Section 7.2 Velocity In the previous chpter, we showed tht velocity is vector becuse it hd both mgnitude (speed) nd direction. In this section, we will demonstrte how two velocities cn be combined to determine their resultnt velocity. EXAMPLE 1 Representing velocity with digrms An irplne hs velocity of (reltive to the ir) when it encounters wind hving velocity of w! v! (reltive to the ground). Drw digrm showing the possible positions of the velocities nd nother digrm showing the resultnt velocity. Solution v w + v, the resultnt velocity w v w Position digrm Vector digrm The resultnt velocity of ny two velocities is their sum. In ll clcultions involving resultnt velocities, it is necessry to drw tringle showing the velocities so there is cler recognition of the resultnt nd its reltionship to the other two velocities. When the velocity of the irplne is mentioned, it is understood tht we re referring to its ir speed. When the velocity of the wind is mentioned, we re referring to its velocity reltive to fixed, the ground. The resultnt velocity of the irplne is the velocity of the irplne reltive to the ground nd is clled the ground velocity of the irplne. EXAMPLE 2 Selecting vector strtegy to determine ground velocity A plne is heding due north with n ir speed of 400 km>h when it is blown off course by wind of 100 km>h from the northest. Determine the resultnt ground velocity of the irplne. CHAPTER 7 365
Solution We strt by drwing position nd vector digrms where represents the velocity of the wind nd v! w! represents the velocity of the irplne in kilometres per hour. NW W SW N S NE E SE w, )w ) = 100 v, )v ) = 400 w, )w ) = 100 v + w, resultnt v, )v ) = 400 Position digrm Vector digrm Use the cosine lw to determine the mgnitude of the resultnt velocity. 0 v! 0 v! w! w! 0 2 0 v! 0 2 0 w! 0 2 2 0 v! 00w! 0 cos u, u 45, 0 w! 0 100, 0 v! 0 400 0 2 400 2 100 2 21100214002cos 45 0 v! w! 0 2 160 000 10 000 80 000 1 V2 b 0 v! w! 0 2 80 000 170 000 0 v! w! V2 0 336.80 To stte the required velocity, the direction of the resultnt vector is needed. Use the sine lw to clculte, the ngle between the velocity vector of the plne nd the resultnt vector. w, )w ) = 100 )v + w* = 336.80 v, )v ) = 400 sin sin 45 100 336.80 100sin 45 sin 0.2099 336.80 12.1 Therefore, the resultnt velocity is pproximtely (or W77.9 N). 336.80 km>h, N12.1 W 366 7.2 VELOCITY
EXAMPLE 3 Using vectors to represent velocities A river is 2 km wide nd flows t 6 km>h. Ann is driving motorbot, which hs speed of 20 km>h in still wter nd she heds out from one bnk in direction perpendiculr to the current. A mrin lies directly cross the river from the strting on the opposite bnk.. How fr downstrem from the mrin will the current push the bot? b. How long will it tke for the bot to cross the river? c. If Ann decides tht she wnts to end up directly cross the river t the mrin, in wht direction should she hed? Wht is the resultnt velocity of the bot? Solution. As before, we construct vector nd position digrm, where w! nd v! represent the velocity of the river nd the bot, respectively, in kilometres per hour. upstrem 2 km upstrem 2 km strting w, )w* = 6 v, )v* = 20 mrin strting v, )v* = 20 v + w mrin w, )w* = 6 The distnce downstrem tht the bot lnds cn be clculted in vriety of wys, but the esiest wy is to redrw the velocity tringle from the vector digrm, keeping in mind tht the velocity tringle is similr to the distnce tringle. This is becuse the distnce trvelled is directly proportionl to the velocity. strting downstrem Position digrm v, )v* = 20 v + w mrin w, )w* = 6 downstrem Vector digrm strting 2 km x mrin d end 6 Using similr tringles,. 20 d, d 0.6 2 The bot will touch the opposite bnk 0.6 km downstrem. b. To clculte the ctul distnce between the strting nd end s, the Pythgoren theorem is used for the distnce tringle, with x being the required distnce. Thus, x 2 2 2 10.62 2 4.36 nd x 2.09, which mens tht the ctul distnce the bot trvelled ws pproximtely 2.09 km. To clculte the length of time it took to mke the trip, it is necessry to clculte the speed t which this distnce ws trvelled. Agin, using similr CHAPTER 7 367
v v + w w!! 20 0v w 0 tringles, Solving this proportion, 0 v! w! 0 20.9, so the 2 2.09. ctul speed of the bot crossing the river ws bout 20.9 km>h. The ctul time tken to cross the river is t d 2.09 0.1 h, or bout 6 min. v 20.9 Therefore, the bot lnded 0.6 km downstrem, nd it took pproximtely 6 min to mke the crossing. c. To determine the velocity with which she must trvel to rech the mrin, we will drw the relted vector digrm. We re given 0w! 0 6 nd 0v! 0 20. To determine the direction in which the bot must trvel, let represent the ngle upstrem t which the bot heds out. sin 6 20 or sin 1 6 20 b 17.5 To clculte the mgnitude of the resultnt velocity, use the Pythgoren theorem. 0 v! 0 2 0 w! 0 2 0 v! w! 2 0 where 0v! 0 20 nd 0w! 0 6 Thus, 20 2 6 2 0 v! w! 2 0 0 v! w! 0 2 400 36 0 v! w! 0 19.08 This implies tht if Ann wnts to trvel directly cross the river, she will hve to trvel upstrem 17.5 with speed of pproximtely 19.08 km>h. The nose of the bot will be heded upstrem t 17.5, but the bot will ctully be moving directly cross the river t wter speed of 19.08 km>h. IN SUMMARY Key Ide Problems involving velocities cn be solved using strtegies involving vectors. Need to Know The velocity of n object is stted reltive to frme of reference. The frme of reference used influences the stted velocity of the object. Air speed/wter speed is the speed of plne/bot reltive to person on bord. Ground speed is the speed of plne or bot reltive to person on the ground nd includes the effect of wind or current.!!! The resultnt velocity v r v1 v2. 368 7.2 VELOCITY
Exercise 7.2 PART A 1. A womn wlks t 4 km>h down the corridor of trin tht is trvelling t 80 km>h on stright trck.. Wht is her resultnt velocity in reltion to the ground if she is wlking in the sme direction s the trin? b. If she wlks in the opposite direction s the trin, wht is her resultnt velocity? 2. An irplne heding north hs n ir speed of 600 km>h.. If the irplne encounters wind from the north t 100 km> h, wht is the resultnt ground velocity of the plne? b. If there is wind from the south t 100 km>h, wht is the resultnt ground velocity of the plne? K PART B 3. An irplne hs n ir speed of 300 km>h nd is heding due west. If it encounters wind blowing south t 50 km>h, wht is the resultnt ground velocity of the plne? 4. Adm cn swim t the rte of 2 km>h in still wter. At wht ngle to the bnk of river must he hed if he wnts to swim directly cross the river nd the current in the river moves t the rte of 1 km> h? 5. A child, sitting in the bckset of cr trvelling t 20 m>s, throws bll t 2 m>s to her brother who is sitting in the front set.. Wht is the velocity of the bll reltive to the children? b. Wht is the velocity of the bll reltive to the rod? 6. A bot heds 15 west of north with wter speed of 12 m>s. Determine its resultnt velocity, reltive to the ground, when it encounters 5 m>s current from 15 north of est. 7. An irplne is heding due north t 800 km>h when it encounters wind from the northest t 100 km>h.. Wht is the resultnt velocity of the irplne? b. How fr will the plne trvel in 1 h? 8. An irplne is heded north with constnt velocity of 450 km>h. The plne encounters wind from the west t 100 km>h.. In 3 h, how fr will the plne trvel? b. In wht direction will the plne trvel? CHAPTER 7 369
A T C 9. A smll irplne hs n ir speed of 244 km>h. The pilot wishes to fly to destintion tht is 480 km due west from the plne s present loction. There is 44 km>h wind from the south.. In wht direction should the pilot fly in order to rech the destintion? b. How long will it tke to rech the destintion? 10. Judy nd her friend Helen live on opposite sides of river tht is 1 km wide. Helen lives 2 km downstrem from Judy on the opposite side of the river. Judy cn swim t rte of 3 km>h, nd the river s current hs speed of 4 km>h. Judy swims from her cottge directly cross the river.. Wht is Judy s resultnt velocity? b. How fr wy from Helen s cottge will Judy be when she reches the other side? c. How long will it tke Judy to rech the other side? 11. An irplne is trvelling N60 E with resultnt ground speed of 205 km>h. The nose of the plne is ctully ing est with n irspeed of 212 km>h.. Wht is the wind direction? b. Wht is the wind speed? 12. Brbr cn swim t 4 km>h in still wter. She wishes to swim cross river to directly opposite from where she is stnding. The river is moving t rte of 5 km>h. Explin, with the use of digrm, why this is not possible. PART C 13. Mry leves dock, pddling her cnoe t 3 m>s. She heds downstrem t n ngle of 30 to the current, which is flowing t 4 m>s.. How fr downstrem does Mry trvel in 10 s? b. Wht is the length of time required to cross the river if its width is 150 m? 14. Dve wnts to cross 200 m wide river whose current flows t 5.5 m>s. The mrin he wnts to visit is locted t n ngle of S45 W from his strting position. Dve cn pddle his cnoe t 4 m>s in still wter.. In which direction should he hed to get to the mrin? b. How long will the trip tke? 15. A stembot covers the distnce between town A nd town B (locted downstrem) in 5 h without mking ny stops. Moving upstrem from B to A, the bot covers the sme distnce in 7 h (gin mking no stops). How mny hours does it tke rft moving with the speed of the river current to get from A to B? 370 7.2 VELOCITY