Problem 2A - Hays High School

NAME ______________________________________ DATE _______________ CLASS ____________________Holt PhysicsProblem 2AAVERAGE VELOCITY AND DISPLACEMENTPROBLEMThe fastest fish, the sailfish, can swim 1.2 × 10 2 km/h. Suppose you havea friend who lives on an island 16 km away from the shore. If you senda message using a sailfish as a messenger, how long will it take for themessage to reach your friend?SOLUTIONGiven: v avg = 1.2 × 10 2 km/h∆x = 16 kmUnknown: ∆t = ?Use the definition of average speed to find ∆t.v avg = ⎯ ∆ ⎯x∆tRearrange the equation to calculate ∆t.∆x∆t = ⎯⎯v avg16 km∆t = ⎯⎯⎯ =1.2 × 10 2 ⎯ k mh⎯1 h⎯⎯ 60 min= 8.0 min16km⎯⎯2.0 km/minHRW material copyrighted under notice appearing earlier in this book.ADDITIONAL PRACTICE1. The Sears Tower in Chicago is 443 m tall. Joe wants to set the world’sstair climbing record and runs all the way to the roof of the tower. If Joe’saverage upward speed is 0.60 m/s, how long will it take Joe to climb fromstreet level to the roof of the Sears Tower?2. An ostrich can run at speeds of up to 72 km/h. How long will it take anostrich to run 1.5 km at this top speed?3. A cheetah is known to be the fastest mammal on Earth, at least for shortruns. Cheetahs have been observed running a distance of 5.50 × 10 2 mwith an average speed of 1.00 × 10 2 km/h.a. How long would it take a cheetah to cover this distance at this speed?b. Suppose the average speed of the cheetah were just 85.0 km/h.What distance would the cheetah cover during the same time intervalcalculated in (a)?Problem 2A 3

NAME ______________________________________ DATE _______________ CLASS ____________________4. A pronghorn antelope has been observed to run with a top speed of97 km/h. Suppose an antelope runs 1.5 km with an average speed of85 km/h, and then runs 0.80 km with an average speed of 67 km/h.a. How long will it take the antelope to run the entire 2.3 km?b. What is the antelope’s average speed during this time?5. Jupiter, the largest planet in the solar system, has an equatorial radius ofabout 7.1 × 10 4 km (more than 10 times that of Earth). Its period of rotation,however, is only 9 h, 50 min. That means that every point onJupiter’s equator “goes around the planet” in that interval of time. Calculatethe average speed (in m/s) of an equatorial point during one periodof Jupiter’s rotation. Is the average velocity different from the averagespeed in this case?6. The peregrine falcon is the fastest of flying birds (and, as a matter of fact,is the fastest living creature). A falcon can fly 1.73 km downward in 25 s.What is the average velocity of a peregrine falcon?7. The black mamba is one of the world’s most poisonous snakes, and witha maximum speed of 18.0 km/h, it is also the fastest. Suppose a mambawaiting in a hide-out sees prey and begins slithering toward it with avelocity of +18.0 km/h. After 2.50 s, the mamba realizes that its prey canmove faster than it can. The snake then turns around and slowly returnsto its hide-out in 12.0 s. Calculatea. the mamba’s average velocity during its return to the hideout.b. the mamba’s average velocity for the complete trip.c. the mamba’s average speed for the complete trip.8. In the Netherlands, there is an annual ice-skating race called the “Tour ofthe Eleven Towns.” The total distance of the course is 2.00 × 10 2 km, andthe record time for covering it is 5 h, 40 min, 37 s.a. Calculate the average speed of the record race.b. If the first half of the distance is covered by a skater moving witha speed of 1.05v, where v is the average speed found in (a), howlong will it take to skate the first half? Express your answer in hoursand minutes.HRW material copyrighted under notice appearing earlier in this book.4Holt Physics Problem Workbook

Motion In One DimensionChapter 2Additional Practice 2AGivensSolutions1. ∆x = 443 mv avg = 0.60 m/s∆x443m∆t = ⎯⎯ = ⎯⎯ = 740 s = 12 min, 20 sv avg0 .60m/s2. v avg = 72 km/h∆x = 1.5 km∆x1.5 km∆t = ⎯⎯ = ⎯⎯ = 75 sv avg 72 ⎯k m ⎯h ⎯ 1 h⎯ 3600s3. ∆x = 5.50 × 10 2 mv avg = 1.00 × 10 2 km/h∆x5.50 × 10 2 ma. ∆t = ⎯⎯ = ⎯⎯⎯⎯ =vavg1.00 × 10 2 ⎯ k mh⎯1 h⎯⎯ 36 00s⎯ 1 000m⎯1 km19.8 sIIv avg = 85.0 km/hb. ∆x =∆v avg ∆t1 h∆x = (85.0 km/h) ⎯⎯ 36 00s⎯ 1 30 m⎯1 km(19.8 s) = 468 mCopyright © by Holt, Rinehart and Winston. All rights reserved.4. ∆x 1 = 1.5 kmv 1 = 85 km/h∆x 1 = 0.80 kmv 2 = 67 km/h5. r = 7.1 × 10 4 km∆t = 9 h, 50 mina. ∆t tot =∆t 1 +∆t 2 = ⎯ ∆ x1⎯ + ⎯ ∆ x2⎯v1v21.5 km0.80 km∆t tot = ⎯⎯ + ⎯⎯ =64 s + 43 s = 107 s 67 ⎯k m ⎯h ⎯ 1 h⎯ 85 ⎯k m ⎯ h ⎯ 1 h⎯ 3600s3600s∆x 1.5 km + 0.80 km 2.3 kmb. v avg = 1 +∆x⎯ 2= ⎯⎯ = ⎯⎯ = 77 km/h∆t1 +∆t 2 (64 s + 43 s) 1 h(107 s) 1 h⎯36⎯ ⎯3⎯ 60000s∆x = 2πr∆x 2π(7.1 × 10 7 m)4.5 × 10 8 mv avg = ⎯ = ⎯⎯⎯⎯ = ⎯⎯⎯∆t60s(540 min + 50 min) ⎯ ⎯ (9 h) ⎯60 min⎯ 1 h+ 50 min ⎯ 60s⎯ 1 min1 min4.5 × 10 8 mv avg = ⎯⎯60s(590 min) ⎯ ⎯ 1 minv avg = 1.3 × 10 4 m/sThus the average speed = 1.3 × 10 4 m/s.On the other hand, the average velocity for this point is zero, because the point’s displacementis zero.Section Two — Problem Workbook Solutions II Ch. 2–1

Givens6. ∆x = –1.73 km∆t = 25 sSolutions∆xv avg = =⎯ –1.73 × 10 3 m⎯ ⎯ = –69 m/s = –250 km/h∆t 25s7. v avg,1 = 18.0 km/h∆t 1 = 2.50 s∆t 2 = 12.0 s1 ha. ∆x 1 = v avg,1 ∆t 1 = (18.0 km/h) ⎯⎯ 36 00s⎯ 1 30 m⎯1 km(2.50s) = 12.5 m∆x 2 = –∆x 1 = –12.5 mv avg,2 = ⎯ ∆ x⎯ 2 = ⎯ – 12.5 m⎯ =∆t 2 12.0 s–1.04 m/sb. v avg,tot = ⎯ ∆ x1+ ∆x2⎯ = ⎯ 12. 5 m + (−12.5m) 0.0 m⎯ = ⎯ =∆t1+ ∆t 2 2.50s + 12.0s 14.5 s0.0 m/sIIc. total distance traveled =∆x 1 – ∆x 2 = 12.5 m – (–12.5 m) = 25.0 mtotal time of travel =∆t 1 +∆t 2 = 2.50 s + 12.0 s = 14.5 saverage speed = ⎯ to taldistance⎯ = ⎯ 2 5.0 m⎯ = 1.72 m/stotaltime14.5 s8. ∆x = 2.00 × 10 2 km∆t = 5 h, 40 min, 37 sa. v avg = ⎯ ∆ x2.00 × 10 5 m⎯ = =⎯ 2.0 0 × 10 5 ⎯⎯⎯⎯⎯ ⎯∆t0ss20437sm 5 h ⎯360 60⎯h + 40 min ⎯ ⎯ m in+ 37 s v avg =9.79 m/s = 35.2 km/hv avg = (1.05)v avg∆x = ⎯ 1 2 ⎯ ∆x ⎯2.00 × 10 5 m⎯ 2∆xb. ∆t = ⎯ = ⎯⎯ =9.73 × 10 3 svavg (1.05) 9.79 ⎯ m s ⎯ ∆t = (9.73 × 10 3 1 hs) ⎯⎯ 36 00s= 2.70 h(0.70 h) ⎯ 60 min⎯1h= 42 min∆t = 2 h, 42 minCopyright © by Holt, Rinehart and Winston. All rights reserved.II Ch. 2–2Holt Physics Solution Manual