Steve Boddeker's Ch4 Homework



Ch 4.2 #8

A particle initially located at the origin has an acceleration of a = 3 j m/s2 and an initial velocity of v = 500 i m/s. Find (a) the vector position and velocity at any time t and (b) the coordinates and speed of the particle at t = 2.00 seconds.

|(a) r(t) = (½ 3j t2 + 500i t) meters |(b) r(2) = (6 j + 1000 i ) meters or (1000, 6) |

|v(t) = (3j t + 500i ) m/s |v(t) = (3j t + 500i ) meters |

| | |

Ch 4.3 #19

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

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

(b) Does the ball approach the crossbar while still rising or while falling?

|(x = 36 m |(a) vx = (x / (t |(b) dy = ½ at2 + vy-init t + dy-init |

|(y = 3.05 m |12 m/s = 36 / t |3.05 = -4.9t2 + 12 t + 0 |

|( = 53((a 3-4-5 () |t = 3 sec |4.9t2 - 16 t + 3.05 = 0 |

|v = 20 m/s | |(16 ( (144 – 4(4.9)3.05)½) / 2(4.9) |

| | |t = (16 ( 14) / 9.8 |

| | |t = 3.06 sec, 0.20 sec |

| | |12 m/s = (x / 3.06 seconds |

| | |(x = 36.72 m |

| | |It barely makes it over the bar; thus descending. |

|vy = sin 53( 20 m |dy = ½ at2 + vy-init t + dy-init | |

|vy = 16 m/s |dy = ½ -9.8(3)2 + 16 (3) | |

| |dy = 3.9 meters @ 3 seconds | |

|vx = cos 53( 20 m | | |

|vx = 12 m/s |3.9 m clears the 3.05 m bar by | |

| |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 above the Earth’s surface, where the free-fall acceleration is 8.21 m/ s2. The radius of the Earth is 6,400 km. Determine the speed of the satellite and the time required to complete one orbit around the earth.

|a = v2 / r |t = 5800 seconds |

|8.21 = (2(*(6.4 + .6)*106)2 / (6.4 + .6)*106t2 |t = 1 hour & 36 2/3 minutes |

|Ch 4.5 #35 |[pic] | [pic] |

|The figure below represents the total acceleration 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) | (b) ar = v2 / r |(c) a2 = ar2 + at2 |

| |ar = cos 30( a |13m/s2 = v2 / 2.5 m |152 = 132 + at2 |

| |ar = 13 m/s2 |v = 5.7 m/s |at = 7.5 m/s2 |

Ch 4.6 #38

Heather in her Corvette accelerates at the rate of (3.00 i – 2.00 j) m/s2, while Jill in her Jaguar accelerates at (1.00 i + 3.00 j) m/s2. The both start from rest 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’s acceleration relative to Jill?

|Integration tutorial ( |∫dt |Try it for |

| |∫ t0 dt |∫ t dt |

|Other hints |raise the index by one |∫ t1 dt |

|v = ∫ a dt |take the new index and place it in the denominator |½ t2 (from ti to tf) |

|( opposite to (a = dv/dt) |t1 / 1 (from ti to tf) | |

|x = ∫ v dt | | |

|( opposite to (v = dx/dt) | | |

|HEA|v(t) = (3∫dt i – 2∫dt j) |(a) |v(5)H| -|v(5)J| = ((15-5) i + (-10-15) j) |a = dv/dt |

|THE|v(t) = (3t i – 2t j) m/s ||v(5)H| -|v(5)J| = ((10) i + (-25) j) m/s |a ∫dt = ∫dv |

|R |v(5) = (15 i – 10 j) m/s ||v(5)H| -|v(5)J| = (102 + -252)1/2 | |

| | ||v(5)H| -|v(5)J| = 26.9 m/s |v = dx/dt |

| | |(b) |x(5)H| -|x(5)J| = ((37.5-12.5)i + (-25-37.5)j) |v ∫dt = ∫dx |

| | ||x(5)H|-|x(5)J| = ((25)i + (-62.5)j) m | |

| | ||x(5)H|-|x(5)J| = (252 + -62.52)1/2 |speed = |v| |

| | ||x(5)H|-|x(5)J| = 67.3 m | |

| |x(t) = (3t∫dt i – 2t∫dt j) | | |

| |x(t) = (1.5 t2 i – t2 j) m/s | | |

| |x(5) = (37.5 i – 25 j) m/s | | |

| |v(t) = (1∫dt i + 3∫dt j) | | |

|JIL|v(t) = (t i + 3t j) m/s | | |

|L |v(5) = (5 i + 15 j) m/s | | |

| | |(c) |

| | |aH - aJ = (3.00 i – 2.00 j) - (1.00 i + 3.00 j) |

| | |aH - aJ = (2 i – 5 j) m/s2 |

| |x(t) = (t∫dt i + 3t∫dt j) | |

| |x(t) = (½ t2 i + 1.5t2 j) m/s | |

| |x(5) = (12.5 i+ 37.5 j) m/s | |

Ch 4 #63

|A car is parked on a steep incline overlooking the ocean, where the incline makes an angle of 37.0° below the |[pic] |

|horizontal. The negligent driver leaves the car in neutral, and the parking brakes are defective. Starting from | |

|rest at t = 0, the car rolls down the incline with a constant acceleration of 4 m/s2, traveling 50 m to the edge of | |

|the vertical cliff. The cliff is 30 m above the ocean. Find (a) the speed of the car when it reaches the edge of | |

|the cliff and the time at which it arrives there, (b) the velocity of the | |

|car when it lands in the ocean, (c) the total time interval that the car 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 |(b) |vx = v sinθ |a = Δv / Δt |

|r = ½at2 + vot + ro |vy = v cosθ |vx = 20 sin53 |-10 = (vy-f – -12) / 1.53 sec |

|rincline = (½(-4)t2 + 50) m |vy = 20 cos53 |vx = 16.0 m/s |vy-f = - 27.3 m/s |

|v = dr / dt |vy = 12.0 m/s | | |

|v = -4t m/s | | | |

| | | |

| | |v = (16 i - 27.3) j ) m/s |

|(a) Use point of view perspective where my zero |(c) dy = ½ gt2 + vy-i t + dy-i |(d) |

|point is the top and it’s speeding up thus (+) |0 = ½-10t2 + -12 t + 30 |dx = 16 (1.53) |

|acceleration |t2 + 2.4 t = 6 |dx = 24.5 meters |

|rincline = ½ a t2 |(t + 1.2)2 = 6 + 1.22 |or |

|50 = ½ 4 t2 |tcliff = 1.53 seconds |r = 24.5 i meters |

|tincline = 5.00 sec |ttotal = 5.00 + 1.53 = 6.53 s | |

|v = 4 t ( from above | | |

|v = 4 (5) | | |

|v = 20 m/s | | |

Ch 4.1 #2

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

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

y = 4 t – 4.9 t2 or with units y = 4m/s * t – 4.9m/s2 * t2

|(a) Write a vector expression for the ball position as a function of time, |Next use unit-vector notation to write expressions for |

|using the unit vectors i and j. |(d) the position r(t) = (18t) i + (4t – 4.9t2) j |

|r(t) = (18t) i + (4t – 4.9t2) j | |

|or |r(3) = (18*3) i + (4*3 – 4.9*32) j = 54 i – 32.1 j |

|r(t) = (18t) i + (4t – ½ g t2) j |r(3) = 54 i – 32.1 j |

|By taking the derivatives, obtain expressions for (b) the velocity vector, |(e) the velocity |

|v, as a function of time. |v(3) = (18) i + (4 – 9.8 *3) j |

|v = dr/dt = 18 i + (4 – 9.8 t) j |v = (18) i - (25.4) j |

|(c) the acceleration vector, a, as a function of time. |(f) the acceleration of the golf ball |

|a = dv/dt = (0) i + (-9.8) j |all at time t = 3.00 seconds. |

|a = dv/dt = -(9.8 m/s2) j (w/units) |a = dv / dt |

| |a = -(9.8 m/s2) j (w/units) |

Ch 4.2 #5

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

|(a) the acceleration of the particle |(b) its coordinates at any time t. |

|a = (v / (t |r = ro + vot + ½ at2 |

|a = (9 – 3)i + (7 - -2)j / 3 |r(t) = 0 + (3i – 2j)*t + ½ (2i +3j)*t2 |

|a = 6i + 9j / 3 | |

|a = 2i +3j m/s2 |x(t) = 3t + t2 y(t) = -2t + 3t2/2 |

Ch 4.3 #22

|A dive bomber has a velocity of 280 m/s at an angle ( below the horizontal. |[pic] |

|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 point for| |

|release of the bomb to the target is 3.25 km. Find the angle, (. | |

|x = ½ at2 + vx-init t + xo |y = ½ at2 + vy-init t + yo |where t = 8.7 / cos( |

|2437 = 0 + 280 cos( t + 0 |2150 = ½ 10t2 + 280 sin( t |430 = 75.8 / cos2( + 487 tan( |

|t = 8.7 / cos( |430 = t2 + 56 sin( t | |

| Plug in excel and increment ( until left side equals right side. |

|0.583 radians or 33.4( |

|[pic] |

Ch 4.4 #31

Young David who slew Goliath experimented with slings before tackling the giant. He found that he could revolve a sling of length 0.6 m at the rated of 8 rps. If he increased the length to 0.9 m, he could revolve the sling only 6 rps. (a) Which rate of rotation gives the greater speed for the stone at the end of the sling? (b)What is the centripetal acceleration of the stone at 8 rps? (c) What is the centripetal acceleration at 6 rps?

|(a) |a = (v / (t = v2 / r |

|v = 8 rev / s * 2(r.6 / rev |a = ((*2(r / t)2 / r |

|v = 16(*0.6 |a = (2()2(2r / t2 |

|v = 9.6( m/s | |

| | |

|v = 6 rev / s * 2(r.9 / rev | |

|v = 12(*0.9 | |

|v = 10.8( m/s | |

| | |

|Thus at 6 rps results in a greater tangential velocity | |

| |(b) a = (2()2(2r / t2 |

| |a = (2*8)2(2(0.6) / 12 |

| |a = 1516 m/s2 |

| |(c) a = (2()2(2r / t2 |

| |a = (2*6)2(2(0.9) / 12 |

| |a = 1279 m/s2 |

Ch 4.6 #41

A river has a steady speed of ½ m/s. A student swims upstream a distance of 1 km and swims back to the starting point. If the student can swim at a speed of 1.2 m/s in still water how long does the trip take?

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

|(a) vnet up = 1.2 m/s – 0.5 m/s |ttotal = tnet up + tnet down |

|vnet up = 0.7 m/s |ttotal = 1428.6 s + 588.2 s |

|tnet up = 1000 m / 0.7 m/s |ttotal = 2017 s |

|tnet up = 1428.6 sec | |

| vnet down = 1.2 m/s + 0.5 m/s |(b) |

|vnet down = 1.7 m/s |total time w/o current = 2000 m / 1.2 m/s |

|tnet down = 1000 m / 1.7 m/s |total time w/o current = 1667 seconds |

|tnet down = 588.2 sec |(2017 – 1667) / 1667 = 21.0% |

| |w/ current takes 21 % longer than w/o current |

|Ch 4 #56 |[pic] |

|A boy can throw a ball a maximum horizontal distance of 40 m on a level playing field. How far| |

|can he throw the same ball vertically upward? | |

|Assume that his muscles give the ball the same speed in each case. vx = cos ( v vy = sin ( v | |

|a = (v / (t |Thrown the ball straight UP |

|-10 = (0 - sin( v) / ttop |…the only force on it is gravity |

|ttop = v sin( / 10 |a = (v / (t |

|ttotal = ttop + tbottom |-10 = (0 - (10r)½) / (t |

|Level Ground: ttop = tbottom |t = (r/10)½ seconds |

|ttotal = v sin( / 5 | |

|vx = (x / ttotal |dy = ½ at2 + vot |

|vcos( = R / (v sin( / 5) |dy = ½ -10(r/10)½*2 + (10r)½* (r/10)½ |

|v2 = 5R / sin(cos( |dy = ½ r |

|v = (10r)½ m/s | |

Let’s solve again but this time the total distance traveled is r instead of 40 meters.

vx = cos ( v

vy = sin ( v

ay = (v / (t -10 = (0 - sin ( v) / t solve for time t = sin ( * v / 10

but this is only half of the time. Thus the ttotal = (2 * sin ( v / 10) = sin ( * v / 5

We also know the vx = dx / ttotal where vx = cos ( * v

cos ( v = r / sin ( v / 5

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

v2 = 5 r / (0.707 * 0.707) = 10 r v = (10r)½

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

-10 = (0 – (10r)½) / t t = (r/10)½ seconds

dy = ½ at2 + vyot = ½ -10(r/10)½*2 + (10r)½* (r/10)½ = ½ r

Ch 4 #57

A stone at the end of a sling is whirled in a vertical circle of radius 1.2 m at a constant speed vi = 1.5 m/s as in Figure P4.57. The center of the string is 1.5 m above the ground. What is the range of the stone if it is released when the sling is inclined at 30.0( with the horizontal (a) at A? (b) at B? What is the acceleration of the stone (c) just before it is released at A? (d) just after it is released at A?

|vy = sin 60(*1.5 vx = cos 60(*1.5 |[pic] |

|vy = 1.3 m/s vx = 0.75 m/s | |

|(a) | |

|ay = (vy / (t | |

|-9.8 = (0-1.3) / t | |

|ttop = 0.133 sec | |

| | |

|dy-init = 0.3+1.2+sin30(*1.2 | |

|dy-init = 2.1 m | |

| | |

|dtop = ½ at2 + vot + dy-init | |

|dtop = 0.75 t – 5 t2 + 2.1 | |

|dtop = 2.185 m | |

| | |

|dbottom = ½ at2 + vy-init + dy-ini | |

|2.185 = ½ 9.8 t2 | |

| | |

|tbottom = 0.668 sec | |

|ttotal = ttop + tbottom | |

|ttotal = 0.133 + 0.668 | |

|ttotal = 0.8 sec | |

|vx = (x / (t | |

|0.75 m/s = (x / 0.8 sec | |

|(x = 0.60 meters | |

| |(b) dabove ground = vo in y compt + ½ at2 |

| |2.1 m = 1.2 t + 5t2 |

| |5t2 + 1.2t – 2.1 = 0 |

| |[-b ( (b2 – 4ac)1/2] / 2a |

| |(-1.2 ( 6.6) / 10 |

| |t = 0.54 seconds |

| |vx = (x / (t |

| |0.75 m/s = (x / 0.54 sec |

| |(x = 0.40 meters |

|(c) |(d) |

|a = (v / (t |immediately after it’s release ANYWHERE, the only force acting on it (neglecting air |

|a = v2 / r |friction) is gravity which has a rate of acceleration of 9.8 m/s2 pointed down |

|a = 2.25 / 1.2 | |

|a= 1.875 m/s2 | |

Ch 4 #65

The determined coyote is out once more to try to capture the elusive roadrunner. The coyote wears a pair of Acme jet-powered roller skates, which provide a constant horizontal acceleration of 15 m/s2. The coyote starts off at rest 70 m 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 have 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 straight ahead. (b) If the cliff is 100 m above the floor of a canyon, determine where the coyote lands in the canyon (assume his skates remain horizontal and continue to operate when he is in flight). (c) Determine the components of the coyote’s impact velocity.

(a) d = do + vot + ½ at2

Coyote distance = ½ at2 70 m = ½ 15 t2 t = 3.055 seconds

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

vave = d / t vroad runner = 70 m / 3.055 s = 22.9 m/s

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

100 m = do y + vo yt + ½ ayt2 = ½ 10t2 = 4.47 seconds

(c) d = do + vot + ½ at2 dx = ½ at2 = ½ 15 (7.52)2 = 424 meters total in the x-direction

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

(d) Total time the coyote was accelerating in the x direction is 3.055 + 4.47 = 7.52 seconds

a = (v / (t = 15 = (v / 7.52 seconds vf ( x dir = 113 m/s

10 = (v / 4.47 vf ( y dir = 44.7 m/s

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