Physics Solution Manual for 1100 and 2101

6 10 m s 110 10 s x vt 91 number ll 914 m

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: __________________________________________ 6. REASONING The distance traveled by the Space Shuttle is equal to its speed multiplied by the time. The number of football fields is equal to this distance divided by the length L of one football field. SOLUTION The number of football fields is ( )( −3 ) 7.6 ×10 m / s 110 ×10 s x vt = = 9.1 Number = = LL 91.4 m ______________________________________________________________________________ 7. 3 REASONING In a race against el-Guerrouj, Bannister would run a distance given by his average speed times the time duration of the race (see Equation 2.1). The time duration of the race would be el-Guerrouj’s winning time of 3:43.13 (223.13 s). The difference between Bannister’s distance and the length of the race is el-Guerrouj’s winning margin. SOLUTION From the table of conversion factors on the page facing the front cover, we find that one mile corresponds to 1609 m. According to Equation 2.1, Bannister’s average speed is Distance 1609 m Average speed = = Elapsed time 239.4 s Had he run against el-Guerrouj at this average speed for the 223.13-s duration of the race, he would have traveled a distance of 1609 m Distance = Average speed × Time = ( 223.13 s ) 239.4 s while el-Guerrouj traveled 1609 m. Thus, el-Guerrouj would have won by a distance of 1609 m 1609 m − ( 223.13 s ) = 109 m 239.4 s 8. REASONING The younger (and faster) runner should start the race after the older runner, the delay being the difference between the time required for the older runner to complete the race and that for the younger runner. The time for each runner to complete the race is equal to the distance of the race divided by the average speed of that runner (see Equation 2.1). SOLUTION The difference between the times for the two runners to complete the race is t50 − t18 , where 48 KINEMATICS IN ONE DIMENSION t50 = Distance ( Average Speed )50-yr-old and t18 = Distance ( Average Speed )18-yr-old (2.1) The difference between these two times (which is how much later the younger runner should start) is Distance Distance t50 − t18 = − ( Average Speed )50-yr-old ( Average Speed )18-yr-old 10.0 × 103 m 10.0 ×103 m = − = 64 s 4.27 m/s 4.39 m/s ______________________________________________________________________________ 9. REASONING In order for the bear to catch the tourist over the distance d, the bear must reach the car at the same time as the tourist. During the time t that it takes for the tourist to reach the car, the bear must travel a total distance of d + 26 m. From Equation 2.1, vtourist = d t (1) and vbear = d + 26 m t (2) Equations (1) and (2) can be solved simultaneously to find d. SOLUTION Solving Equation (1) for t and substituting into Equation (2), we find vbear = d + 26 m (d + 26 m)vtourist = d / vtourist d 26 m vbear = 1 + v d tourist Solving for d yields: 26 m 26 m = = 52 m vbear 6.0 m/s −1 −1 4.0 m/s vtourist ______________________________________________________________________________ d= 10. REASONING AND SOLUTION Let west be the positive direction. The average velocity of the backpacker is x +x e v= w t +t we where t x =w wv w and x t=e ev e Chapter 2 Problems 49 Combining these equations and solving for xe (suppressing the units) gives xe = – (1 – v / vw ) xw (1 – v / ve ) = – 1 – (1.34 m/s ) / ( 2.68 m/s ) ( 6.44 km ) 1 – (1.34 m/s ) / ( 0.447 m/s ) = –0.81 km The distance traveled is the magnitude of xe, or 0.81 km . ______________________________________________________________________________ 11. SSM WWW REASONING Since the woman runs for a known distance at a known constant speed, we can find the time it takes for her to reach the water from Equation 2.1. We can then use Equation 2.1 to determine the total distance traveled by the dog in this time. SOLUTION The time required for the woman to reach the water is Elapsed time = d woman vwoman 4.0 km 1000 m = = 1600 s 2.5 m/s 1.0 km In 1600 s, the dog travels a total distance of ddog = vdog t = (4.5 m/s)(1600 s) = 7.2 × 103 m ______________________________________________________________________________ 12. REASONING The definition of average velocity is given by Equation 2.2 as Average velocity = Displacement/(Elapsed time). The displacement in this expression is the total displacement, which is the sum of the displacements for each part of the trip. Displacement is a vector quantity, and we must be careful to account for the fact that the displacement in the first part of the trip is north, while the displacement in the second part is south. SOLUTION According to Equation 2.2, the displacement for each part of the trip is the average velocity for that part times the corresponding elapsed time. Designating north as the positive direction, we find for the total displacement that Displacement = m/s ( 27 m/s ) t 3 + 1442444 ( −174 ) tSouth 144 North 244 3 Northward Southward where tNorth and tSouth denote, respectively, the times for each part of the trip. Note that the minus sign indicates a direction due south. Noting that the total elapsed time is tNorth + tSouth, we can use Equation 2.2 to find the average velocity for the entire trip as follows: 50 KINEMATICS IN...
View Full Document

This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

Ask a homework question - tutors are online