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# Note for PHYS 1321 with Professor Hor at UH

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This 12 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Houston taught by a professor in Fall. Since its upload, it has received 36 views.

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Date Created: 02/06/15

Chapter 2 Concept Question 3 Figure 216 shows four paths along which objects move from a starting point all in the same time interval The paths pass over a grid of equally spaced straight lines In the following questions you will need to rank the paths If multiple paths rank equally use the same rank for each then exclude the intermediate ranking ie if objects A B and C must be ranked and Aand B must both be ranked first the ranking would be Al Bl C3 If all paths rank equally rank each as 39139 Rank the paths according to the average velocity of the objects greatest first Solution Velocity is a vector quantity the displacements of object 1234 are equal all in the same time so object 1234 have the same velocity So their ranks are Object 1 1 Object 2 1 Object 3 1 Object 4 1 Rank the paths according to the average speed of the objects greatest first Solution Speed is a scalar quantity Object l234 have the same motion time So the greater the path length is the greater its speed is We can get their ranks as Object l 2 Object 2 2 Object 3 4 Object 4 l Chapter 2 Concept Question 4 Figure2l7 is a graph of a particle s position along an x axis versus time At time t 0 what is the sign of the particle s position Solution negative Is the particle s velocity positive negative or 0 at t 1 s Solution positive Is the particle s velocity positive negative or 0 at t 2 s Solution zero Is the particle s velocity positive negative or 0 at t 3 s Solution negative How many times does the particle go through the point X 0 Solution 2 at t1 s and t3 5 Chapter 2 Concept Question 5 Figure 218 gives the velocity of a particle moving along an axis Point 1 is at the highest point on the curve point 4 is at the lowest point and points 2 and 6 are at the same height v What is the direction of travel at time t 0 Solution positive What is the direction of travel at point 4 Solution negative At which of the six numbered points does the particle reverse its direction of travel Multiple answers may be correct Solution Point 3 Point 5 Rank the six points according to the magnitude of the acceleration greatest first If multiple points rank equally use the same rank for each then exclude the intermediate ranking ie if objects A B and C must be ranked and A and B must both be ranked first the ranking would be Al Bl C3 If all points rank equally rank each as 39139 v Solution a d we can see from the graph the derlvatlves of v have the greatest t magnitude at point 2 and point 6 the second greatest magnitude at point 3 and point 5 the least great magnitude at point 1 and point 4 So we get Point 1 5 Point 2 1 Point 3 3 Point 4 5 Point 5 3 Point 6 1 Chapter 2 Concept Question 7 Hanging over the railing of a bridge you drop an egg no initial velocity as you throw a second egg downward Curves A and B are parallel so are C D and E so are F and G Which curve in Fig 219 give the velocity vt for the dropped egg Solution D v00 and then V increases in the negative direction when t grows Which curve in Fig 219 give the velocity vt for the dropped egg Solution E v0lt0 and then V increases in the negative direction when t grows Chapter 2 Problem 15 A particle39s position is given by X 400 1200t 312 in which X is in meters and tis in seconds a What is its velocity at t 1 s b Is it moving in the positive or negative direction of gtltjust then c What is its speed just then d Is the speed increasing or decreasing just then Try answering the next two questions without further calculation e Is there ever an instant when the velocity is zero If so give the time t if not answer quot0quot f Is there a time after t 3 s when the particle is moving in the negative direction of X If so give the time t if not answer quot0quot Solution We use Eq 24 to solve the problem a The velocity of the particle is d 2 v 4 12t3t 126t dt Thus at t l s the velocity is v 712 6l 4 ms b Since v lt 0 it is moving in the 7x direction at t l s c At t l s the speed is lvl 6 ms d For 0 lt tlt 2 s lvl decreases until it vanishes For 2 lt tlt 3 s lvl increases from zero to the value it had in part c Then lvl is larger than that value for t gt 3 s e Yes since v smoothly changes from negative values consider the t 1 result to positive note that as t gt 00 we have v gt 00 One can check that v 0 when t 2 s t No In fact from v 712 6t we know that v gt 0 for t gt 2 s Chapter 2 Problem 28 On a dry road a car with good tires may be able to brake with a constant deceleration of 492 ms2 a How long does such a car initially traveling at 246 ms take to stopb How far does it travel in this time Solution We take x in the direction of motion so v0 246 ms and a 7 492 msz We also take x0 0 a The time to come to a halt is found using Eq 211 0v0at 2 t LmSzziom 492ms b Although several of the equations in Table 21 will yield the result we choose Eq 2 16 since it does not depend on our answer to part a 246 ms2 2 492 msz61395m39 0v 2ax 3 x Chapter 2 Problem 37 The figure depicts the motion of a particle moving along an X axis with a constant acceleration The figure39s vertical scaling is set by X 600 m What are the a magnitude and b direction of the particle39s acceleration x denote 1 0 otherwise x i 1111 Solution a From the gure we see that x0 720 m From Table 21 we can apply x7x0v0t at2 with t 10 s and then again with t 20 s This yields two equations for the two unknowns v0 and a 00 20 mv0 10 sal0 s2 60 m 20 mv0 20 sa20 s2 Solving these simultaneous equations yields the results v0 0 and a 40 msz b The fact that the answer is positive tells us that the acceleration vector points in the x direction Chapter 2 Problem 53 A key falls from a bridge that is 45 m above the water It falls directly into a model boat moving with constant velocity that is 12 m from the point of impact when the key is released What is the speed of the boat S olution The speed of the boat is constant given by vb dt Here d is the distance of the boat from the bridge when the key is dropped 12 m and t is the time the key takes in falling To calculate t we put the origin of the coordinate system at the point where the key is dropped and take the y aXis to be positive in the downward direction Taking the time to be zero at the instant the key is dropped we compute the time I when y 45 m Since the initial velocity of the key is zero the coordinate of the key is given by y fgtz Thus I 2y lm3o3s g 98ms Therefore the speed of the boat is 12 m 303 s v 40ms b Chapter 2 Problem 54 A stone is dropped into a river from a bridge 439 m above the water Another stone is thrown vertically down 1 s after the first is dropped Both stones strike the water at the same time What is the initial speed of the second stone S olution We neglect air resistance which justi es setting a 7g 798 ms2 taking down as the 7 y direction for the duration of the motion We are allowed to use Eq 215 with Ay replacing Ax because this is constant acceleration motion We use primed variables except I with the rst stone which has zero initial velocity and unprimed variables with the second stone with initial downward velocity ivo so that v0 is being used for the initial speed SI units are used throughout Ay39 0 th Since the problem indicates Ay Ay 4139 In we solve the rst equation for t nding t 299 s and use this result to solve the second equation for the initial speed of the second stone i 439 m v0l99 s 98ms2l99 s2 which leads to v0 123 ms Chapter 2 Problem 64 A ball is shot vertically upward from the surface of another planet A plot of y versus tfor the ball is shown in the figure where y is the height of the ball above its starting point and t 0 at the instant the ball is shot The figure39s vertical scaling is set by y 250 m What are the magnitudes of a the freefall acceleration on the planet and b the initial velocity of the ball J lt y m Solution The graph shows y 25 m to be the highest point where the speed momentarily vanishes The neglect of air friction or whatever passes for that on the distant planet is certainly reasonable due to the symmetry of the graph a To nd the acceleration due to gravity gP on that planet we use Eq 215 with y up yyo vt gpt2 2 25 m 0025sgp25 s2 so that gP 80 msz b That same max point on the graph can be used to nd the initial velocity yy0v0vt 3 25m 0v0025s Therefore v0 20 ms Chapter 2 Problem 69 How far does the runner whose velocity time graph is shown in the figure travel in 16 s The figure39s vertical scaling is set by S 800 ms s 1 m 0 J 12 143 i M Solution Since v dx dt Eq 24 then Ax Iv dt which corresponds to the area under the v vs I graph Dividing the total area A into rectangular base gtlt height and triangular base gtlt height areas we have A A0lttlt2 A2lttlt10 A10lttlt12 A12lttlt16 lt2gtlt8gtlt8gtlt8gtlt2gtlt4gt lt2gtlt4gtlt4gtlt4gt with SI units understood In this way we obtain Ax 100 m Chapter 2 Problem 77 A hot rod can accelerate from 0 to 60 kmh in 54 s a What is its average acceleration in ms2 during this time b How far will it travel during the 54 s assuming its acceleration is constant c From rest how much time would it require to go a distance of 025 km if its acceleration could be maintained at the value in a S olution Since the problem involves constant acceleration the motion of the rod can be readily analyzed using the equations in Table 21 We take x in the direction of motion so v60kmh W l67ms 3600sh and a gt 0 The location where it starts from rest v0 0 is taken to be x0 0 a Using Eq 27 we nd the average acceleration to be am V v0W309ms2z3lms2 3 At t to 54s 0 b Assuming constant acceleration a czan 309 ms2 the total distance traveled during the 54s time interval is xx0 v0t at2 00309ms254 s2 45 m c Using Eq 215 the time required to travel a distance ofx 250 m is 2 250 leat2 3 t E 2135 2 lla 31ms2 Note that the displacement of the rod as a function of time can be written as xt 609 ms2t2 Also we could have chosen Eq 217 to solve for b xan taM7m 649 m Chapter 2 Problem 100 A parachutist bails out and freely falls 50 m Then the parachute opens and thereafter she decelerates at 2 ms2 She reaches the ground with a speed of 30 ms a How long is the parachutist in the air b At what height does the fall begin S olution During free fall we ignore the air resistance and set a 7g 798 ms2 where we are choosing down to be the 7y direction The initial velocity is zero so that Eq 215 becomes Ay gt2 where Ay represents the negative of the distance d she has fallen Thus we can write the equation as d 3 gt2 for simplicity a The time t1 during which the parachutist is in free fall is using Eq 215 given by 1 d 50m gt12 980ms2t12 which yields t1 32 s The speed of the parachutist just before he opens the parachute is given by the positive root v12 2 gal1 or v12ghl 2980ms250 m 3lm s If the nal speed is vz then the time interval t2 between the opening of the parachute and the arrival of the parachutist at the ground level is 3lms 30ms l4s a 2ms2 This is a result of Eq 211 where speeds are used instead of the negativevalued velocities so that finalvelocity minus initialvelocity turns out to equal initialspeed minus finalspeed we also note that the acceleration vector for this part of the motion is positive since it points upward opposite to the direction of motion 7 which makes it a deceleration The total time of ight is therefore II t l7 s b The distance through which the parachutist falls after the parachute is opened is given by V V 3lms2 30m s2 61 2a 220ms2 z 240m In the computation we have used Eq 216 with both sides multiplied by 71 which changes the negativevalued Ay into the positive d on the lefthand side and switches the order of v1 and v on the righthand side Thus the fall begins at a height of h 50 d m 290 m Chapter 2 Problem 101 A ball is thrown down vertically with an initial speed of 29 fts from a height of 30 ft a What is its speed in fts just before it strikes the ground b How long does the ball take to reach the ground What would be the answers to c part a and d part b if the ball were thrown upward from the same height and with the same initial speed Before solving any equations decide whether the answers to c and d should be greater than less than or the same as in a and b S olution We neglect air resistance which justifies setting a 7g 798 ms2 3215 fts2 taking down as the 7y direction for the duration of the motion We are allowed to use Table 21 with Ay replacing Ax because this is constant acceleration motion The ground level is taken to correspond to y 0 a With yo h and v0 replaced with ivo Eq 216 leads to V v02 2gy y0 Vvd 2537 so we get v 29fts2 2gtlt3215fts2 X30 5263 ft s The positive root is taken because the problem asks for the speed the magnitude of the velocity b We use the quadratic formula to solve Eq 215 for t with v0 replaced with ivo VmV22gAy Ay V0t lgt2 3 1 2 g where the positive root is chosen to yield tgt 0 With y 0 and yo h this becomes v Zgh v0 g From part awe know quotv 2gh v so we get t5263 s 29 s 0735 s 3215 ft s2 c If it were thrown upward with that speed from height h then in the absence of air friction it would return to height h with that same downward speed and would therefore yield the same final speed before hitting the ground as in part a An important perspective related to this is treated later in the book in the context of energy conservation d Having to travel up before it starts its descent certainly requires more time than in part b The calculation is quite similar however except for now having v0 in the equation where we had put in ivo in part b The details follow v0 quotv ZgAy 1 Ay vot gt2 3 t 2 g with the positive root again chosen to yield t gt 0 With y 0 and yo h we obtain 1v 2gh v0 g So t5263 s29fts 2539 5 3215 ft 52

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