HW2Sol - Physics 112 HW#2 Solutions 2-23 Use vx2 = vox2 2a(x xo to relate velocity and position with 103 m 1 hr vox = 105 km/hr = 29 m/s k m 3600 s

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Physics 112 - HW #2 Solutions 2-23 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: 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."] 2-31 (a) Use a x (t) = dv x (t) dt = slope of v x (t) vs. t graph. At t = 3 s: a x = 0 At t = 7 s: a x = 45 m/s - 20 m/s 9 s - 5 s = 6.2 m/s 2 At t = 11 s: a x = 0 m/s - 45 m/s 13 s - 9 s = - 11 m/s 2 (b) Use x = t 1 t 2 v x (t) dt = area "under" v x (t) vs. t graph from t 1 to t 2 . In the first 5 s: x = (20 m/s)(5 s) = 100 m From t = 5 s to 9 s: x = 45 m/s + 20 m/s 2 (9 s - 5 s) = 130 m, so. .. In the first 9 s: x = 100 m + 130 m = 230 m From t = 9 s to 13 s: x = 45 m/s 2 (13 s - 9 s) = 90 m, so. .. In the first 13 s: x = 230 m + 90 m = 320 m 2-37 (a) x(t) t Car Motorcycle 0 0 d t 1
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v x (t) The conceptually easy way to show that the motorcycle's speed v M1 when it overtakes the car is twice the car's speed v C is to use a graph of the vehicles' velocities vs. time. When the motorcycle overtakes the car at time t = t 1 , the two vehicles will have traveled the same displacement d from where they first met. Since displacement = area under the velocity vs. time graph, this means that the area under the Motorcycle's velocity graph
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This note was uploaded on 04/02/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell University (Engineering School).

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HW2Sol - Physics 112 HW#2 Solutions 2-23 Use vx2 = vox2 2a(x xo to relate velocity and position with 103 m 1 hr vox = 105 km/hr = 29 m/s k m 3600 s

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