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HW_Solns_2_F05

# HW_Solns_2_F05 - Physics 112 HW#2 Solutions...

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- 1 - Physics 112 - HW #2 Solutions Fall 2005 2-23 [ Acceleration and the Human Body] Use v x 2 = v o x 2 + 2a(x - x o ) to relate velocity and position with: v o x = 105 km/hr 10 3 m km 1 hr 3600 s = 29 m/s v x = 0 and a x - 250 m/s 2 (or |a x | 250 m/s 2 ) [NOTE: The "-" sign in a x denotes a direction opposite to the velocity v o x .] Solve for x - x o = - v o x 2 2a - (29 m/s) 2 2(-250 m/s 2 ) = 1.7 m . [NOTE: This is the minimum distance your body must move relative to the road in coming to a stop. It is unlikely to be achieved with an airbag alone if your car stopped suddenly in a hard collision. But if the car's front end crumples in the collision (without letting the passenger compartment be crushed), then it may be possible to achieve this distance of 1.7 m relative to the road by crumple + airbag. This is why motor vehicles now have "crumple zones."] 2-82 [ Model Rocket] (a) Graphs shown at the right ( ): (b) Engine running: y 1 = v o t 1 + 1 2 a( t 1 ) 2 = (0)(2.50 s) - 1 2 (40.0 m/s 2 )(2.50 s) 2 = 125 m Maximum velocity: v 1 = v o + a t 1 = 0 + (40.0 m/s 2 )(2.50 s) = 100 m/s After engine stops: v top 2 = v 1 2 - 2g y 2 = 0 y 2 = v 1 2 2g = (100 m/s) 2 2(9.8 m/s 2 ) = 510 m Maximum height = 125 m + 510 m = 635 m How long to get there? v top = 0 = v 1 - g t 2 t 2 = v 1 g = 100 m/s 9.8 m/s 2 = 10.2 s t = 2.5 s + 10.2 s = 12.7 s (c) Just before reaching ground: v f 2 = v top 2 - 2g y 3 v f = - v top 2 - 2g y 3 = - (0) 2 - 2(9.8 m/s 2 )(-635 m) = - 112 m/s = v top - g t 3 t 3 = - v f g = - -112 m/s 9.8 m/s 2 = 11.4 s t = 12.7 s + 11.4 s = 24.1 s Speed |v f | = 112 m/s [NOTE: Take the - for v f since we know it's down ( ).] (d) No. The rocket's acceleration is not the same constant value throughout its flight.

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- 2 - 2-88 [ Free Fall With Sound] (a) Time taken for rock to reach ground is t R = 2H g , where H is the height of the cliff. Time taken for sound to reach top again t S = H v S , where v S = speed of sound Total time from dropping rock to hearing sound is t = t R + t S = 10.0 s. So 2H g + H v S = t, or H v S + 2H g - t = 0 This is a quadratic in the variable H . We can solve it analytically, and we get H = 19.6 m 1/2 , or H = 383 m (b) If you ignore the time it takes for the sound to reach you, you will overestimate the height of the cliff. You will think the rock reaches the bottom later than it really does, and you will say the cliff must be higher to account for that extra time. 1-47 [Vector Components] (d) (a) | A |= (4.00) 2 + (3.00) 2 = 5.00 | B |= (5.00) 2 + (2.00) 2 = 5.39 (b) A - B = -1.00 i + 5.00 j (c) | A - B |= (1.00) 2 + (5.00) 2 = 5.10 1-51 [Dot Product & Angles] (a) A B =(4.00 × 5.00+3.00 × (-2.00))=14.0 (b) A B =| A || B |cos( θ ) cos θ = A B | A || B | = 14.0 (5.00) (5.39) = 0.519 θ = 59°
- 3 - 3-81 [Relative Velocity/Reference Frames] Let "p" = plane, "a" = air, and "g" = ground: (a) In x-y (E-N) components: v p/a = (-220 km/h , 0) v p/g = -120 km 0.50 h , -20 km 0.50 h = (-240 km/h , -40 km/h) v p/a v p/g v a/g v p/a + v a/g = v p/g v a/g = v p/g - v p/a = (-240 km/h - (-220 km/h) , -40 km/h) = (-20 km/h , -40 km/h) wind speed = | v a/g | = (-20 km/s) 2 + (-40 km/h) 2 = 45 km/h tan θ = 20 km/h 40 km/h = 0.50 θ

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