 University Physics Volume 1

# Chapter 4

4.1

(a) Taking the derivative with respect to time of the position function, we have $v→(t)=9.0t2i^andv→(3.0s)=81.0i^m/s.v→(t)=9.0t2i^andv→(3.0s)=81.0i^m/s.$ (b) Since the velocity function is nonlinear, we suspect the average velocity is not equal to the instantaneous velocity. We check this and find
$v→avg=r→(t2)−r→(t1)t2−t1=r→(4.0s)−r→(2.0s)4.0s−2.0s=(188i^−20i^)m2.0s=84i^m/s,v→avg=r→(t2)−r→(t1)t2−t1=r→(4.0s)−r→(2.0s)4.0s−2.0s=(188i^−20i^)m2.0s=84i^m/s,$
which is different from $v→(3.0s)=81.0i^m/s.v→(3.0s)=81.0i^m/s.$

4.2

The acceleration vector is constant and doesn’t change with time. If a, b, and c are not zero, then the velocity function must be linear in time. We have $v→(t)=∫a→dt=∫(ai^+bj^+ck^)dt=(ai^+bj^+ck^)tm/s,v→(t)=∫a→dt=∫(ai^+bj^+ck^)dt=(ai^+bj^+ck^)tm/s,$ since taking the derivative of the velocity function produces $a→(t).a→(t).$ If any of the components of the acceleration are zero, then that component of the velocity would be a constant.

4.3

(a) Choose the top of the cliff where the rock is thrown from the origin of the coordinate system. Although it is arbitrary, we typically choose time t = 0 to correspond to the origin. (b) The equation that describes the horizontal motion is $x=x0+vxt.x=x0+vxt.$ With $x0=0,x0=0,$ this equation becomes $x=vxt.x=vxt.$ (c) Equation 4.16 through Equation 4.18 and Equation 4.19 describe the vertical motion, but since $y0=0andv0y=0,y0=0andv0y=0,$ these equations simplify greatly to become $y=12(v0y+vy)t=12vyt,y=12(v0y+vy)t=12vyt,$$vy=−gt,vy=−gt,$$y=−12gt2,y=−12gt2,$ and $vy2=−2gy.vy2=−2gy.$ (d) We use the kinematic equations to find the x and y components of the velocity at the point of impact. Using $vy2=−2gyvy2=−2gy$ and noting the point of impact is −100.0 m, we find the y component of the velocity at impact is $vy=44.3m/s.vy=44.3m/s.$ We are given the x component, $vx=15.0m/s,vx=15.0m/s,$ so we can calculate the total velocity at impact: v = 46.8 m/s and $θ=71.3°θ=71.3°$ below the horizontal.

4.4

The golf shot at $30°.30°.$

4.5

134.0 cm/s

4.6

Labeling subscripts for the vector equation, we have B = boat, R = river, and E = Earth. The vector equation becomes $v→BE=v→BR+v→RE.v→BE=v→BR+v→RE.$ We have right triangle geometry shown in Figure 04_05_BoatRiv_img. Solving for $v→BEv→BE$, we have
$vBE=vBR2+vRE2=4.52+3.02vBE=vBR2+vRE2=4.52+3.02$
$vBE=5.4m/s, θ=tan−1(3.04.5)=33.7°.vBE=5.4m/s, θ=tan−1(3.04.5)=33.7°.$ ### Conceptual Questions

1 .

straight line

3 .

The slope must be zero because the velocity vector is tangent to the graph of the position function.

5 .

No, motions in perpendicular directions are independent.

7 .

a. no; b. minimum at apex of trajectory and maximum at launch and impact; c. no, velocity is a vector; d. yes, where it lands

9 .

They both hit the ground at the same time.

11 .

yes

13 .

If he is going to pass the ball to another player, he needs to keep his eyes on the reference frame in which the other players on the team are located.

15 . ### Problems

17 .

$r → = 1.0 i ^ − 4.0 j ^ + 6.0 k ^ r → = 1.0 i ^ − 4.0 j ^ + 6.0 k ^$

19 .

$Δ r → Total = 472.0 m i ^ + 80.3 m j ^ Δ r → Total = 472.0 m i ^ + 80.3 m j ^$

21 .

$Sum of displacements = −6.4 km i ^ + 9.4 km j ^ Sum of displacements = −6.4 km i ^ + 9.4 km j ^$

23 .

a. $v→(t)=8.0ti^+6.0t2k^,v→(0)=0,v→(1.0)=8.0i^+6.0k^m/sv→(t)=8.0ti^+6.0t2k^,v→(0)=0,v→(1.0)=8.0i^+6.0k^m/s$,
b. $v→avg=4.0​i^+2.0k^m/sv→avg=4.0​i^+2.0k^m/s$

25 .

$Δr→1=20.00mj^,Δr→2=(2.000×104m)(cos30°i^+sin30°j^)Δr→1=20.00mj^,Δr→2=(2.000×104m)(cos30°i^+sin30°j^)$
$Δr→=1.700×104mi^+1.002×104mj^Δr→=1.700×104mi^+1.002×104mj^$

27 .

a. $v→(t)=(4.0ti^+3.0tj^)m/s,v→(t)=(4.0ti^+3.0tj^)m/s,$$r→(t)=(2.0t2i^+32t2j^)mr→(t)=(2.0t2i^+32t2j^)m$,
b. $x(t)=2.0t2m,y(t)=32t2m,t2=x2⇒y=34xx(t)=2.0t2m,y(t)=32t2m,t2=x2⇒y=34x$ 29 .

a. $v→(t)=(6.0ti^−21.0t2j^+10.0t−3k^)m/sv→(t)=(6.0ti^−21.0t2j^+10.0t−3k^)m/s$,
b. $a→(t)=(6.0i^−42.0tj^−30t−4k^)m/s2a→(t)=(6.0i^−42.0tj^−30t−4k^)m/s2$,
c. $v→(2.0s)=(12.0i^−84.0j^+1.25k^)m/sv→(2.0s)=(12.0i^−84.0j^+1.25k^)m/s$,
d. $v→(1.0s)=6.0i^−21.0j^+10.0k^m/s,|v→(1.0s)|=24.0m/sv→(1.0s)=6.0i^−21.0j^+10.0k^m/s,|v→(1.0s)|=24.0m/s$
$v→(3.0s)=18.0i^−189.0j^+0.37k^m/s,v→(3.0s)=18.0i^−189.0j^+0.37k^m/s,$$|v→(3.0s)|=190m/s|v→(3.0s)|=190m/s$,
e. $r→(t)=(3.0t2i^−7.0t3j^−5.0t−2k^)mr→(t)=(3.0t2i^−7.0t3j^−5.0t−2k^)m$
$v→avg=9.0i^−49.0j^+3.75k^m/sv→avg=9.0i^−49.0j^+3.75k^m/s$

31 .

a. $v→(t)=−sin(1.0t)i^+cos(1.0t)j^+k^v→(t)=−sin(1.0t)i^+cos(1.0t)j^+k^$, b. $a→(t)=−cos(1.0t)i^−sin(1.0t)j^a→(t)=−cos(1.0t)i^−sin(1.0t)j^$

33 .

a. $t=0.55st=0.55s$, b. $x=110mx=110m$

35 .

a. $t=0.24s,d=0.28mt=0.24s,d=0.28m$, b. They aim high. 37 .

a., $t=12.8s,x=5619mt=12.8s,x=5619m$ b. $vy=125.0m/s,vx=439.0m/s,|v→|=456.0m/svy=125.0m/s,vx=439.0m/s,|v→|=456.0m/s$

39 .

a. $vy=v0y−gt,t=10s,vy=0,v0y=98.0m/s,v0=196.0m/svy=v0y−gt,t=10s,vy=0,v0y=98.0m/s,v0=196.0m/s$,
b. $h=490.0m,h=490.0m,$
c. $v0x=169.7m/s,x=3394.0m,v0x=169.7m/s,x=3394.0m,$
d. $x=169.7m/s(15.0s)=2550m y=(98.0m/s)(15.0s)–4.9(15.0s)2=368m sr=2550mi^+368mj^x=169.7m/s(15.0s)=2550m y=(98.0m/s)(15.0s)–4.9(15.0s)2=368m sr=2550mi^+368mj^$

41 .

$−100m=(−2.0m/s)t−(4.9m/s2)t2,−100m=(−2.0m/s)t−(4.9m/s2)t2,$ $t=4.3s,t=4.3s,$$x=86.0mx=86.0m$

43 .

$R M o o n = 48 m R M o o n = 48 m$

45 .

a. $v0y=24m/sv0y=24m/s$$vy2=v0y2−2gy⇒h=29.3mvy2=v0y2−2gy⇒h=29.3m$,
b. $t=2.4sv0x=18m/sx=43.2mt=2.4sv0x=18m/sx=43.2m$,
c. $y=−100my0=0y=−100my0=0$ $y−y0=v0yt−12gt2−100=24t−4.9t2y−y0=v0yt−12gt2−100=24t−4.9t2$ $⇒t=7.58s⇒t=7.58s$,
d. $x=136.44mx=136.44m$,
e. $t=2.0sy=28.4mx=36mt=2.0sy=28.4mx=36m$
$t=4.0sy=17.6mx=72mt=4.0sy=17.6mx=72m$
$t=6.0sy=−32.4mx=108mt=6.0sy=−32.4mx=108m$

47 .

$v0y=12.9m/sy−y0=v0yt−12gt2−20.0=12.9t−4.9t2v0y=12.9m/sy−y0=v0yt−12gt2−20.0=12.9t−4.9t2$
$t=3.7sv0x=15.3m/s⇒x=56.7mt=3.7sv0x=15.3m/s⇒x=56.7m$
So the golfer’s shot lands 13.3 m short of the green.

49 .

a. $R=60.8mR=60.8m$,
b. $R=137.8mR=137.8m$

51 .

a. $vy2=v0y2−2gy⇒y=2.9m/svy2=v0y2−2gy⇒y=2.9m/s$
$y=3.3m/sy=3.3m/s$
$y=v0y22g=(v0sinθ)22g⇒sinθ=0.91⇒θ=65.5°y=v0y22g=(v0sinθ)22g⇒sinθ=0.91⇒θ=65.5°$

53 .

$R = 18.5 m R = 18.5 m$

55 .

$y = ( tan θ 0 ) x − [ g 2 ( v 0 cos θ 0 ) 2 ] x 2 ⇒ v 0 = 16.4 m / s y = ( tan θ 0 ) x − [ g 2 ( v 0 cos θ 0 ) 2 ] x 2 ⇒ v 0 = 16.4 m / s$

57 .

$R = v 0 2 sin 2 θ 0 g ⇒ θ 0 = 15.9 ° R = v 0 2 sin 2 θ 0 g ⇒ θ 0 = 15.9 °$

59 .

(a) It takes the wide receiver 1.1 s to cover the last 10 m of his run.
$Ttof=2(v0sinθ)g⇒sinθ=0.27⇒θ=15.6°Ttof=2(v0sinθ)g⇒sinθ=0.27⇒θ=15.6°$
(b)
Therefore, the ball will be overthrown, and the receiver will not be able to catch it.

61 .

$a C = 40 m / s 2 a C = 40 m / s 2$

63 .

$aC=v2r⇒v2=raC=78.4,v=8.85m/saC=v2r⇒v2=raC=78.4,v=8.85m/s$
$T=5.68s,T=5.68s,$ which is $0.176rev/s=10.6rev/min0.176rev/s=10.6rev/min$

65 .

Venus is 108.2 million km from the Sun and has an orbital period of 0.6152 y.
$r=1.082×1011mT=1.94×107sr=1.082×1011mT=1.94×107s$
$v=3.5×104m/s,aC=1.135×10−2m/s2v=3.5×104m/s,aC=1.135×10−2m/s2$

67 .

$360rev/min=6rev/s360rev/min=6rev/s$
$v=3.8m/sv=3.8m/s$ $aC=144.m/s2aC=144.m/s2$

69 .

a. $O′(t)=(4.0i^+3.0j^+5.0k^)tmO′(t)=(4.0i^+3.0j^+5.0k^)tm$,
b. $r→PS=r→PS′+r→S′S, r→PS=r→PS′+r→S′S,$ $r→(t)=r→′(t)+(4.0i^+3.0j^+5.0k^)tmr→(t)=r→′(t)+(4.0i^+3.0j^+5.0k^)tm$,
c. $v→(t)=v→′(t)+(4.0i^+3.0j^+5.0k^)m/sv→(t)=v→′(t)+(4.0i^+3.0j^+5.0k^)m/s$, d. The accelerations are the same.

71 .

$v → P C = ( 2.0 i ^ + 5.0 j ^ + 4.0 k ^ ) m / s v → P C = ( 2.0 i ^ + 5.0 j ^ + 4.0 k ^ ) m / s$

73 .

a. A = air, S = seagull, G = ground
$v→SA=9.0m/sv→SA=9.0m/s$ velocity of seagull with respect to still air
$v→AG=?v→SG=5m/sv→AG=?v→SG=5m/s$ $v→SG=v→SA+v→AG⇒v→AG=v→SG−v→SAv→SG=v→SA+v→AG⇒v→AG=v→SG−v→SA$
$v→AG=−4.0m/sv→AG=−4.0m/s$
b. $v→SG=v→SA+v→AG⇒v→SG=−13.0m/sv→SG=v→SA+v→AG⇒v→SG=−13.0m/s$
$−6000m−13.0m/s=7 min 42 s−6000m−13.0m/s=7 min 42 s$

75 .

Take the positive direction to be the same direction that the river is flowing, which is east. S = shore/Earth, W = water, and B = boat.
a. $v→BS=11km/hv→BS=11km/h$
$t=8.2mint=8.2min$
b. $v→BS=−5km/hv→BS=−5km/h$
$t=18mint=18min$
c. $v→BS=v→BW+v→WSv→BS=v→BW+v→WS$ $θ=22°θ=22°$ west of north d. $|v→BS|=7.4km/h|v→BS|=7.4km/h$ $t=6.5mint=6.5min$
e. $v→BS=8.54km/h,v→BS=8.54km/h,$ but only the component of the velocity straight across the river is used to get the time $t=6.0mint=6.0min$
Downstream = 0.3 km

77 .

$v→AG=v→AC+v→CGv→AG=v→AC+v→CG$
$|v→AC|=25km/h|v→CG|=15km/h|v→AG|=29.15km/h|v→AC|=25km/h|v→CG|=15km/h|v→AG|=29.15km/h$ $v→AG=v→AC+v→CGv→AG=v→AC+v→CG$
The angle between $v→ACv→AC$ and $v→AGv→AG$ is $31°,31°,$ so the direction of the wind is $14°14°$ north of east. 79 .

$a C = 39.6 m / s 2 a C = 39.6 m / s 2$

81 .

$90.0km/h=25.0m/s,9.0km/h=2.5m/s,90.0km/h=25.0m/s,9.0km/h=2.5m/s,$ $60.0km/h=16.7m/s60.0km/h=16.7m/s$
$aT=−2.5m/s2,aC=1.86m/s2,a=3.1m/s2aT=−2.5m/s2,aC=1.86m/s2,a=3.1m/s2$

83 .

The radius of the circle of revolution at latitude $λλ$ is $REcosλ.REcosλ.$ The velocity of the body is $2πrT.aC=4π2REcosλT22πrT.aC=4π2REcosλT2$ for $λ=40°,aC=0.26%gλ=40°,aC=0.26%g$

85 .

$aT=3.00m/s2aT=3.00m/s2$
$v(5s)=15.00m/saC=150.00m/s2θ=88.8°v(5s)=15.00m/saC=150.00m/s2θ=88.8°$ with respect to the tangent to the circle of revolution directed inward. $|a→|=150.03m/s2|a→|=150.03m/s2$

87 .

$a→(t)=−Aω2cosωti^−Aω2sinωtj^a→(t)=−Aω2cosωti^−Aω2sinωtj^$
$aC=5.0mω2ω=0.89rad/saC=5.0mω2ω=0.89rad/s$
$v→(t)=−2.24m/si^−3.87m/sj^v→(t)=−2.24m/si^−3.87m/sj^$

89 .

$r → 1 = 1.5 j ^ + 4.0 k ^ r → 2 = Δ r → + r → 1 = 2.5 i ^ + 4.7 j ^ + 2.8 k ^ r → 1 = 1.5 j ^ + 4.0 k ^ r → 2 = Δ r → + r → 1 = 2.5 i ^ + 4.7 j ^ + 2.8 k ^$

91 .

$vx(t)=265.0m/svx(t)=265.0m/s$
$vy(t)=20.0m/svy(t)=20.0m/s$
$v→(5.0s)=(265.0i^+20.0j^)m/sv→(5.0s)=(265.0i^+20.0j^)m/s$

93 .

$R = 1.07 m R = 1.07 m$

95 .

$v 0 = 20.1 m / s v 0 = 20.1 m / s$

97 .

$v=3072.5m/sv=3072.5m/s$
$aC=0.223m/s2aC=0.223m/s2$

### Challenge Problems

99 .

a. $−400.0m=v0yt−4.9t2359.0m=v0xtt=359.0v0x−400.0=359.0v0yv0x−4.9(359.0v0x)2−400.0m=v0yt−4.9t2359.0m=v0xtt=359.0v0x−400.0=359.0v0yv0x−4.9(359.0v0x)2$
$−400.0=359.0tan40−631,516.9v0x2⇒v0x2=900.6v0x=30.0m/sv0y=v0xtan40=25.2m/s−400.0=359.0tan40−631,516.9v0x2⇒v0x2=900.6v0x=30.0m/sv0y=v0xtan40=25.2m/s$
$v=39.2m/sv=39.2m/s$, b. $t=12.0st=12.0s$

101 .

a. $r→TC=(−32+80t)i^+50tj^,|r→TC|2=(−32+80t)2+(50t)2r→TC=(−32+80t)i^+50tj^,|r→TC|2=(−32+80t)2+(50t)2$
$2rdrdt=2(−32+80t)(80)+5000tdrdt=160(−32+80t)+5000t2r=02rdrdt=2(−32+80t)(80)+5000tdrdt=160(−32+80t)+5000t2r=0$
$17800t=5184⇒t=0.29 hr17800t=5184⇒t=0.29 hr$,
b. $|r→TC|=17km|r→TC|=17km$

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