from what horizontal distance must this kick be made if it is to score?

Ch iv.2 ������������ #8

A particle initially located at the origin has an dispatch of a = 3 j m/s² and an initial velocity of 5 = 500 i m/south.� Detect (a) the vector position and velocity at any fourth dimension t and (b) the coordinates and speed of the particle at t = 2.00 seconds.

(a)����� r(t) = (� 3 j �t2 + 500 i �t) meters ��������� v(t) = (3 j �t + 500 i ) m/south

(b)� �� r(ii) = (6 j + 1000 i ) meters or (one thousand, 6) ��������� v(t) = (3 j �t + 500 i ) meters

Ch four.3 ������������ #19

A place kicker must kick a football from a point 36 grand from the goal.� Half the crowd hopes the ball volition articulate the crossbar, which is 3.05 meters high.� When kicked, the brawl leaves the ground with a speed of 20 m/s and an angle of 53 degrees higher up the horizontal.

(a) By how much does the ball articulate or fall short of the crossbar?

(b) Does the ball approach the batten while all the same rising or while falling?

D ten = 36 k D y = 3.05 m q = 53 � (a 3-4-5 D ) v = xx k/s	(a)����� 510 = D x / D t ��������� 12 m/south =� 36 / t ��������� t = 3 sec	(b) dy = � at2 + fivey-init t + dy-init ��������� iii.05 = -4.9tii + 12 t + 0 ��������� 4.9ttwo - sixteen t + three.05 = 0 (16 � (144 � iv(4.nine)3.05)�) / 2(iv.nine) ��������� t = (16 � fourteen) / nine.8 ��������� t = 3.06 sec, 0.20 sec 12 k/s =� D 10 / 3.06 seconds ��������� D x = 36.72 m Information technology barely makes information technology over the bar; thus descending.
vy = sin 53 � 20 thou fivey = sixteen m/s vx = cos 53 � 20 m vten = 12 chiliad/s	dy = � at2 + 5y-init t + dy-init dy = � -9.8(3)2 + 16 (3) dy = 3.9 meters @ 3 seconds iii.9 m clears the iii.05 m bar by 0.85 meters

Ch four.4 ������������ #32

The astronaut orbiting the Earth in Figure P4.32 is preparing to dock with a Westar VI satellite.� The satellite is in a circular orbit 600 km above the Earth�s surface, where the gratuitous-autumn acceleration is 8.21 yard/ s^two.� The radius of the World is 6,400 km.� Determine the speed of the satellite and the time required to complete one orbit effectually the globe.

a������� = ������ five2 ������������������������ / ������ r ���������������� 8.21 � = (2 p *(6.four + .6)*ten6)2 ����� / (6.4 + .6)*106ttwo

t = 5800 seconds t = i hr & 36 ii/3 minutes

Ch four.5 ������������ #35 The effigy beneath represents the full acceleration of a particle moving clockwise in a circle of radius 2.l meters at a sure instant of time.� At this instant, detect (a) the radial acceleration (b) the speed of the particle (c) its tangential acceleration			���������
	(a) ar = cos 30 � a ar = 13 k/s2�	�(b)� ar� = v2 / r 13m/due southii� = 5two / two.5 g v ������ ��= 5.7 m/s		(c)� a2 = ar 2 + at ii fifteen2 ��� = 13ii + at ii at ����� = seven.five m/southwardii�

Ch 4.6 ������������ #38

Heather in her Corvette accelerates at the charge per unit of (3.00 i � ii.00 j ) m/s², while Jill in her Jaguar accelerates at (ane.00 i + 3.00 j ) m/sⁱⁱ.� The both start from residuum at the origin of the xy coordinate system.� After v.00 south, (a) what is Heather�south speed with respect to Jill, (b) how far apart are they, and (c) what is Heather�s acceleration relative to Jill?

Integration tutorial� � Other hints 5 = ∫ a dt ��� � reverse to (a = dv/dt) 10 = ∫ 5 dt ��� � opposite to (v = dx/dt)

∫dt ∫ t0 dt � raise the index by one � take the new index and place it in the denominator t1 / 1� (from ti to tf)

Attempt information technology for ∫ t dt ∫ t1 dt � tii � (from ti to tf)

HEATHER	v(t) = (three∫dt i � 2∫dt j ) v(t) = (3t i � 2t j ) m/s v(5) = (15 i � 10 j ) g/s	(a) � |v(v)H| -|v(5)J| = ((15-five) i � + (-10-15) j ) |v(five)H| -|five(5)J| = ((10) i � + (-25) j ) chiliad/due south |v(five)H| -|five(v)J| = (102 + -252)1/2 |v(five)H| -|5(five)J| = 26.9 g/due south (b) � |10(5)H| -|ten(5)J| = ((37.v-12.five) i � + (-25-37.5) j ) |x(5)H|-|x(v)J| = ((25) i � + (-62.v) j ) k |ten(5)H|-|x(v)J| = (252 + -62.52)1/2 |ten(v)H|-|x(5)J| = 67.3 m	a = dv/dt a ∫dt = ∫dv� v = dx/dt v ∫dt = ∫dx speed = |five|
	x(t) = (3t∫dt i � 2t∫dt j ) x(t) = (one.5 tii i � t2 j ) m/s ten(five) = (37.5 i � 25 j ) one thousand/southward
�JILL	v(t) = (1∫dt i + iii∫dt j ) v(t) = (t i + 3t j ) one thousand/southward 5(v) = (5 i + 15 j ) g/southward
		(c) aH - aJ = (3.00 i � 2.00 j ) - (1.00 i + three.00 j ) aH - aJ = (2 i � 5 j ) thousand/s2
	ten(t) = (t∫dt i + 3t∫dt j ) x(t) = (� ttwo i + i.5t2 j ) chiliad/s x(5) = (12.five i+ 37.5 j ) m/south

Ch four ������� ��������� #63

A car is parked on a steep incline overlooking the ocean, where the incline makes an angle of 37.0� below the horizontal.� The negligent driver leaves the motorcar in neutral, and the parking brakes are defective.� Starting from residuum at t = 0, the car rolls down the incline with a constant acceleration of four m/s2, traveling 50 g to the border of the vertical cliff.� The cliff is 30 k above the ocean.� Discover (a) the speed of the machine when information technology reaches the edge of the cliff and the fourth dimension at which information technology arrives there, (b) the velocity of the

automobile when it lands in the bounding main, (c) the total time interval that the machine is in motion, and (d) the position of the car when it lands in the ocean, relative to the base of the cliff.

Given:��������� a = 4 m/s2 r ������ = �at2 + vot + ro rincline = (�(-four)ttwo + 50) m v = dr / dt 5 = -4t m/south	(b) vy = v cosθ fivey = 20 cos53 fivey = 12.0 1000/s	5ten = v sinθ vten = twenty sin53 vx = 16.0 m/due south	a = Δv / Δt -10 = (vy-f � -12) / 1.53 sec vy-f = - 27.iii m/s
		��������� 5 = (16 i � - 27.3) j ) m/s
(a) Use point of view perspective where my zero bespeak is the summit and it�southward speeding up thus (+) acceleration � rincline �������� = � a t2 ��������� 50����� = � 4 t2� ��������� tincline�� = 5.00 sec 5 = 4 t������������������ � from above five = 4 (5) v = 20 m/south	(c)� dy = � gt2�� + vy-i t +� dy-i�� ������ 0 = �-10t2� + -12 t +� 30 t2� + 2.4 t =� half-dozen (t + i.2)2 = 6 + 1.2two�� tcliff = one.53 seconds ttotal = five.00 + 1.53 = half-dozen.53 s		(d) dten = sixteen (1.53) dten = 24.5 meters or r = 24.five i �� meters

Ch 4.1 ������������� #two

A golf game ball is hit off a tee at the edge of a cliff.� Its ten & y coordinates as functions of time are given by the following expressions:

x = xviii t���������������� or with units �������� x = 18m/s * t

y = 4 t � 4.9 t^two ����� or with units������������������ y = 4m/s * t � 4.9m/s² * t^two

(a) Write a vector expression for the brawl position equally a function of time, using the unit vectors i and j. r(t) = (18t) i + (4t � 4.9t2) j ����� or r(t) = (18t) i + (4t � � 1000 t2) j	Adjacent use unit-vector notation to write expressions for (d) the position����� r(t) = (18t) i + (4t � four.9ttwo) j r(3) = (18*3) i + (four*3 � four.9*threetwo) j = 54 i � 32.1 j r(3) = 54 i � 32.1 j
By taking the derivatives, obtain expressions for (b) the velocity vector, five, as a part of fourth dimension. v = dr/dt = 18 i + (four � 9.8 t) j	(due east) the velocity v(iii) �� = (18) i + (4 � 9.8 *3) j���� v ������ = (18) i - (25.iv) j
(c) the acceleration vector, a, as a role of time. a = dv/dt = (0) i + (-9.8) j a = dv/dt =� -(nine.8 m/due south2) j (w/units)	(f) the acceleration of the golf ball all at time t = 3.00 seconds. a =� dv / dt a =� -(9.eight m/south2) j �� (due west/units)

Ch four.2 ������������ #5

At t = 0, a particle moving in the xy plane with a constant acceleration has a velocity of 5 _i = 3i � 2j chiliad/s and is at the origin.� At t = iii seconds, the particles velocity is v = (9i + 7j) m/s.� Detect

(a) the acceleration of the particle a = D 5 / D t a = (9 � three)i + (7 - -2)j / 3 a = 6i + 9j / 3 a = 2i +3j m/southwardtwo

(b) its coordinates at whatever time t. r = ro + vot + � at2� r(t) = 0 + (3i � 2j)*t + � (2i +3j)*ttwo���� x(t) = 3t + ttwo������������������ y(t) = -2t + 3t2/two

Ch 4.three ������������ #22

A dive bomber has a velocity of 280 m/s at an bending q below the horizontal.� When the altitude of the aircraft is 2.15 km, it releases a bomb, which hits a target on the ground.� The magnitude of the displacement from the betoken for release of the bomb to the target is 3.25 km.� Notice the angle, q .
ten ������ = � at2 + vx-init t + xo 2437� = 0� + 280 cos q t + 0 t = viii.seven / cos q	y ������ = � at2 + vy-init t + yo 2150 = � 10t2 + 280 sin q t 430��� =� t2 + 56 sin q t		where t = 8.7 / cos q 430 = 75.eight / cos2 q + 487 tan q
��������� Plug in excel and increment q until left side equals right side. ��������� 0.583 radians or 33.4 �

Ch 4.iv ������������ #31

Young David who slew Goliath experimented with slings earlier tackling the giant.� He constitute that he could revolve a sling of length 0.half dozen thousand at the rated of viii rps.� If he increased the length to 0.9 m, he could circumduct the sling only six rps.� (a) Which rate of rotation gives the greater speed for the stone at the terminate of the sling?� (b)What is the centripetal acceleration of the stone at 8 rps? (c) What is the centripetal dispatch at 6 rps?

(a) ���� v = eight rev / south * 2 p r.6 / rev v = sixteen p *0.6 v = 9.6 p k/s��������� v = 6 rev / s * 2 p r.ix / rev v = 12 p *0.9 five = 10.eight p thou/s�������� Thus at 6 rps results in a greater tangential velocity	a = D v / D t = vtwo / r a = ( due west *2 p r / t)2 / r �������� a = (two w )2 p 2 r / t2
	(b) ������������� a = (2 westward )2 p ii r / t2 a = (2*8)2 p 2 (0.6) / 1ii a = 1516 m/s2
	(c) ���� ��������� a = (2 w )two p 2 r / t2 a = (2*half-dozen)two p two (0.9) / 1ii a = 1279 m/stwo

Ch four.vi ������������ #41

A river has a steady speed of � thousand/s.� A student swims upstream a distance of 1 km and swims back to the starting indicate.� If the pupil tin can swim at a speed of 1.ii m/s in still h2o how long does the trip take?

Compare this with the time the trip would take if the h2o were still.

(a)����� vinternet up = one.2 yard/s � 0.five m/s ��������� vnet up = 0.7 m/s tnet up = 1000 m / 0.seven chiliad/south tnet upwards = 1428.vi sec	ttotal = tinternet up + tinternet down ttotal = 1428.6 s + 588.2 s tfull = 2017 s
��������� fivenet down = i.2 thousand/south + 0.5 m/s ��������� vcyberspace down = 1.vii m/s tinternet down = yard m / 1.vii yard/south tnet down = 588.two sec	(b) total time w/o current = 2000 m / 1.2 m/southward full time w/o current = 1667 seconds (2017 � 1667) / 1667 = 21.0% w/ current takes 21 % longer than w/o current

Ch 4 ���������������� #56 A boy can throw a ball a maximum horizontal altitude of 40 m on a level playing field.� How far can he throw the same ball vertically upwards?� Assume that his muscles give the ball the same speed in each case. ��������� vx = cos q 5�� vy = sin q v

a = D v / D t ����������� -10 = (0 - sin q five) / ttop����� ttop = five sin q / 10 ������������������ ttotal = tmeridian + tlesser ������������������ Level Footing: ttop = tbottom ������������������ ttotal = v sin q / 5	Thrown the ball straight UP �the just forcefulness on it is gravity a ������ = D v ����������� / D t -10 ��� = (0 - (10r)�)��������� / D t������������ t ������ = (r/10)�� seconds
vx ����� = D x / ttotal vcos q = R / (v sin q / v) 52 ����� = 5R / sin q cos q five ������ = (10r)� m/southward	dy ����� = � atii ����������������� + ������ vot dy ����� = � -ten(r/ten)�*2 ���� + (10r)�* (r/x)� dy ����� = � r

Allow�s solve again simply this time the total distance traveled is r instead of forty meters.

v_x = cos q v

v_y = sin q v

a_y = D v / D t ���������� -10 = (0 - sin q v) / t������� solve for fourth dimension ������ t = sin q * v / 10

but this is simply half of the time.� Thus the t_total =� (2 * sin q v / 10) = sin q * 5 / 5

We also know the v_x ������� = d_x �� /���� t_total �������������� where v_ten = cos q * 5

������������������ cos q v ������� = r ��� / sin q � v / 5

v^two = 5 r / (sin q cos q )����� Where q must be 45 � since we know information technology was thrown at the maximum altitude.

fiveⁱⁱ = v r / (0.707 * 0.707)��������� = 10 r���������� v = (10r)^���

So now we know the maximum velocity the ball can be thrown, which is now pointed up.

-10 = (0 � (10r)^�) / t����������������� t = (r/10)^�� seconds

d_y = � at² + v_yot = � -10(r/10)^�*2 + (10r)^�* (r/x)^�� = � r

Ch 4 ���������������� #57

A rock at the end of a sling is whirled in a vertical circle of radius 1.2 m at a constant speed v_i = 1.5 m/south as in Effigy P4.57.� The eye of the string is one.v m above the ground.� What is the range of the stone if information technology is released when the sling is inclined at xxx.0 � with the horizontal (a) at A? (b) at B?� What is the acceleration of the stone (c) just before information technology is released at A? (d) just subsequently it is released at A?

vy = sin sixty � *1.5 ���� vx = cos sixty � *i.five fivey = i.3 chiliad/due south ���������� vx = 0.75 m/s
(a)����� ay ����� = D vy / D t -9.8 � = (0-i.3) / t tpeak ��� = 0.133 sec dy-init = 0.3+one.two+sin30 � *1.2 dy-init = 2.ane thousand delevation = � at2 + 5ot + dy-init dtop = 0.75 t � five t2 + 2.one dmeridian = ii.185 m dbottom = � at2 + vy-init + dy-ini ii.185 = � 9.8 t2 tbottom = 0.668 sec ��������� ttotal = ttop + tbottom ttotal = 0.133 + 0.668 ttotal = 0.eight sec vx = D x / D t 0.75 k/s = D x / 0.eight sec D 10 = 0.60 meters
	(b)����� dabove ground �� =� fiveo in y compt� + � at2 ��������� 2.i grand ���������� =����� ane.two��� t + 5t2��� 5tii + 1.2t � two.1 = 0�� ���������������������������� [-b � (b2 � 4ac)1/2] / 2a ������������������������������������� (-1.2 � 6.6) / 10 ������������������������������������� t = 0.54 seconds 5ten = D x / D t 0.75 yard/s = D x / 0.54 sec D ten = 0.xl meters
(c) a = D v / D t a = vii / r a = two.25 / ane.2 a= 1.875 m/s2	(d) immediately subsequently information technology�s release ANYWHERE, the only forcefulness acting on it (neglecting air friction) is gravity which has a rate of acceleration of 9.8 m/due south2 pointed down

Ch iv ���������������� #65

The determined coyote is out once more to try to capture the elusive roadrunner.� The coyote wears a pair of Meridian jet-powered roller skates, which provide a abiding horizontal dispatch of 15 chiliad/s².� The coyote starts off at remainder 70 g from the edge of a cliff at the instant the roadrunner zips past him in the direction of the cliff. (a) if the roadrunner moves with a constant speed, determine the minimum speed he must accept to reach the cliff before the coyote.� At the brink of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues direct ahead.� (b) If the cliff is 100 1000 above the floor of a canyon, decide where the coyote lands in the canyon (presume his skates remain horizontal and continue to operate when he is in flight). (c) Make up one's mind the components of the coyote�southward impact velocity.

(a)�������������� d = d_o + five_ot + � at²�

Coyote distance = � at² ����������� seventy m = � 15 t² ������ t = three.055 seconds

Thus the route runner must also cover the same distance at a constant velocity in 3.055 seconds.

five_ave = d / t������������ v_{road runner} = seventy grand / three.055 s = 22.ix m/southward

(b) The canyon is 100 thou deep, thus the coyote is in costless fall for 100 m

100 m = d_{o y} + v_{o y}t + � a_yt² = � 10t² = 4.47 seconds

(c) d = d_o + five_ot + � atⁱⁱ����� d_x = � atⁱⁱ = � xv (seven.52)² = 424 meters total in the 10-direction

seventy k was prior to the coyote leaving the cliff, so the coyote landed 354 meters into the coulee.

(d) Full time the coyote was accelerating in the x management is 3.055 + four.47 = 7.52 seconds

a = D v / D t = 15 = D v / vii.52 seconds ������������ v_{f � � x dir} = 113 m/s

10 = D five / 4.47 ������ ������������������������������������� 5_{f � � y dir} = 44.7 m/southward

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Source: https://www.cpp.edu/~skboddeker/131/131hw/ch4h.htm

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