homework 4 solutions - CHAPTER 4 8 Our coordinate system...

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Unformatted text preview: CHAPTER 4 8. Our coordinate system has i pointed east and pointed north. The first displacement is 7,, = (483 km)i and the second is PM = (—966 km}. (a) The net displacement is PAC :7“, +81% =(483 km)i—(966 hmfi which yields 1?“ |=,/(483 km)2+(—966 km)2 =1.08><103 km. (b) The angle is given by = —63.4°. 6 = tan‘1 (L66 483 km We observe that the angle can be alternatively expressed as 63.4° south of east, or 26.6° east of south. (c) Dividing the magnitude of FAC by the total time (2.25 h) gives 9 _ (483 km)i—(966 km} “E 2.25 h with a magnitude [$an i=,/(215 km/h)2 +(—429 km/h)2 =480 km/h. ((1) The direction of i7an is 26.6° eastvo‘f south, same as in part (b). In magnitude-angle notation, we would have 17,“ = (480 km/h 4 —63.4°). =(215 km/h)i—(429 km/h)i (e) Assuming the AB trip was a straight one, and similarly for the BC trip, then Jim] is the distance traveled during the AB trip, and [17%| is the distance traveled during the BC trip. Since the average speed is the total distance divided by the total time, it equals 483 km + 966 km 2.25 h = 644 km/h. 15. Since the acceleration, 21' : axi + ay3=(—1.0 m/s2)i"+ (—0.50 m/s2 )3 , is constant in both x and y directions, we may use Table 2—1 for the motion along each direction. This can be handled individually (for x and y) or together with the unit-vector notation (for AF). The particle started at the origin, so the coordinates of the particle at any time t are given by 17 = 170t +§c7t2. The velocity of the particle at any time t is given by 17 = 170 +Ezt , where 170 is the initial velocity and c? is the (constant) acceleration. Along the x—direction, we have 1 x(t) = v0xt + Eaxt2 , vx (t) = VOX + axt Similarly, along the y-direction, we get 1 y(t) = voyt +—2—ayt2, vy (t) = voy + ayt (a) Given that v0x 23.0 m/s, v0y =0, ax =—l.0 m/sz, ay =—0.5 m/s2 , the components of the velocity are vx(t) = VOX + axt = (3.0 m/s) —(1 .0 m/s2)t vy(t) = voy + ayt =_"—(0.50 m/s2)t When the particle reaches its maximum‘x coordinate at t = tm, we must have vx = 0. Therefore, 3.0 — 1.023,, = 0 or t,,, = 3.0 s. The y component of the velocity at this time is vy(t=3.0$)=—(0.50m/s2)(3.0)=—l.5 m/s Thus, 17," =(—1.5 m/s)j. (b) At t = 3.0 s , the components of the position are x(t =3.0s) =v0xt+éaxt2 =(3.0m/s)(3.0s)+%(—l.0 m/s2)(3.03)2 =4.5 m y(t=3.0s)=v0yz‘+%ayt2 =0+%(—0.5m/sz)(3.05)2 =—2.25 m Using unit-vector notation, the results can be written as 17," = (4.50 m) — (2.25 m)3. 28. (a) Using the same coordinate system assumed in Eq. 4-22, we solve for y = h: . . l h=y0 +v0s1n60t— 5g!2 which yields h = 51.8 m foryo = 0, v0 = 42.0 m/s, «90 = 60.0°, and t= 5.50 s. (b) The horizontal motion is steady, so vx = v0x = v0 cos 4%, but the vertical component of velocity varies according to Eq. 4—23. Thus, the speed at impact is V: (v0 cosl90)2 + (vosin6?0 — gt)2 =27.4 m/s. (c) We use Eq. 4-24 with vy = 0 and y = H: . 2 H=M=67jm 2g 70. We use Eq. 4-44, noting that the upstream corresponds to the +i direction. (a) The subscript b is for the boat, w is for the water, and g is for the ground. vbg =vbw +vwg =(14 km/h) i+(—9 km/h) i: (5 km/h) i. Thus, the magnitude is liz’bg I: 5 km/h.\' (b) The direction of ng is +x, or upstream. (c) We use the subscript c for the child, and obtain veg :17“, +ng :(—6 km/h)i+(5km/h)i=(—1km/h)i. The magnitude is [flag [=1 km/h. ((1) The direction of $62; is —x, or downstream. ...
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