Problem 2A - Hays High School

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