Problem 2A - Hays High School

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