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General Physics

by: Nichole Keebler

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General Physics PHYSICS 201

Nichole Keebler
UW
GPA 3.95

Staff

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Popular in Physics 2

This 5 page Class Notes was uploaded by Nichole Keebler on Thursday September 17, 2015. The Class Notes belongs to PHYSICS 201 at University of Wisconsin - Madison taught by Staff in Fall. Since its upload, it has received 28 views. For similar materials see /class/205227/physics-201-university-of-wisconsin-madison in Physics 2 at University of Wisconsin - Madison.

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Date Created: 09/17/15
Physics 201 Discussion 2 57 A car has an initial velocity 3900 when the driver sees an obstacle in the road in front of him His reaction time is At and the braking acceleration of the car is a Show that the total stopping distance is sStop voAt v02 25 Remember that a is a negative number 58 The yellow caution light on a traffic signal should stay on long enough to allow a driver to either pass through the intersection or safely stop before reaching the intersection A car can stop if its distance from the intersection is greater than the stopping distance found in the previous problem If the car is less than this stopping distance from the intersection the yellow light should stay on long enough to allow the car to pass entirely through the intersection a Show that the yellow light should stay on for a time interval Atlight Atr vs2 Si3900 where At is the driver s reaction time 3900 is the velocity of the car approaching the light at the speed limit a is the braking acceleration and si is the width of the intersection b As city traffic planner you expect cars to approach an intersection 160 m wide with a speed of 600 kin h Be cautious and assume a reaction time of 110 s to allow for a driver s indecision Find the length of time the yellow light should remain on Use a braking acceleration of 200 m s2 P257 The distance the car travels at constant velocity 3900 during the reaction time is Ax1 voAt The time for the car to come to rest from initial velocity 3900 after the brakes are applied is 390 vi 0 v0 1 2 Slot L1 L1 and the distance traveled during this braking period is vfvi 0U0 U0 Uri Axt t 2 v2 2 2 2 a 2a Thus the total distance traveled before coming to a stop is 2 STOP Ax1 Ax2 2 v I 258 a If a car is a distance sstop ngt 0 See the solution to Problem 257 from the intersection of length 5 when the light turns yellow the distance the car must travel before the light turns red is 2 3900 Axsstopsiv0At 2us Assume the driver does not accelerate in an attempt to beat the light an extremely dangerous practice The time the light should remain yellow is then the time required for the car to travel distance Ax at constant velocity 00 This is 2 Ax UOAtr Si 00 51 Athght Atr 00 00 211 00 b With 51 16 m 7260 kmh 12 20 ms2and At ii s 60 kmh 0278 ms 16m 1 kmh Athght 1391 S2 2o mszk 1 kmlh 60 kmh 0278 mls39 71 To protect his food from hungry bears a boy scout raises his food pack with a rope that is thrown over a tree limb at height h above his hands He walks away from the vertical rope with constant velocity vboy holding the free end of the rope in his hands Fig P271 a Show that the speed 0 of the food pack is given by xx2 If 2 vboy where x is the distance he has walked awa from the vertical rope b Show that the acceleration a of the food pack is l12x2 hz 3 Uzboy c What values do the acceleration and velocity U have shortly after he leaves the point under the pack x 0 d What values do the pack s velocity and acceleration approach as the distance x continues to increase P271 a In walking a distance Ax in a time At the length of rope is only increased by Axsin 6 Ax The pack lifts at a rate Esme Ax sin6 x x v v At boy E TJboy x2 hz d7 vboy dx d 1 b 1 dt r dt vboyx dtir vboy vboy x but x 1 O vboy r 2 dt dt v vb y r C Tr 0 FIG P271 d vboy 0 72 In problem 71 let the height 11 equal 600 rm and the speed vboy equal 200 ms Assume that the food pack starts from rest a Tabulate and graph the speed time graph b Tabulate and graph the accelerationtime graph Let the range of time be from 0 s to 500 s and the time intervals be 0500 s AX X vboyx P272 h600m vboy 200 ms Uthm9vbOyZMTyp X 2 vboytL 4t However x vboyt v vioytz h21392 412 36 2 39 FF A C V a vms 0 032 063 089 111 128 141 152 160 166 171 01 b1 0 4 llllllJJ ll1 I I 1 1 1 1 1 1 1 1 T wwNNHHoo 3901 U1HgtHgt 01 112711230 112721230 144 x2 112 3 2 b From problem 271 above a ts ams2 0 067 05 064 1 057 15 048 2 038 25 030 3 024 35 018 4 014 45 011 5 009 FIG P272b 1 1 1 1 1 1 1 1 1 1 1 1 1 74 Astronauts on a distant planet toss a rock into the air With the aid of a camera that takes pictures at a steady rate they record the height of the rock as a function of time as given in the Table P274 a Find the average velocity of the rock in the time interval between each measurement and the next b Using these average velocities to approximate instantaneous velocities at the midpoints of the time intervals make a graph of velocity as a function of tune Does the rock move with constant acceleration If so plot a straight line of best fit on the graph and calculate its slope to find the acceleration TABLE P274 Height of a Rock versus Time Time Height Time Height 8 In S In 000 500 275 762 025 575 300 725 050 640 325 677 075 694 350 620 100 738 375 552 125 772 400 473 150 796 425 385 175 810 450 286 200 813 475 177 225 807 500 058 250 790 P274 Time Height Ah At 5 midpt time t S h m m S InS t S 000 500 075 025 300 013 025 575 065 025 260 038 050 640 054 025 216 063 075 694 044 025 176 088 100 738 034 025 136 113 125 772 024 025 096 138 150 796 014 025 056 163 175 810 003 025 012 188 200 813 006 025 024 213 225 807 017 025 068 238 250 790 028 025 112 263 275 762 037 025 148 288 300 725 048 025 192 313 325 677 057 025 228 338 350 620 068 025 272 363 375 552 079 025 316 388 400 473 088 025 352 413 425 385 099 025 396 438 450 286 109 025 436 463 475 177 119 025 476 488 500 058 acceleration slope of line is constant 52 163 ms2 2 163 ms2 downward FIG P274

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