# 250 h 170 h a v x 900 km 529 k m h t 170 h b

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Unformatted text preview: 200 h + 0.750 h + 0.250 h = 1.70 h . (a) v= ∆x 90.0 km = = 52.9 k m h ∆t 1.70 h (b) ∆x = 90.0 k m (see above) 2.2 (a) v= 1 yr ∆x 20 ft ⎛ 1 m ⎞ ⎛ ⎞ −7 = ⎜ ⎟⎜ ⎟ = 2 × 10 m s , 7 ∆t 1 yr ⎝ 3.281 ft ⎠ ⎝ 3.156 × 10 s ⎠ or in particularly windy times v= 1 yr ∆x 100 ft ⎛ 1 m ⎞ ⎛ ⎞ −6 = ⎜ ⎟⎜ ⎟ = 1 × 10 m s 7 ∆t 1 y r ⎝ 3.281 ft ⎠ ⎝ 3.156 × 10 s ⎠ (b) The time required must have been ∆t = 2.3 ∆x 3 × 103 m i ⎛ 1609 m ⎞ ⎛ 103 m m ⎞ 8 = ⎜ ⎟⎜ ⎟ = 5 × 10 yr . 10 m m y r ⎝ 1 m i ⎠ ⎝ 1 m ⎠ v (a) Boat A requires 1.0 h to cross the lake and 1.0 h to return, total time 2.0 h. Boat B requires 2.0 h to cross the lake at which time the race is over. Bo a t A w in s, b ein g 60 k m a h ea d o f B when the race ends. (b) Average velocity is the net displacement of the boat divided by the total elapsed time. The winning boat is back where it started, its displacement thus being zero, yielding an average velocity of zero . 2.4 (a) v= 1 yr ∆x 20 ft ⎛ 1 m ⎞ ⎛ ⎞ -11 = ⎜ ⎟⎜ ⎟ = 5 × 10 m s 7 ∆t 4000 yr ⎝ 3.281 ft ⎠ ⎝ 3.156 × 10 s ⎠ 21 CHAPTER (b) v = 2.5 (a) 2 ∆x 2 ft ⎛ 1 m ⎞ ⎛ 1 d a y ⎞ -6 = ⎜ ⎟⎜ ⎟ = 7 × 10 m s ∆t 1 d a y ⎝ 3.281 ft ⎠ ⎝ 8.64 × 104 s ⎠ ⎛ 35.0 ⎞ Disp la cem en t = ( 85.0 k m h ) ⎜ h + 130 k m ⎝ 60.0 ⎟ ⎠ ∆x = ( 49.6 + 130) k m = 180 k m (b) Av er a ge v elocit y = 2.6 (a) v= Disp la cem en t ( 49.6 + 130) km = = 63.4 k m h ela p sed t im e ⎡ ( 35.0 + 15.0) ⎤ + 2.00 ⎥ h ⎢ 60.0 ⎣ ⎦ ∆x 10.0 m = = 5.00 m s ∆t 2.00 s x(m) 10 8 ∆x 5.00 m = = 1.25 m s (b) v = ∆t 4.00 s (c) v= (d) v = ∆x 5.00 m − 10.0 m = = − 2.50 m s ∆t 4.00 s − 2.00 s ∆x − 5.00 m − 5.00 m = = − 3.33 m s ∆t 7.00 s − 4.00 s 2.7 v= (a) v 0,1 = (b) v 0, 4 = (c) v 1, 5 = (d) v 0,5 = 2 0 –2 –4 1234567 8 t(s) –6 ∆x x 2 − x 1 0−0 = = =0 ∆t t 2 − t1 8.00 s − 0 (e) 6 4 x 1 − x 0 4.0 m − 0 = = + 4.0 m s 1.0 s ∆t x 4 − x 0 − 2...
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