if not, from what horizontal distance must this kick be made if it is to score?

Ch 4.ii ������������ #8

A particle initially located at the origin has an acceleration of a = iii j thou/s² and an initial velocity of v = 500 i m/s.� Find (a) the vector position and velocity at any fourth dimension t and (b) the coordinates and speed of the particle at t = two.00 seconds.

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

(b)� �� r(two) = (6 j + 1000 i ) meters or (1000, half dozen) ��������� v(t) = (iii j �t + 500 i ) meters

Ch 4.3 ������������ #19

A place kicker must kick a football from a point 36 m from the goal.� One-half the oversupply hopes the brawl will clear the crossbar, which is three.05 meters high.� When kicked, the ball leaves the ground with a speed of 20 one thousand/southward and an bending of 53 degrees in a higher place the horizontal.

(a) By how much does the ball clear or fall short of the batten?

(b) Does the brawl approach the batten while still rise or while falling?

D x = 36 yard D y = 3.05 yard q = 53 � (a three-four-5 D ) five = 20 m/s	(a)����� five10 = D x / D t ��������� 12 m/due south =� 36 / t ��������� t = iii sec	(b) dy = � at2 + vy-init t + dy-init ��������� three.05 = -four.9t2 + 12 t + 0 ��������� 4.9t2 - sixteen t + three.05 = 0 (xvi � (144 � four(4.9)3.05)�) / 2(iv.9) ��������� t = (16 � 14) / ix.8 ��������� t = 3.06 sec, 0.20 sec 12 chiliad/s =� D ten / 3.06 seconds ��������� D x = 36.72 1000 It barely makes it over the bar; thus descending.
fivey = sin 53 � xx m vy = 16 m/s vx = cos 53 � xx thou vx = 12 m/s	dy = � attwo + 5y-init t + dy-init dy = � -nine.8(3)2 + 16 (3) dy = 3.9 meters @ 3 seconds 3.9 1000 clears the three.05 m bar past 0.85 meters

Ch 4.4 ������������ #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 in a higher place the Earth�south surface, where the gratis-autumn acceleration is viii.21 chiliad/ s².� The radius of the World is half dozen,400 km.� Determine the speed of the satellite and the time required to complete one orbit around the earth.

a������� = ������ 5two ������������������������ / ������ r ���������������� eight.21 � = (2 p *(6.4 + .6)*106)2 ����� / (6.four + .6)*10half dozentii

t = 5800 seconds t = one 60 minutes & 36 2/3 minutes

Ch iv.5 ������������ #35 The effigy below represents the full dispatch of a particle moving clockwise in a circle of radius 2.50 meters at a certain instant of time.� At this instant, find (a) the radial acceleration (b) the speed of the particle (c) its tangential acceleration			���������
	(a) ar = cos xxx � a ar = xiii m/s2�	�(b)� ar� = vii / r 13m/stwo� = v2 / ii.v m v ������ ��= 5.vii yard/south		(c)� a2 = ar 2 + at ii 15two ��� = 132 + at 2 at ����� = 7.five m/southtwo�

Ch iv.6 ������������ #38

Heather in her Corvette accelerates at the rate of (3.00 i � ii.00 j ) 1000/s², while Jill in her Jaguar accelerates at (one.00 i + 3.00 j ) m/sⁱⁱ.� The both start from balance at the origin of the xy coordinate system.� After 5.00 s, (a) what is Heather�s speed with respect to Jill, (b) how far apart are they, and (c) what is Heather�due south dispatch relative to Jill?

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

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

Endeavour it for ∫ t dt ∫ tone dt � t2 � (from ti to tf)

HEATHER	v(t) = (3∫dt i � 2∫dt j ) v(t) = (3t i � 2t j ) m/s v(5) = (15 i � 10 j ) thou/due south	(a) � |5(5)H| -|v(5)J| = ((xv-five) i � + (-10-xv) j ) |v(5)H| -|5(5)J| = ((10) i � + (-25) j ) yard/due south |v(5)H| -|v(five)J| = (102 + -25two)ane/2 |5(5)H| -|five(v)J| = 26.nine grand/s (b) � |x(5)H| -|x(5)J| = ((37.5-12.v) i � + (-25-37.5) j ) |10(five)H|-|10(5)J| = ((25) i � + (-62.5) j ) thou |x(5)H|-|x(v)J| = (25two + -62.52)i/ii |ten(v)H|-|x(5)J| = 67.3 grand	a = dv/dt a ∫dt = ∫dv� v = dx/dt 5 ∫dt = ∫dx speed = |five|
	x(t) = (3t∫dt i � 2t∫dt j ) x(t) = (1.5 t2 i � t2 j ) 1000/s ten(v) = (37.5 i � 25 j ) m/s
�JILL	5(t) = (1∫dt i + 3∫dt j ) v(t) = (t i + 3t j ) m/s five(5) = (5 i + xv j ) m/s
		(c) aH - aJ = (3.00 i � 2.00 j ) - (ane.00 i + 3.00 j ) aH - aJ = (2 i � 5 j ) thou/south2
	x(t) = (t∫dt i + 3t∫dt j ) ten(t) = (� t2 i + ane.5tii j ) grand/south 10(5) = (12.five i+ 37.5 j ) m/s

Ch iv ������� ��������� #63

A car is parked on a steep incline overlooking the body of water, where the incline makes an angle of 37.0� beneath the horizontal.� The negligent driver leaves the car in neutral, and the parking brakes are defective.� Starting from residuum at t = 0, the car rolls down the incline with a constant dispatch of 4 grand/sii, traveling l k to the edge of the vertical cliff.� The cliff is thirty m above the body of water.� Find (a) the speed of the automobile when it reaches the edge of the cliff and the fourth dimension at which it arrives in that location, (b) the velocity of the

motorcar when it lands in the ocean, (c) the total fourth dimension interval that the automobile is in move, and (d) the position of the car when information technology lands in the ocean, relative to the base of the cliff.

Given:��������� a = iv grand/due south2 r ������ = �at2 + vot + ro rincline = (�(-iv)t2 + 50) m 5 = dr / dt five = -4t thousand/s	(b) vy = v cosθ vy = twenty cos53 5y = 12.0 m/due south	5ten = v sinθ vx = 20 sin53 vx = 16.0 chiliad/s	a = Δv / Δt -10 = (vy-f � -12) / i.53 sec 5y-f = - 27.3 grand/s
		��������� 5 = (16 i � - 27.iii) j ) k/south
(a) Use indicate of view perspective where my nil point is the top and it�s speeding upward thus (+) acceleration � rincline �������� = � a t2 ��������� 50����� = � 4 tii� ��������� tincline�� = 5.00 sec v = 4 t������������������ � from above v = four (5) 5 = xx m/s	(c)� dy = � gt2�� + 5y-i t +� dy-i�� ������ 0 = �-10t2� + -12 t +� 30 t2� + 2.4 t =� half dozen (t + 1.2)2 = 6 + one.22�� tcliff = ane.53 seconds ttotal = five.00 + 1.53 = 6.53 s		(d) dx = 16 (i.53) d10 = 24.v meters or r = 24.v i �� meters

Ch 4.1 ������������� #2

A golf game brawl is hit off a tee at the edge of a cliff.� Its x & y coordinates every bit functions of fourth dimension are given by the post-obit expressions:

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

y = four t � iv.9 tⁱⁱ ����� or with units������������������ y = 4m/s * t � 4.9m/s² * t^two

(a) Write a vector expression for the ball position as a part of time, using the unit vectors i and j. r(t) = (18t) i + (4t � iv.9t2) j ����� or r(t) = (18t) i + (4t � � g t2) j	Next use unit-vector notation to write expressions for (d) the position����� r(t) = (18t) i + (4t � 4.9tii) j r(3) = (18*iii) i + (4*iii � four.nine*32) j = 54 i � 32.one j r(iii) = 54 i � 32.1 j
Past taking the derivatives, obtain expressions for (b) the velocity vector, v, as a function of fourth dimension. v = dr/dt = 18 i + (4 � ix.8 t) j	(e) the velocity v(3) �� = (18) i + (four � ix.viii *3) j���� v ������ = (18) i - (25.4) j
(c) the acceleration vector, a, equally a function of time. a = dv/dt = (0) i + (-ix.viii) j a = dv/dt =� -(ix.8 thousand/s2) j (westward/units)	(f) the dispatch of the golf game ball all at time t = 3.00 seconds. a =� dv / dt a =� -(9.8 m/due south2) j �� (w/units)

Ch 4.ii ������������ #5

At t = 0, a particle moving in the xy airplane with a constant acceleration has a velocity of v _i = 3i � 2j k/southward and is at the origin.� At t = three seconds, the particles velocity is 5 = (9i + 7j) yard/s.� Discover

(a) the dispatch of the particle a = D five / D t a = (9 � 3)i + (seven - -ii)j / iii a = 6i + 9j / 3 a = 2i +3j one thousand/s2

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

Ch iv.3 ������������ #22

A dive bomber has a velocity of 280 m/s at an angle q below the horizontal.� When the altitude of the aircraft is 2.xv km, information technology releases a bomb, which hits a target on the ground.� The magnitude of the deportation from the point for release of the bomb to the target is 3.25 km.� Notice the angle, q .
x ������ = � attwo + vx-init t + xo 2437� = 0� + 280 cos q t + 0 t = 8.seven / cos q	y ������ = � atii + vy-init t + yo 2150 = � 10t2 + 280 sin q t 430��� =� t2 + 56 sin q t		where t = 8.7 / cos q 430 = 75.8 / cos2 q + 487 tan q
��������� Plug in excel and increment q until left side equals right side. ��������� 0.583 radians or 33.4 �

Ch 4.4 ������������ #31

Immature David who slew Goliath experimented with slings before tackling the behemothic.� He found that he could revolve a sling of length 0.six m at the rated of eight rps.� If he increased the length to 0.9 m, he could revolve the sling only six rps.� (a) Which charge per unit of rotation gives the greater speed for the stone at the end of the sling?� (b)What is the centripetal dispatch of the stone at 8 rps? (c) What is the centripetal acceleration at 6 rps?

(a) ���� v = eight rev / s * two p r.6 / rev v = sixteen p *0.6 five = 9.six p m/south��������� five = vi rev / s * 2 p r.9 / rev v = 12 p *0.ix v = ten.eight p m/s�������� Thus at 6 rps results in a greater tangential velocity	a = D v / D t = 52 / r a = ( w *two p r / t)two / r �������� a = (2 westward )2 p 2 r / ttwo
	(b) ������������� a = (2 w )two p 2 r / t2 a = (2*8)2 p 2 (0.half dozen) / ane2 a = 1516 m/s2
	(c) ���� ��������� a = (ii due west )ii p ii r / t2 a = (two*half-dozen)2 p 2 (0.9) / 1two a = 1279 1000/s2

Ch 4.six ������������ #41

A river has a steady speed of � one thousand/s.� A student swims upstream a altitude of 1 km and swims back to the starting betoken.� If the student tin swim at a speed of 1.2 m/s in still water how long does the trip take?

Compare this with the fourth dimension the trip would take if the h2o were even so.

(a)����� vcyberspace upwardly = one.ii m/s � 0.v m/s ��������� vnet up = 0.7 m/s tinternet upwards = 1000 1000 / 0.seven grand/s tcyberspace upwards = 1428.6 sec	ttotal = tnet up + tnet down ttotal = 1428.6 s + 588.two south ttotal = 2022 due south
��������� vnet down = 1.2 m/s + 0.5 m/south ��������� vnet downwards = 1.7 m/s tnet down = 1000 1000 / ane.7 m/south tnet down = 588.ii sec	(b) total time w/o current = 2000 thousand / 1.2 yard/s total time w/o current = 1667 seconds (2017 � 1667) / 1667 = 21.0% west/ current takes 21 % longer than west/o current

Ch 4 ���������������� #56 A boy can throw a ball a maximum horizontal distance of 40 one thousand on a level playing field.� How far can he throw the same ball vertically upward?� Presume that his muscles give the ball the same speed in each case. ��������� v10 = cos q v�� fivey = sin q v

a = D five / D t ����������� -10 = (0 - sin q five) / ttop����� tpinnacle = v sin q / 10 ������������������ tfull = ttop + tlesser ������������������ Level Ground: televation = tbottom ������������������ ttotal = v sin q / 5	Thrown the ball straight UP �the only forcefulness on it is gravity a ������ = D five ����������� / D t -ten ��� = (0 - (10r)�)��������� / D t������������ t ������ = (r/10)�� seconds
v10 ����� = D x / tfull vcos q = R / (five sin q / 5) v2 ����� = 5R / sin q cos q five ������ = (10r)� m/southward	dy ����� = � at2 ����������������� + ������ fiveot dy ����� = � -10(r/10)�*2 ���� + (10r)�* (r/10)� dy ����� = � r

Let�south solve once more but this fourth dimension the total distance traveled is r instead of xl meters.

v_x = cos q v

5_y = sin q v

a_y = D 5 / D t ���������� -x = (0 - sin q v) / t������� solve for time ������ t = sin q * v / 10

only this is just half of the time.� Thus the t_total =� (2 * sin q five / ten) = sin q * 5 / 5

We also know the v₁₀ ������� = d_x �� /���� t_total �������������� where 5_x = cos q * v

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

v² = 5 r / (sin q cos q )����� Where q must be 45 � since we know it was thrown at the maximum altitude.

v² = 5 r / (0.707 * 0.707)��������� = 10 r���������� v = (10r)^���

So now we know the maximum velocity the brawl tin be thrown, which is at present pointed upwardly.

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

d_y = � at² + five_yot = � -10(r/10)^�*2 + (10r)^�* (r/ten)^�� = � r

Ch 4 ���������������� #57

A stone at the stop of a sling is whirled in a vertical circle of radius 1.2 1000 at a constant speed v_i = 1.5 thousand/s as in Figure P4.57.� The center of the string is 1.v chiliad to a higher place 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 earlier it is released at A? (d) just after it is released at A?

5y = sin lx � *one.five ���� five10 = cos 60 � *1.5 vy = 1.3 g/s ���������� fivex = 0.75 m/s
(a)����� ay ����� = D vy / D t -9.8 � = (0-i.3) / t tpinnacle ��� = 0.133 sec dy-init = 0.3+1.2+sin30 � *1.two dy-init = 2.one m dtop = � at2 + vot + dy-init dtop = 0.75 t � 5 t2 + 2.one dtop = 2.185 thousand dbottom = � at2 + 5y-init + dy-ini 2.185 = � 9.viii t2 tbottom = 0.668 sec ��������� ttotal = ttop + tbottom ttotal = 0.133 + 0.668 ttotal = 0.8 sec vten = D ten / D t 0.75 1000/due south = D x / 0.8 sec D x = 0.lx meters
	(b)����� dabove footing �� =� 5o in y compt� + � at2 ��������� 2.1 m ���������� =����� one.2��� t + 5t2��� 5t2 + 1.2t � 2.1 = 0�� ���������������������������� [-b � (b2 � 4ac)1/2] / 2a ������������������������������������� (-one.two � half-dozen.half dozen) / 10 ������������������������������������� t = 0.54 seconds v10 = D x / D t 0.75 thousand/s = D x / 0.54 sec D x = 0.40 meters
(c) a = D 5 / D t a = fivetwo / r a = 2.25 / ane.two a= 1.875 m/south2	(d) immediately after information technology�s release ANYWHERE, the only force acting on it (neglecting air friction) is gravity which has a charge per unit of dispatch of ix.8 m/southward2 pointed down

Ch 4 ���������������� #65

The determined coyote is out once more to effort to capture the elusive roadrunner.� The coyote wears a pair of Acme jet-powered roller skates, which provide a abiding horizontal dispatch of fifteen m/s².� The coyote starts off at residuum 70 thou from the border of a cliff at the instant the roadrunner zips by him in the direction of the cliff. (a) if the roadrunner moves with a abiding speed, determine the minimum speed he must have to reach the cliff before the coyote.� At the brink of the cliff, the roadrunner escapes by making a sudden plow, while the coyote continues straight ahead.� (b) If the cliff is 100 m above the floor of a canyon, determine where the coyote lands in the coulee (presume his skates remain horizontal and continue to operate when he is in flying). (c) Determine the components of the coyote�s touch on velocity.

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

Coyote distance = � at² ����������� 70 m = � fifteen t^two ������ t = 3.055 seconds

Thus the road runner must also cover the aforementioned altitude at a abiding velocity in iii.055 seconds.

v_ave = d / t������������ v_{road runner} = lxx m / 3.055 s = 22.nine m/south

(b) The coulee is 100 m deep, thus the coyote is in complimentary fall for 100 m

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

(c) d = d_o + 5_ot + � at²����� d_x = � at^two = � 15 (vii.52)² = 424 meters full in the x-management

70 m was prior to the coyote leaving the cliff, and then the coyote landed 354 meters into the canyon.

(d) Total time the coyote was accelerating in the x direction is 3.055 + iv.47 = 7.52 seconds

a = D v / D t = 15 = D v / 7.52 seconds ������������ v_{f � � 10 dir} = 113 m/due south

10 = D v / 4.47 ������ ������������������������������������� five_{f � � y dir} = 44.7 1000/s

yorksarasked.blogspot.com

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

Share this post