d R. Subtitute = 75 + 4. the equation of otion that relate diplaceent, initial velocity, acceleration, 3.8 + (0.075 ) and the tie interval for Michael for d = 75 + 4. M. The tie interval fro tie zero i the 3.8 4. ae for the two runner when Michael + ( 0.075 ) 75 = 0 catche up with Robert. Solve for, the 0.4 + ( 0.075 ) unknown, after ubtituting the known 75 = 0 ( 0.075 ) value into the equation. 0.4 75 = 0 19
Ue the quadratic forula to olve for, ince it cannot be eaily factored. = b ± b 4ac a = ( 0.4) ± ( 0.4) 4(0.075)( 75) (0.075) = 0.4 ± 0.16 +.5 0.15 = 0.4 ±.66 0.15 = 0.4 ± 4.76 0.15 0.4 + 4.76 Exclude 9.07, ince a negative tie = 0.15 ha no eaning in thi ituation. = 5.16 0.15 = 34.4 It will take Michael 34 to catch Robert. The tie i poitive, and it ee to be reaonable. 6. Frae the Proble - Sketch and label the ituation. = 0.4 4.76 0.15 = 4.36 0.15 = 9.07 V i = 00 k h = 55.6 a = 5.0 = 8.0 - Since the acceleration of the race car i contant, the equation of otion that relate diplaceent, initial velocity, acceleration, and tie interval decribe it otion. How far the car ha travelled (it diplaceent) after 8 v i = 00 k h = 8.0 d a = 5.0 Firt, convert 00 k/h to /, ince acceleration and tie are given in thee unit. 00 k h 1 h 3600 1000 1 k = 55.6 Select the equation of otion that relate d = v i + 1 a the unknown variable, d, to the three known variable, a, v i, and. All of the needed quantitie are known, d = 55.6 (8.0 ) o ubtitute the into the equation. + 1 (5.0 )(8.0 ) 0
Siplify. d = 444.8 + 160 d = 604.8 = 605 d = 604.8 = 600 The race car travelled 6.0 10 during the 8.0 tie interval. The unit cancelled to give etre for diplaceent, which i correct, and the diplaceent ee to be reaonable. 7. Frae the Proble - Sketch and label the diagra of otion. V i = 0 a V f 150 = 10 - The otorit ut low down and cover the 150 in 10 without topping, and continue travelling when the light turn green. - Since the acceleration of the car i contant, the equation of otion that relate to acceleration, diplaceent, initial velocity, and tie interval decribe the otion. (a) The acceleration of the car (b) The peed of the car jut a it pae the green light v i = 0.0 d = 1.50 10 a = 10.0 Select the equation of otion that relate the unknown variable, acceleration, to the three known variable, tie, initial velocity, and diplaceent. Subtitute and iplify. (a) The otorit acceleration i 10.0 / Select the equation that relate final velocity to the initial velocity, acceleration, and tie interval. v f 1 Rearrange d = v + a to olve for a: d v i ( 150 ) 0 ( ) 10 10 a = a = a = ( ) a = 30 40 a = 10 300 40 100 10 v f = v i + a i 1
All of the needed quantitie are known, o ubtitute the into the equation. v f = 0 v f = 0 + ( 1 )10 + 10 Siplify. v f = 10 (b) The final velocity of the car jut a it pae the green light i 10.0 /. The negative value of the acceleration indicated the otorit lowed down to a reaonable final velocity of 10 /. Practice Proble Solution Student Textbook page 93 94 8. Frae the Proble - Make a cale diagra of the proble. Chooe a cale of 1.0 c : 1.0 10 k. - The airplane trip conited of two eparate diplaceent. - The vector u of the two diplaceent vector yield one reultant vector that give the airplane final diplaceent. (a) The diplaceent, d R, of the firt two leg of the airplane trip (b) The [direction] the airplane need in order to fly traight back v v d A = 618. 10 k[n58.0 W] d B = 361 k[e35 S] Meaure the length of the reultant vector in the cale diagra. Multiply the length of the vector by the cale factor. With a protractor, eaure the angle between the horizontal axi and the reultant vector. d v R [direction] v d = 6. c R v 10. 10 d R 6. c v d R = 6. 10 k θ = 8 k = ( 1.0 c ) (a) The reultant diplaceent wa.6 10 k [W8 N]. (b) To return directly back to Sydney, the airplane ut fly in a direction [E8 S]. The total ditance travelled wa 979 k (618 k + 361 k). However, the airplane path wa not traight. The two leg of the trip, plu the final return leg, for a triangle. Any ide of a triangle ut be horter than the u of the other two ide. Becaue the econd leg of the trip i directed partially toward the tarting point, you would expect the reultant vector to be uch le than the u of the other two ide. In fact, it i the hortet ide.
9. Frae the Proble - Make a cale diagra of the proble. Scale 1.0 c =.0 k d B = 4.0 k[w] y[n] d R = 5.0 k[w37 N] d A = 3.0 k[n] θ = 37 x[e] - The canoeit trip conited of two eparate tep repreented by the two diplaceent vector in the diagra. - The vector u of the two vector yield one reultant vector that how the canoeit final diplaceent. - To travel traight hoe the canoeit will have to travel a diplaceent that i equal in agnitude to her reultant vector and oppoite in direction. (a) The diplaceent d R, of the two part of the canoeit trip (b) The direction he would have to head her canoe to paddle traight hoe d A = 3.0 k[n] d B = 4.0 k[w] d R [direction] Meaure the length of the reultant diplaceent vector in the cale diagra. Multiply the length of the vector by the cale factor. With a protractor, eaure the angle between the horizontal axi and the reultant vector. (a) The reultant diplaceent i 5.0 k [W37 N]. d R =.5 c ( ) d R =.5 c.0 k 1.0 c d R = 5.0 k θ = 37 (b) The direction he would have to head her canoe to travel traight hoe i oppoite to [W37 N], or [E37 S]. The total ditance the canoeit paddled wa 7.0 k (3.0 k + 4.0 k). However, her path wa not traight. The path the canoe travelled fored a triangle. Since any ide of a triangle ut be horter than the u of the other two, you would expect her trip will be horter than 7.0 k. In fact it wa.0 k horter. 10. Frae the Proble - Make a cale diagra of the proble. Chooe a cale of 1.0 c : 1.0 k. - The hiker trip conited of two eparate diplaceent. 3
- The vector u of the two diplaceent vector yield one reultant vector that give the hiker final diplaceent. The diplaceent, d R, of the firt two leg of the hiker trip. d A = 4.5 k d R d B = 6.4 k θ = 60.0 Meaure the length of the reultant vector in the cale diagra. Multiply the length of the vector by the cale factor. With a protractor, eaure the angle between the horizontal axi and the reultant vector. The reultant diplaceent wa 5.8 k [18 away fro the horizontal fro the lookout]. d R = 5.8 c ( ) d R = 5.8 c 1.0 k 1.0 c d R = 5.8 k θ = 18 The total ditance travelled wa 10.9 k (4.5 k + 6.4 k). However, the hiker path wa not traight. The two leg of the trip, plu the final return leg for a triangle. Any ide of a triangle ut be horter than the u of the other two ide. Becaue the econd leg of the trip i directed partially toward the tarting point (i.e. 60 away), you expect the reultant vector to be uch le than the u of the other two ide. In fact, it i between the longet and hortet ide. 11. Frae the Proble - Make a cale diagra of the proble. Chooe a cale of 1.0 c : 6.0 k. - The boat trip conited of three eparate diplaceent. - The vector u of the three diplaceent vector yield one reultant vector that give the boat final diplaceent. (a) The diplaceent, d R, of the three leg of the boat trip (b) The [direction] back to the boat hoe port d A = 1.0 k[n] d B = 31.0 k[w30.0 S] d C = 36.0 k[w10.0 N] d R [direction] Meaure the length of the reultant vector in the cale diagra. d R = 10.5 c 4
Multiply the length of the vector by the cale factor. With a protractor, eaure the angle θ = 11.0 between the horizontal axi and the reultant vector. (a) The reultant diplaceent wa 63.0 k [W11.0 N]. ( ) d R = 10.5 c 6.0 k 1.0 c d R = 63.0 k (b) To return directly back to the hoe port, the boat ut travel in a direction [E11.0 S]. The total ditance the boat travelled wa 87.0 k (1.0 k + 30.0 k + 36.0 k). However, the boat path wa not traight. The firt two egent of the trip ake a triangle with the hoe port. But the third egent of the trip i in a direction further away fro the hoe port. Therefore, you expect the reultant vector to be longer than any of the individual egent of the trip, though horter than the total ditance travelled. In fact, it i nearly twice a long a the longet egent. Practice Proble Solution Student Textbook page 98 99 1. (a) Frae the Proble - Etablih a coordinate yte and chooe a cale. Ue 1.0 c :.0 k. - The firt vector will be drawn fro the origin and the negative of the econd vector will be drawn fro the tip of the firt vector. - The reultant vector will be drawn fro the origin to the tip of the econd vector. Three vector: P Q, R Q, Q R P = 1 k[n] Q = 15 k[s] R = 10 k[n30 E] P Q R Q Q R (i) Meaure the length of the reultant vector in the cale diagra. Multiply the length of the vector by the cale factor. With a protractor, eaure the angle between the horizontal axi and the reultant vector. (ii) Repeat the above trategy for the next pair of vector. P Q = 13.5 c ( ) P Q = 13.5 c.0 k 1.0 c P Q = 7 k θ = 0 R Q = 1.0 c ( ) R Q = 1.0 c.0 k R Q = 4 k θ = 1 1.0 c 5
(iii) Repeat the above trategy for the lat pair of vector. Q R = 1.0 c ( ) Q R = 1.0 c.0 k 1.0 c Q R = 4 k θ = 1 (a) The reultant vector are: (i) 7 k[n] (ii) 4 k[1 E] and (iii) 4 k[s1 W] (b) Frae the Proble - Etablih a coordinate yte and chooe a cale. Ue 1.0 c :.0 k. - Both the firt and vector will be drawn fro the origin. - The reultant vector will be drawn fro the tip of the econd vector to the tip of the firt vector. Three vector: P Q, R Q, Q R P = 1 k[n] Q = 15 k[s] R = 10 k[n30 E] P Q R Q P R (i) Meaure the length of the reultant vector in the cale diagra. Multiply the length of the vector by the cale factor. P Q = 13.5 c ( ) P Q = 13.5 c.0 k P Q = 7 k 1.0 c With a protractor, eaure the angle between the horizontal axi and the reultant vector. (ii) Repeat the above trategy for the next pair of vector. (iii) Repeat the above trategy for the lat pair of vector. θ = 0 R Q = 1.0 c ( ) R Q = 1.0 c.0 k 1.0 c R Q = 4 k θ = 1 P R = 3.0 c ( ) P R = 3.0 c.0 k 1.0 c P R = 6.0 k θ = 34 The reultant vector are: (i) 7 k[n] (ii) 4 k[n1 E] and (iii) 6.0 k[w34 N] The different ethod give the ae olution for part (i) and (ii). Particularly, for (a) (i), (ii) and (iii), ince the vector are pointed in oppoite direction, you expect their difference to be larger than either of the, which it i. For (b)(iii), ince the vector are pointed in nearly the ae direction, you expect their difference to be aller than either of the, which it i. 6
13. Frae the Proble - Make a cale diagra of the car initial and final velocitie. Ue 1 c : 5 k. - The car change both it agnitude and direction of velocity. - Either of the graphical vector ubtraction ethod can be ued. The car change in velocity, v. v1 = 45 k h [E] v = 50 k h [N] v Write the atheatical definition for v = v v 1 the change in velocity Draw a coordinate yte and chooe a cale. Ue 1.0 c = 5 k/h. Put v in the coordinate yte. Place the tail of vector v 1 at the tip of v and draw v. Meaure the agnitude and direction of v. Multiply the agnitude of the vector by the cale factor 1.0 c = 5 k/h. v = 13.4 c, θ = 48 v = 13.4 c v = 67 k h The car change in velocity wa 67 k/h[w48 N] ( 5 k ) h 1.0 c The agnitude of the change in the car velocity i larger than the car initial or final velocity, which it hould be becaue the initial and final velocitie are 90 away fro one another. In thi cae, the change in velocity ake up the hypotenue of a right triangle, and the hypotenue i the longet ide of a right triangle. The olution can be checked uing the other graphical vector ubtraction ethod. 14. Frae the Proble - Make a cale diagra of the airplane initial and final velocitie. Ue 1 c : 0 k/h. - The airplane change both it agnitude and direction of velocity. - Either of the graphical vector ubtraction ethod can be ued. The airplane change in velocity, v. v1 =.00 10 k h [S30.0 W] v =.00 10 k h [E] v Write the atheatical definition for v = v v 1 the change in velocity 7
Draw a coordinate yte and chooe a cale. Ue 1.0 c = 0 k/h. Put v in the coordinate yte. Place the tail of vector v 1 at the tip of v and draw v. Meaure the agnitude and direction of v. Multiply the agnitude of the vector by the cale factor 1.0 c = 0 k/h. v = 17.3 c, θ = 30 v = 17.3 c v = 346 k h The airplane change in velocity wa 346 k/h[e30.0 N] ( 0 k ) h 1.0 c The agnitude of the change in the airplane velocity i larger than the airplane initial or final velocity, which it hould be becaue the initial and final velocitie are ore than 90 away fro one another. The olution can be checked uing the other graphical vector ubtraction ethod. 15. Frae the Proble - Make a cale diagra of the puck initial and final velocitie. Ue 1.0 c : /. - The puck change both it agnitude and direction of velocity. - Either of the graphical vector ubtraction ethod can be ued. The puck change in velocity, v. v1 = 1 [30 to the board] v = 10 [5 to the board] v Write the atheatical definition for v = v v 1 the change in velocity Draw a coordinate yte and chooe a cale. Ue 1.0 c = /. Put v in the coordinate yte. Place the tail of vector v 1 at the tip of v and draw v. Meaure the agnitude and direction of v. Multiply the agnitude of the vector by the cale factor 1.0 c = /. v = 5. c, θ = 7 v = 5. c v = 10.4 ( ) 1.0 c The puck change in velocity wa 10 /[7 ] away fro the noral to the board, toward the puck initial direction. The agnitude of the change in the puck velocity i between the puck initial and final velocity. Thi i reaonable becaue you don t expect the puck to loe uch energy a it deflect off the board. The olution can be checked uing the other graphical vector ubtraction ethod. 8
16. Frae the Proble - Make a cale diagra of the runner velocity at the four point along the route. Ue 1.0 c : 1.0 / - The runner change both hi agnitude and direction of velocity at each point along the route. - Either of the graphical vector ubtraction ethod can be ued. The runner change in velocity, v 1, v 31, v 41 at three point along the route v1 = 3.5 [S] v 1 v = 5.0 [N1 W] v 31 v3 = 4. [W] v 41 v4 =.0 [S76 E] Write the atheatical definition for v = v v 1 the change in velocity Draw a coordinate yte and chooe a cale. Ue 1.0 c = 1.0/. Put v in the coordinate yte. Place the tail of vector v 1 at the tip of v and draw v 1. Meaure the agnitude and direction of v 1. Multiply the agnitude of the vector by the cale factor 1.0 c = 1.0 /. Repeat the above procedure for v 31. Repeat the above procedure for v 41. v 1 = 8.4 c, θ = 7 ( v 1 1.0 ) = 8.4 c 1.0 c v 1 = 8.4 v 31 = 5.5 c, θ = 40 ( v 31 1.0 ) = 5.5 c 1.0 c v 31 = 5.5 v 41 = 3.6 c, θ = 57 ( v 41 1.0 ) = 3.6 c 1.0 c v 41 = 3.6 (a) The runner change in velocity wa 8.4 /[N7 W]. (b) The runner change in velocity wa 5.5 /[N40 E]. (c) The runner change in velocity wa 3.6 /[E57 N]. (a) The individual velocitie are in oppoite direction, o the agnitude of their difference hould be larger than either of the, which it i. (b) The change in velocity for the hypotenue of a right triangle, which i larger than either of the ide of the triangle, a i calculated. (c) The change in velocity ut be le than the u of the two ide of the triangle, which it i. 9
Practice Proble Solution Student Textbook page 10 103 17. Frae the Proble - Make a diagra of the proble. - The hang-glider change coure while flying; diplaceent i a vector that depend on the initial and final poition. - Velocity i the vector quotient of diplaceent and the tie interval. - The direction of the hang-glider average velocity i therefore the ae a the direction of her diplaceent. The average velocity of the trip, v ave. d A = 6.0 k[s] d B = 4.0 k[nw] = 45 in = 0.75 h d R vave Chooe a coordinate yte; ue a cale of 1 c : 1.0 k. Draw the firt diplaceent vector. Draw the econd diplaceent vector fro the tip of the firt diplaceent vector. Meaure the length and direction of the reultant vector. Multiply the length of the reultant vector by the cale factor. d R = 4.3 c, θ = 43 ( d R = 4.3 c 1.0 k 1.0 c d R = 4.3 k ) Ue the equation that define average vave = d velocity to calculate the agnitude of vave = the average velocity The average velocity for the trip wa 5.7 k/h[s4 W] 4.3 k 0.75 h = 5.7 k/h Becaue you don t know the tie for the individual egent of the trip, you have nothing to which to copare the average velocity. The wind caued a ignificant change of coure. Since the hang-glider ade a harp turn, you expect the total diplaceent, 4.3 k, to be horter than the u of the two egent of the trip (6.0 k + 4.0 k), which it i. 18. Frae the Proble - Make a diagra of the proble. - The plane flight i in two egent which are given a diplaceent; diplaceent i a vector that depend on the initial and final poition. - Velocity i the vector quotient of diplaceent and the tie interval. - The direction of the plane average velocity i the ae a the direction of it diplaceent. 30
(a) The average velocity of the trip, v ave. (b) The velocity needed to return in 1.0 h, v R v d A = 195 k[n15 W] v d B = 149 k[n33 E] = 375. h d v R vave vr Chooe a coordinate yte; ue a cale of 1 c : 10.0 k. Draw the firt diplaceent vector. Draw the econd diplaceent vector fro the tip of the firt diplaceent vector. Meaure the length and direction of the reultant vector. Multiply the length of the reultant vector by the cale factor. Ue the equation that define average velocity to calculate the agnitude of the average velocity To find the velocity required to return in 1.0 h, ue the equation for velocity. (a) The average velocity for the trip wa 84 k/h[e84 N]. (b) To fly back in 1.0 h, the pilot need to fly at 3. 10 k/h[w84 S] The average velocity i all for a plane, but conidering the ditance travelled in the rather long period of tie, the value i reaonable. 19. Frae the Proble - Make a diagra of the proble. - The canoeit trip i in two egent which are given in ter of velocitie and tie interval. - The total diplaceent i the vector u of the two diplaceent vector. The diplaceent of the canoeit, d total A = 30.0 in B = 15.0 in vaave = 3.0 [N] vbave =.5 [W] d A d B d total v d R = 31. 5 c θ = 84 10.0 k d R = 31. 5 c ( 1.0 c ) d R = 315 k v d vave = 315 k vave = = 84 k / h 375. h v d vr = 315 k v = = 315 k / h R 1.0 h 31
Ue the velocity for egent A to calculate the diplaceent for egent A. Multiply by the nuber of econd in 1 inute to convert inute to econd. Chooe a coordinate yte; ue a cale of 1 c : 1000. Draw the diplaceent vector, d A with it tail at the origin. Ue the velocity for egent B to calculate the diplaceent for egent B. Multiply by the nuber of econd in 1 inute to convert inute to econd. Draw the diplaceent vector, d B, with it tail at the tip of vector A. Draw the reultant diplaceent vector, d total, fro the tail of A to the tip of B. Meaure the agnitude and direction of the total diplaceent vector. Multiply the agnitude of the vector by the cale factor. d A = v A A ( ) d A = 3.0 [N](30.0 in) 60 1 in d A = 5400 [N] d B = v B B ( ) d B =.5 [N](15.0 in) 60 1 in d B = 50 [W] d total = 5.8 c, θ = 3 ( ) d total = 5.8 c 1000 1.0 c d total = 5800 The diplaceent for the canoe trip i 5.8 10 3 [N3 W]. In thi cae, the total diplaceent i the hypotenue of a right triangle. The hypotenue i longer than either of the two individual ide of the triangle, a i the cae here. 0. Frae the Proble - Make a diagra of the proble. - The hiker trip i in three egent which are given in ter of velocitie and tie interval. - The total diplaceent i the vector u of the three diplaceent vector. (a) The diplaceent of the hiker, d total (b) The hiker average velocity, v ave A = 48.0 in B = 40.0 in C = 1.5 h vaave = 5.0 k h [N35 E] vbave = 4.5 k h [W] d C = 6.0 k[n30 W] d A d B d C d total vave 3
Ue the velocity for egent A to calculate the diplaceent for egent A. Multiply by the nuber of inute in 1 hour to convert inute to hour. Chooe a coordinate yte; ue a cale of 1 c : 1 k. Draw the diplaceent vector, d A with it tail at the origin. Ue the velocity for egent B to calculate the diplaceent for egent B. Multiply by the nuber of inute in 1 hour to convert inute to hour. Draw the diplaceent vector, d B, with it tail at the tip of vector A. Draw the diplaceent vector C, with it tail at the tip of vector B. Draw the reultant diplaceent vector, d total, fro the tail of A to the tip of C. Meaure the agnitude and direction of the total diplaceent vector. d A = v A A ( ) d A = 5.0 k h [N35 E](48.0 in) 1 h 60 in d A = 4.0 k[n35 E] d B = v B B ( ) d B = 4.5 k h [W](40 in) 1 h 60 in d B = 3.0 k[w] d total = 9. c, θ = 4 ( ) d total = 9. c 1.0 k 1.0 c d total = 9. k Multiply the agnitude of the vector by the cale factor. Ue the equation that define vave = d average velocity to find the average velocity. Convert all the tie to hour. vave = (a) The diplaceent i 9. k[n4 W]. 9. k 1h 48 in( ) + 40 in( 1h ) + 1.5 h 60 in 60 in vave = 3.1 k/h (b) The average velocity i 3.1 k/h[n4 W]. On the third egent of the trip, the hiker travelled further away fro the tarting point than on the other egent, o the total diplaceent hould be larger than any individual diplaceent, which it i. Note that average velocity i different fro average peed, o it cannot be deterined fro the peed on the individual egent. For the average velocity, a agnitude of 3.1 k/h ee reaonable conidering the data. 33
Practice Proble Solution Student Textbook page 110 1. Frae the Proble - Make a diagra of the proble. - The kayak i oving relative to the river and the river i oving relative to the hore. - The velocity of the kayak relative to the hore i the u of the velocity of the kayak relative to the river and the velocity of the river relative to the hore. - The vector for the velocity of the kayak relative to the river and the river relative to the hore are in oppoite direction. The velocity of the river relative to the hore, v r. vkr = 3.5 /[uptrea] vk = 1.7 /[uptrea] vr Write an equation that decribe the ituation. Subtitute known value. vk = v kr + v r 1.7 = 3.5 + v r Solve for the unknown, v r vr = 1.7 3.5 vr = 1.8 The velocity of the current i 1.8 /[downtrea]. Becaue the kayak and the river are travelling oppoite direction, they ut have oppoite ign, which they do. Oberver on the hore ee the kayak paddling about half a fat a he think he going. The other half of the peed ut be due to the river.. Frae the Proble - Make a diagra of the proble. - The jet-ki i heading due outh, but the river i carrying it to the wet. - The velocity of the jet-ki relative to the hore, v j. will be the hypotenue of a right triangle whoe other ide are the velocity of the river relative to the hore, v r and the velocity of the jet-ki relative to the river, v jr. The velocity of the jet-ki relative to the hore, v j vjr = 11 /[S] vr = 5.0 /[W] vj 34
Ue the Pythagorean theore to find the agnitude of the velocity of the jet-ki relative to the hore. Find the direction of thi velocity fro the tangent. vj = vjr + vr vj = vj = 1.1 (11 ) + (5.0 ) tan θ = v r v jr = 5.0 11.0 = 0.4545 θ = tan 1 0.4545 θ = 4.4 The velocity of the jet-ki relative to the hore i 1 /[S4 W]. Becaue the jet-ki i being carried by the current in the river, it velocity relative to the hore hould be larger than it velocity relative to the river and the velocity of the river relative to the hore, which it i. 3. Frae the Proble - Make a ketch of the proble. y[n] Scale 1 c = 40 k/ vpw = 10 k/h v wg = 40 k/h[e] α 60 x[e] vpg θ ϕ 30 - The pilot will be carried eat by the wind, o he will ai her plane lightly wet of the lake. - Her ground peed, v pg,will be greater than her air peed ( v pw = 10 k/h ), becaue, in a ene, he going with the wind ( v wg = 40 k/h ) (a) The direction that the pilot hould head her plane, θ (b) The velocity of the plane relative to the ground, v pg, for that direction (c) The tie it will take her to reach the lake, vpw = 10 k/h vpg vwg = 40 k/h [E] [direction, θ] 35
A the vector do not for a right triangle, the ine law can be ued to olve for the unknown direction, θ.. To get the heading fro North, note that θ + ϕ = 30 Apply the ine law again to deterine the velocity of the plane relative to the ground, v pg. The angle, α can be deterined fro τhe other two angle in the triangle, α + θ + 60 = 180 vpw in 60 = vwg in θ vwg in 60 in θ = vpw in 60 in θ = 40 k h 10 k h in θ = 0.16496 θ = in 1 0.16496 θ = 9.49 ϕ=30 θ ϕ=30 9.49 ϕ=0.51 vpw in 60 = vpg in α vpg = vpw in α in 60 vpg = 10 k h vpg = 7.1 k h in(180 60 9.49) in 60 Ue the definition of velocity to find the tie required to coplete the journey. vave = d = d vave = 50 k 7.1 k h = 1.10 h (a) The direction he hould head her plane i [N0.5 E]. (b) With thi heading, her velocity relative to the ground will be 7 k/h[n30.0 E]. (c) It will take her 1.10 h to reach her detination. The wind peed i not that trong copared to her velocity with repect to the air o you don t expect her heading to be too far away fro her detination. In fact, he only ha to point her plane 9.5 away fro her detination, (or 0.5 away fro North), which ee reaonable. Her velocity with repect to the ground i larger than her velocity with repect to the air, a it hould be. The unit cancelled in the tie deterination to give the reult in hour. 4. Frae the Proble - Make a diagra of the proble. - The airplane trip i in three egent which are given in ter of diplaceent and tie interval. - The total diplaceent i the vector u of the three diplaceent vector. - Velocity i the vector quotient of the diplaceent and the tie interval. (a) The diplaceent of the airplane, d total (b) The velocitie, v Aave, v Bave, v Cave, for each egent of the trip. (c) The average velocity for the total trip, v ave. 36
A = 0.0 in B = 40.0 in C = 1.0 in d A = 1.0 10 k[n] d B = 1.5 10 k[w] d C = 5.0 10 k[s] d total vaave vbave vcave vave Chooe a coordinate yte; ue a cale of 1 c : 50 k. Draw the diplaceent vector, d A with it tail at the origin. Draw the diplaceent vector, d B, with it tail at the tip of vector A. Draw the diplaceent vector C, with it tail at the tip of vector B. Draw the reultant diplaceent vector, d total, fro the tail of A to the tip of C. Meaure the agnitude and d total = 3. c/ θ = 18 direction of the total diplaceent vector. Multiply the agnitude of the ( ) d total = 3. c 50 k 1.0 c vector by the cale factor. d total = 160 k Calculate the velocity for each egent fro the diplaceent and the tie interval. Convert the tie interval to hour. va = d A A va = d A A va = 1.0 10 k[n] 0.0 in( 1h ) 60 in va = 3.0 10 k[n] vb = d B B vb = 1.5 10 k[w] 40.0 in( 1h ) 60 in vb =.5 10 k[w] vc = d C C vc = 5.0 101 k[s] 1.0 in( Ue the equation that define vave = d average velocity to find the average velocity. Convert all the tie to hour. vave = 1h ) 60 in vc =.50 10 k[s] 160 k 1h 0.0 in( ) + 40.0 in( 1h ) + 1.0 in 60 in 60 in vave = 133.33 k/h 37
(a) The diplaceent i 1.6 10 k[w18 N]. (b) The velocitie for the three egent are: 3.0 10 k/h[n],. 10 k/h[w],.5 10 k/h[s] (c) The average velocity for the total trip i: 1.3 10 k/h. The diplaceent, 160 k, i le than the total ditance travelled (100 k + 150 k + 50 k), a it hould be. The individual velocitie ee reaonable. The pilot went both north and outh, o you expect the average velocity for the trip to be le than the individual velocitie for thoe egent, which it i. The total diplaceent can be checked uing the other graphical ethod. 5. Frae the Proble - Make a diagra of the proble. - The wier ut wi at an angle uptrea o that between hi effort and the river, he travel traight acro. - The velocity of the wier relative to the river, v r will be the hypotenue of a right triangle whoe other ide are the velocity of the river relative to the ground (or the hore), v rg, and the velocity of the wier relative to the ground, v g. - The velocity of the river relative to the ground, v rg, can be deterined fro the ditance and tie interval that the tick travel. - The velocity of the wier relative to the ground, v g, need to be deterined before the tie to cro can be deterined. (a) The direction the wier hould head, θ, in order to land directly acro. (b) The tie it will take the wier to cro the river,. vr = 1.9 / d river = 10 [N] d tick = 4 [W] tick = 30.0 θ vrg vg Ue the definition of velocity to find the velocity of the current fro the data for the tick. vrg = d tick vrg = Now, the agnitude of the hypotenue, in θ = vr, and oppoite ide of the triangle, vrg, are known. Find the angle, θ. tick 4 [W] 30.0 vrg = 0.8 [W] vrg vr in θ = 0.8 1.9 in θ = 0.4105 θ = in 1 0.4105 θ = 4.9 38
Ue the Pythagorean theore to find the agnitude of the velocity of the wier relative to the ground, v g vr = vg + vrg vg = vr vrg (1.9 ) (0.8 ) vg = vg = 1.73 Find the tie to cro the river fro the definition of velocity. vg = d river = d river vg = 10 [N] 1.73 [N] = 69.63 (a) The wier hould head in a direction [N5 E] to arrive directly acro the river. (b) The trip would take 69 econd. In each cae, the unit cancelled to give the correct unit for the deired quantity. You expect the wier velocity relative to the ground to be le than the wier velocity relative to the river becaue he ha to direct hielf uptrea. Thee value were oberved. 6. Frae the Proble - Make a diagra of the proble. - The hiker trip i in three egent which are given in ter of diplaceent. - The total diplaceent i the vector u of the three diplaceent vector. - Speed i the calar quotient of the ditance and the tie interval. It can be ued to find the total tie of the trip. (a) The diplaceent of the hiker, d total. (b) The direction, θ, the hiker would have to head to return to her tarting point. (c) The total tie of the trip. d A = 4.0 k[n40.0 W] d B = 3.0 k[e10.0 N] d C =.5 k[s40.0 W] v = 4.0 k/h d total θ Chooe a coordinate yte; ue a cale of 1c:1.0 k. Draw the diplaceent vector, d A with it tail at the origin. Draw the diplaceent vector, d B, with it tail at the tip of vector A. Draw the diplaceent vector C, with it tail at the tip of vector B. Draw the reultant diplaceent vector, d total, fro the tail of A to the tip of C. 39
Meaure the agnitude and direction of the total diplaceent vector. Multiply the agnitude of the vector by the cale factor. d total =.1 c, θ = 54 ( ) d total =.1 c 1.0 k 1.0 c d total =.1 k Calculate the total tie fro the total ditance and the peed. (a) The diplaceent i.1 k[w54 N]. v = d = d v = (4.0 + 3.0 +.5) k 4.0 k h =.375 h (b) To head traight back, he hould walk in the direction [S54 E]. (c) The total tie taken for the trip i.4 h. In each cae, the unit cancelled to give the correct unit for the deired quantity. The diplaceent,.1 k, i le than the total ditance travelled (4.0 k +3.0 k +.5 k), a it hould be. Given the peed that he walk, hiking about 10 k in jut over h ee reaonable. 7. Frae the Proble - Make a diagra of the proble. - The canoeit i heading due eat, but the river i carrying the canoe to the outh. - The velocity of the canoeit relative to the hore, v c will be the hypotenue of a right triangle whoe other ide are the velocity of the river relative to the hore, vr, and the velocity of the canoeit relative to the river, v cr. The velocity of the canoe relative to the hore, v c vcr = 1.5 /[E] vr = 0.50 /[S] vc Ue the Pythagorean theore to find the agnitude of the velocity of the canoe relative to the hore. Find the direction of thi velocity fro the tangent. vc = vcr + vr vc = (1.5 ) + (0.50 ) vc = 1.58 tan θ = v r v jr = 0.50 = 0.3333 1.5 θ = tan 1 0.3333 θ = 18.4 The velocity of the canoeit relative to the hore i 1.6 /[E18 S]. In each cae, the unit cancelled to give the correct unit for the deired quantity. Becaue the canoeit i being carried by the current in the river, the canoe velocity relative to the hore hould be larger than it velocity relative to the river and the velocity of the river relative to the hore, which it i. 40
Anwer to Proble for Undertanding Student Textbook page 116 117 11. The initial velocity of the runner wa 3 /. v f = v i + a Solve for v i v f = v i + a = 6.4 (0.3 )(1 ) =.8 = 3 1. The baeball would have a velocity of 1.9 /[down] after 4.0. v f = v i + a v f = 4.5 + ( 1.6 )(4.0 ) = 4.5 6.4 = 1.9 13. (a) The final velocity of the car i 17 /. d = ( v i + v f ) 50 = ( 0 + v f )(6.0 ) 50 = (v f )3.0 v f = 16.7 (b) The acceleration of the car i.8 /. a = v f v i = 16.7 0 6.0 =.8 14. (a) The cyclit travel 7 during the 4.0 interval. d = v i + 1 a = (5.6 )(4.0 ) + 1 (0.6 )(4.0 ) =.4 + 4.8 = 7. (b) The cyclit attain a velocity of 8.0 /. v f = v i + a = 5.6 + 0.6 (4.0 ) = 5.6 +.4 = 8.1 15. (a) The acceleration of the truck i a = 1. /. To derive an equation that relate initial and final velocitie and ditance to acceleration, tart with the following. a = v f v i Solve for and ubtitute it into the following. v f = v i + a 41
Rearrange the reulting equation to obtain the following (ee Think It Through on page 8 of the textbook). v f = v i + ad ( 14 ) ( = ) + a(15 ) 196 = 484 + a(50 ) a = (196 484) 50 a = 1.15 (b) It took 6.9 for the truck driver to change hi peed. Solve for in the following. a = v f v i = v f v i a = 14 1.15 = 6.94 16. Find the initial velocity. v f = v i + a v i = v f a v i = ( 30 ) ( 3. )8.0 v i = ( 30 ) + 5.6 v i = 4.4 Find the diplaceent. d = v i + 1 a = d ( 4.4 )8.0 + 1 ( 3. )(8.0 ) = d ( 4.4 ) 10.4 = d = 106.8 The ky diver diplaceent wa 1.1 10 or 1.1 10 [down]. 17. When the police car catche up with the peeder, both vehicle will have travelled the ae ditance during the ae tie interval. (a) It will take the police car 3 to catch up to the peeder. d peeder = d police v = 1 a 4 = 1 (.1 ) 1 (.1 ) 4 = 0 1.05 4 = 0 ( 1.05 4 ) = 0 = 0 or 1.05 4 = 0 1.05 = 4 =.86 Note that the olution, = 0, repreent the fact that the tie at which the peeder paed the police car the firt tie. = 3 i the econd tie that the peeder and the police car were at the ae poition, that i, when the police car caught up with the peeder. 4
(b) Each vehicle will travel 5.50 10. Ue either forula d = v for the peeder or d = 1 a for the police car to obtain the ditance. d = v d = 4 (.86 ) = 548.57 18. (a) Refer to the vector diagra and calculation below. Scale 1.0 c = 0 k/h y[n] k 71 [SW] h 45 45 k 50 [W] h k 50 [N] h x[e] v = v x + vy v ( = v = 500 v = 5000 v = 71 k h ) + ( 50 k h ( k h ( ) k h 50 k h ) ) + 500 ( k tan θ = 50 k h =+1 50 k h θ = tan 1 1 = 45 v = 71 k h [W45 S] or 71 k h [SW] (b) The acceleration during the turn i 3.9 / [SW], a hown. 71 k 1h 1000 h 3600 1k = 19.7 a = v 19.7 a = [SW] 5.0 a = 3.9 [SW] 19. (a) Refer to the vector diagra below. Scale 1.0 c = 1.0 k y[n] h ) 3.6 k[s34 W] 34 3.0 k[s].0 k[w] x[e] (b) The total diplaceent fro the cale diagra above i 3.6 k[s34 W]. 43
0. Refer to the cale vector diagra and calculation below. Scale 1.0 c =.0 k y[n] 3.4 k[w] 5.6 k[n] d = 6.6 k[n31 W] 31 x[e] d = dx + d y d = ( 3.4 k) + (5.6 k) d = 11.56 k + 31.36 k d = 4.9 k d = 6.6 k 3.4 k tan θ = +5.6 k = 0.607 θ = tan 1 0.607 = 31 d = 6.6 k[n31 W] (b) total = 1h+ 0.5 h = 1.5 h v = d v = 6.6 k[n31 W] 1.5 h v = 4.4 k h [N31 W ] 1. (a) The cale vector diagra below repreent the change in velocity. Scale 1.0 c = / y [N] = 5.6 [W] = 5.6 [E] vi v = 7.9 [N45 W] 7.9 or [NW] vf = 5.6 [N] x [E] 44
(b) The following i a calculation for the change in velocity. v = v x + vy v ( ) = 5.6 ( ) + 5.6 v ( ) ( ) = 31.36 + 31.36 v = 6.7 v = 7.9 ( ) tan θ = 5.6 = 1 +5.6 θ = tan 1 1 = 45 v = 7.9 [N45 W] or 7.9 [NW]. (a) Refer to the cale vector diagra and the calculation below. y[n] Scale: 1 c = 4 k 16. k[w] 4 7. k[s] d = 18 k[w 4 S] x d x = v x d x = 6.0 [W] 45 in 60 d x = 16. k[w] 1 in d y = v y d y = 4.0 [S] 30 in 60 1 in d x = 7. k[s] d = d x + d y d = ( 16. k) + ( 7. k) d = 6.4 k + 51.8 k d = 314.8 k d = 17.7 k = 18 tan θ = d y d x 7. k tan θ = 16. k θ = tan 1 0.444 θ = 4 d = 18 k[w4 S] 45
(b) total = 45 in + 30 in = 75 in 75 in 1h 60 in = 1.5 h v = d v = 18 k[w4 S] 1.5 h v = 14.4 k h [W4 S] = 14 k h [W4 S] 3. (a) Refer to the cale vector diagra and calculation below. Scale 1 c = 1 / (a) v = 1. WS [S] v TS = 1.3 [N] v TW =.5 [N] (b) v TW =.5 [S] v TS = 3.7 [S] vw = 1. [S] vtw =.5 [N] vw = 1. [S] vt = v Tw + v w vt =.5 [N] + 1. [S] vt = 1.3 [N] (b) If Thao wi downtrea, her velocity relative to the hore i 3.7 /[S] a hown in the above diagra and in the calculation below. vtw =.5 [S] v w = 1. [S] vt = v Tw + v w vt =.5 [S] + 1. [S] vt = 3.7 [N] 4. (a) Refer to the cale vector diagra below. 4. [S] vw = 5.0 vw = 0.84 [S] vpw = 1.9 in θ = 0.84 1.9 in θ = 0.441 θ = in 1 0.441 θ = 6 Hi direction hould be [E6 N]. He hould wi at 1.9 /[E6 N]. 46
(b) vpw = vp + vw ( ) 1.9 ( ) = vp + 0.84 ( ) vp ( = 3.61 0.7056 ( ) vp =.9044 ) (c) vp = 1.7 Hi velocity relative to the hore i 1.7 /[E]. = d vp = 4800 1.7 [E] = 84 1 in 60 = 47 in It will take the phyic teacher 47 in to cro the lake. Scale 1 c = 0.5 / y [N] vpw = 1.9 [E6 N] 6 vp = 1.7 [E] vw = 0.84 [N] x [E] 5. Refer to the cale vector diagra and the calculation below. Scale 1 c = 10 / y[n] vw = 1. [Ν0.0 E] vcwy = +3. y[n] vcw = 3. [N] 4.4 /[N5.4 E] x[e] vwy = 1.13 θ w = 70 x[e] vwx = 0.410 vc = v cw + v w vcw = 3. [N] vw = 1. [N0.0 E] vcw (y) =+3. θ w = 90 0.0 = 70 47
v w (x) = vw co θ w v w (x) = 1. co 70 v w (x) = 1. (0.340) v w (x) = 0.410 v w (y) = vw in θ w v w (y) = 1. in 70 v w (y) = 1. (0.9397) v w (y) = 1.13 Vector x-coponent y-coponent vcw 0.0 / 3. / vw 0.410 / 1.13 / vc 0.410 / 4.33 / vcg = vcg (x) + v cg (y) ( ) vcg = 0.410 ( ) + 4.33 ) ( + 18.75 vcg = 0.1681 ( ( vcg = 18.9 vcg = 4.4 ) tan θ cg = 4.33 0.410 = 10.56 θ cg = tan 1 10.56 θ cg = 84.6 The velocity of the canoeit relative to the hore i 4.4 /[N5.4 E]. 6. Refer to the cale vector diagra and the calculation below. ) Scale 1 c = k y[n] y[n] d x =.88 k d y = 4.99 k x[e] θ θ = 60 d = 5.76 k[n30.0 W] d 1 = 8.4 k[w] x[e] 1 k[w4 N] d 1 = 8.4 k[w] 48
1 = 50 in 60 1 in = 3000 d 1 = v 1 1 d 1 =.8 [W] 3000 d 1 = 8400 [W] = 8.4 k[w] = 30 in 60 1 in = 1800 d = v d = 3. [N30.0 W] 1800 d = 5760 [N30.0 W] d = 5.76 k[n30.0 W] θ = 90 30.0 = 60 d (x) = d co θ d (x) = 5.76 k co 60 d (x) = 5.76 k(0.5) d (x) =.88 k d (y) = d in θ d (y) = 5.76 k in 60 d (y) = 5.76 k(0.8660) d (y) = 4.99 k Vector x-coponent y-coponent d 1 8.4 k 0.0 k d.88 k 4.99 k d T 11.8 k 4.99 k d T = ( dt (x) ) + ( d T (y) ) d T = ( 11.8 k) + (4.99 k) d T = 17.4 k + 4.90 k d T = 15.14 k d T = 1.33 k = 1 k tan θ = d T (y) d T (x) 4.99 k tan θ = 11.8 k = 0.44 θ = tan 1 0.44 θ = 3.86 = 4 The diplaceent of the runner i 1 k[w4 N]. 49
7. Refer to the cale vector diagra and calculation below: y[n] Scale 1 c = 4 k d 1y = 1.3 k θ = 65 y[n] d y = 6.80 k θ 1 = 55 x[e] d 1x = 8.60 k θ d x = 3.17 k x[e] y[n] d = 7.5 k[n5 W] 0 k[n16 E] 16 d 1 = 15 k[n35 E] x[e] d 1 = 15 k[n35 E] d = 7.5 k[n5 W] θ 1 = 90 35 = 55 d 1 (x) = d 1 co θ 1 d 1 (x) = 15 k co 55 d 1 (x) = 15 k(0.5736) d 1 (x) = 8.60 k d 1 (y) = d 1 in θ 1 d 1 (y) = 15 k in 55 d 1 (y) = 15 k(0.819) d 1 (y) = 1.3 k θ = 90 5 = 65 d (x) = d co θ d (x) = 7.5 k co 65 d (x) = 7.5 k(0.46) d (x) = 3.17 k d (y) = d in θ d (y) = 7.5 k in 65 d (y) = 7.5 k(0.9063) d (y) = 6.80 k Vector x-coponent y-coponent d 1 8.60 k 1.3 k d 3.17 k 6.80 k d T 5.43 k 19.1 k 50
d T = ( dt (x) ) + ( d T (y) ) d T = (5.43 k) + (19.1 k) d T = 9.49 k + 364.81 k d T = 394.3 k d T = 19.86 k = 0 k tan θ = d T (y) d T (x) 19.1 k tan θ = 5.43 k = 3.5 θ = tan 1 3.5 θ = 74 (a) The jogger diplaceent i 0 k[n16 E]. (b) v ave = d v ave = 19.86 k[n16 E].0 h vave = 9.9 k h [N16 E] The jogger average velocity i 9.9 k[n16 E]. 8.Refer to the cale vector diagra and calculation below. Scale 1 c =.5 / y[n] θ f = 50 x[e] v i = 11.0 [W] y[n] x[e] v fx = 7.71 vfy = 9.19 v i = 11.0 [E] 19.7 v f = 1.0 [S40.0 E] vi = 11.0 [E] v i = 11.0 [W] vf = 1.0 [S40.0 E] v = 9.76 [S91.7 W] 51
θ f = 90 40 = 50 v f (x) = vf co θ f v f (x) = 1.0 co 50 v f (x) = 1.0 (0.648) v f (x) = 7.71 v f (y) = vf in θ f v f (y) = 1.0 in 50 v f (y) = 1.0 (0.7660) v f (y) = 9.19 Vector x-coponent y-coponent v i 11.0 / 0.0 vf 7.71 / 9.19 / v 3.9 / 9.19 / v = v x + v y v = v = 10.8 v = 95.8 v = 9.76 ( ) 3.9 ( + 9.19 ( ) ( + 84.46 ( ) tan θ = 9.19 =.79 3.9 θ = tan 1 (.79) θ = 70.3 a = v v = 9.76 [S19.7 W] = 45.0 9.76 a = [S19.7 W] 45.0 a = 0.17 [S19.7 W] The acceleration of the ailboat i 0.17 / [S19.7 W]. 9. (a) Refer to the cale vector diagra below. vw = 3.0 k h [downtrea] vcw = 4.0 k h in θ = v w vcw ) ) in θ = 3.00 k h 4.0 k h in θ = 0.75 θ = in 1 0.75 θ = 49 He hould ai uptrea at an angle of 41 with repect to the river bank. 5
(b) ( 4.0 k h vcw = vc + vw ) ( = vc + vc = 16 ( k h vc = 7 ( k h ) 3.0 k h ) ) 9 ( k h ) v c =.65 k h Hi velocity relative to the hore i.65 k/h. = d vc = 0.10 k.65 k h = 0.0377 h 60 in 1h =.3 in The trip will take hi.3 in. Scale 1.0 c = 1 k/h y[n] k v w= 3.0 h v cw = 4.0 k h 49 41 v c =.65 k h x[e] 53