HW__3_Solutions_S06_try2

# HW__3_Solutions_S06_try2 - Physics 112 — HW#3 Solutions...

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Unformatted text preview: Physics 112 — HW #3 Solutions Spring 2006 2-23 [Automobile Airbags] Use vx 2 = v ox 2 + 2a(x - x o ) to relate velocity and position with: v ox = 105 km/hr 10 3 m km 1 hr 3600 s = 29 m/s vx = 0 and ax ≥- 250 m/s 2 (or |ax| ≤ 250 m/s 2 ) [NOTE: The "-" sign in ax denotes a direction opposite to the velocity v ox .] Solve for x - x o ≥- v ox 2 2a = - (29 m/s) 2 2(-250 m/s 2 ) = 1.7 m . [NOTE: Keep in mind that 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 were stopped suddenly in a hard collision. However, if the car's front end crumples in the collision (without allowing the passenger compartment to be crushed), then it may be possible to achieve this total distance of 1.7 m relative to the road by crumple + airbag. This is why modern motor vehicles are designed to have "crumple zones."] 2-74 Non-Constant Acceleration (a) x(t) = ∫ v x dt = ∫ ( α- β t 2 )dt = α t – β t 3 3 = (4.00 m / s ) t " (2/3 m / s 3 ) t 3 a x (t) = dv x dt = d dt ( α- β t 2 ) = - 2 β t = (-4.00 m/s 3 ) t (b) x is maximum (or minimum) when v x = 0: v x (t) = ( α- β t 2 ) = 0 so t = ± α β = ± 4.00 m/s 2.00 m/s 3 = ± 2 s Now x(t) = α t – β t 3 3 = (4.00 m/s) t - (0.667 m/s 3 ) t 3 = (4.00 m/s) ( ± 2 s) - (0.667 m/s 3 ) ( ± 2 s) 3 = ± 3.77 m We want the maximum positive displacement, so must take the +ve sign: x max = ± 3.77 m 2-82 [Model Rocket] (a) Graphs shown at the right ( → ): (b) Engine running: Δ y 1 = v o Δ t 1 + 1 2 a( Δ 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 Δ t 1 = 0 + (40.0 m/s 2 )(2.50 s) = 100 m/s After engine stops: v top 2 = v 1 2- 2g Δ y 2 = 0 ⇒ Δ 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 Δ t 2 ⇒ Δ 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 Δ y 3 v f = - v top 2- 2g Δ y 3 = - (0) 2- 2(9.8 m/s 2 )(-635 m) = - 112 m/s = v top- g Δ t 3 ⇒ Δ 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. 3-16 We define our coordinate system such that at t=0, x=y=0, and y is positive in the upward direction....
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## This homework help was uploaded on 04/21/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell.

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HW__3_Solutions_S06_try2 - Physics 112 — HW#3 Solutions...

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