Question

e. Once we choose a set of coordinatea, we can how much of a given vector liea along one of the coordinate axes. The portion of a vector that lies along an axia ia called 최 component vector and it too ia vector. However, vector's component is acalar which . The poaitive or negative aign indicatea whether the aasociated component veotor pointa in the agreed upon poaitive or negative coordinate direotion. Confuaed? Here ia an example. Fill one word into each blank. Choose from the worda in italicsa can be either poative or negative A poaition vector ia 10.0 cm long and makea a COW 143.13° angle with the poaitive a- axia. To three aignificant figurea, the -axcia 1a 8.00 cm2 while the 2-axi (Fortunately, we usually focus on 최 vector's component and lees often on vector of thia poaition vector ia -8.00 om. component- 8. A car movea along a atraight atreet. The car'a initial poaition at time ti ia given by a poaition vector , The car's later poation vector ia x, at time t2. Suppose → = 700 mi and x, = 400 ms a. Between t1 and勾, did the oar move to the right or to the left? Customarily, right is in the positive a-direction. diaplacement vector uaing a component and a unit vector The next two parta relate to the firat section of Chapter 2. You can wait until the next b. The car's displacement is the difference vector Δ→ 2,-, write the car's rk aaaignment to anawer them. omewor 9. Tarzan and Jane have an argument and aimultaneoualy walk away from where they were atanding. Tarzan atomps off for 42 metera in a direotion 50 weat of north and then atopa Jane walka 24 metera in a direotion 80° south of eaat and then atopa. After aome aulking Tarzan realizea he waa at fault and turna to face Jane a. Draw a vector addition diagram for thia aituation to some relative acale b. How far muat Tarzan walk in a atraight line to get to Jane? Solve thia queation uaing veotor addition and veotor componenta. Show all atepa. Anawer: 62.3 m o. In what direction muat he walk? Anawer: 54.3° south of eaat

Answer 1

(8)
Initial position (X₁) = 700 i
Final position(X₂) = 400 i
(a) Since the final position is near origin then initial one therefore the car moves toward left.
(b)
Now the car displacement
$\Delta \vec{X} = \vec{X_{2}} - \vec{X_{1}}$
$\Delta \vec{X} = \left ( 400 i - 700 i \right )$
$\Delta \vec{X} = -300 i$
-ve sign indicate that car has move left wards.


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