(1) v_avg  =  distance/time  =  230 km/3.25 h  =  70.8 km/h  =  70.8×10³ m/3600 s  =  19.7 m/s
(3) (110 ×10³m/3600 s) × 2 s  =  61.1 m
(13) a  =  [(95000/3600)/6.2] m/s²  =  4.3 m/s²
(15) (10 m/s)/1.35 s  =  7.41 m/s²  ×10^-3km/m×(3600 s)²/h²  =  9.60×10⁴ km/h²
(19)(v_f-v_i)/Dt  =  [(25-12)m/s] / 6 s  =  2.2 m/s²
Dx  =  (v_f²-v_i²)/(2a)  =  (25²-12²)/(2×2.17)  =  110 m
(21) Dx  =  v²/(2a)  =  (30 m/s)²/2/3 m/s²  =  150 m
(25) (a) Dx  =  - v₀²/(2a)  =  -(45000/3600)²/2/(-.5) m  =  160 m
(b) t  =  (2Dx/a)^1/2  =  25 s
(c) Dx₁  =  (45000/3600)*1 m/s  =  12 m
v(5 s)  =  [(45000/3600)m/s + 5 s×(-.5 m/s²)]  =  10 m/s
Dx₅  =  10 m/s ×1 s  =  10 m
(27) (a) Dx  =  1 s×(90000/3600)m/s + 25²/2/4 m  =  100 m
(b) Dx  =  25 m/s×1 s + 25²/2/8 m  =  64 m
(33) 90 km/h  =  25 m/s. t  =  v/g  =  25 m/s/9.8 m/s²  =  2.6 s
(35) (b) v  =  (2×9.8 ×380)^1/2  =  86 m/s
(a) t  =  y/v_avg  =  380/43  =  8.8 s.
(43) At t  =  0, v  =  v₀, Later, when y  =  0 again
0  =  v₀ t + g t²/2. Solving for the time, t  =  -2 v₀/g
Substituting this time into the kinematics equation v  =  v₀ + gt yields
v  =  v₀ - 2 v₀ g/g  =  - v₀; i.e., the speed is the same as the launch speed. (QED)
(51) Keep in mind that the instantaneous velocity is obtained from the slope of the x(t) curve. The average is obtained from the slope of the straight line that connects the starting and ending points.
(53) Note concerning (d): The magnitude of the acceleration [since the curve is v(t)] is maximum where the curve is steepest.
(61) h  =  0.5×v₀²/|g|® 0.5×v₀²/(|g|/6)  =  6×0.5×v₀²/|g|; i.e., it goes 6 times higher.

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File translated from T_EX by T_TH, version 1.95.
On 31 Aug 1999, 10:13.