Physics 1111 Exam 1 2 30 A Solutions 8 February 2019 Name Physics 1111 Exam 1 2 30 A Solutions Instructions This is a closed book closed notes exam You will be given a separate sheet of formulas and numerical data that you may consult Nothing written on the formula sheet will be graded There is space after each question to show your work if you need more space you may use the back of the page or request more paper Please clearly indicate where your work for each problem is Underline or draw a box around your final answer The exam consists of five sections I recommend that you read all the questions at the start so that you can allocate your time wisely You may use a scientific calculator for arithmetic only your calculator must be non graphing non programmable and non algebraic You are not allowed to share your calculator The use of cell phones PDAs or any other electronic devices besides calculators is forbidden All such gadgets must be turned off and put away throughout the exam Do not open the exam until told to begin You have one class period 50 minutes to finish the exam Make sure every page of the exam has your last name You must provide explanations and or show work legibly to receive full credit for Parts III V Make sure that your answers include appropriate units and significant digits Assume that air resistance is negligible unless otherwise stated By signing below you indicate that you understand the instructions for this exam and agree to abide by them You also certify that you will personally uphold the university s standards of academic honesty for this exam and will not tolerate any violations of these standards by others Unsigned exams will not be graded Signature UGA ID 81x Row Seat Copyright c 2019 University of Georgia Unauthorized duplication or distribution prohibited Physics 1111 Exam 1 2 30 A Solutions 8 February 2019 Part I Score 12 II 27 III 40 Last Name IV B 21 Total Score 100 Grade 100 I Multiple Choice Questions 12 points For each question below choose the single best response and write the corresponding capital letter in the box provided There is no penalty for guessing 1 Marie wants to kayak across a river The current flows from north to south at speed v and Marie can paddle at a speed 2v in still water In what general direction should she head to cross the river to the east bank in the least amount of time C A B C D North of East South of East Due East The three directions take the same amount of time 2 A kicker kicks a football from ground level at the 10 yard line to the 50 yard line Where along the trajectory is the football s speed a minimum D A B C D E F At the 10 yard line just after the football leaves the kicker s foot At the 50 yard line just before the football lands At the 20 yard line At the 30 yard line At the 40 yard line The speed is the same throughout 3 A car s brakes produce a constant magnitude acceleration For some initial speed the stopping distance of the car is d If the initial speed doubles what will be the new stopping distance D A B C D E d 2 d 2d 4d 8d Copyright c 2019 University of Georgia 2 Physics 1111 Exam 1 2 30 A Solutions 8 February 2019 Last Name I Multiple Choice Questions Explanations 1 To get across the river in the shortest time we want to maximize the Easterly component of the kayak s velocity relative to the shore The river will only contribute a Southerly component and will not influence the motion across the river The component pointed across the river will come entirely from the velocity of the kayak through the water In order to maximize the Easterly component of that velocity we should point the kayak straight across the river 2 Since we ignore air resistance the horizontal speed of the ball will remain constant throughout the flight Thus the minimum overall speed will occur when the vertical component is smallest This happens at the top of the motion when the vertical velocity is zero Since the football comes back to the same height at which it started the top of the motion is halfway between the starting and ending points 3 This is constant acceleration motion which means we can use the constant acceleration equations to answer this First let s look at the original case We want to relate the distance traveled to the initial speed We also know that our final speed will be zero in both cases and the acceleration will also be the same in both cases We have an equation that relates these four quantities together vf2 vi2 2a x Let s apply that to this case letting v be the initial speed and setting the forward direction to be positive so the acceleration will be in the negative direction 0 v 2 2 a d d v2 2 a Now we can use the same relationship for the second case to find the new distance D 2 2v 2 v 2 0 2v 2 a D D 4 4d 2 a 2 a Copyright c 2019 University of Georgia 3 Physics 1111 Exam 1 2 30 A Solutions 8 February 2019 Last Name II Order Choice Questions 27 points For each question below write your responses in capital letters in the space provided 1 Each of the graphs below shows position velocity or acceleration plotted versus time for an object in motion along a straight line A B C v x a t t D t E x F a v t t t In the space provided identify in capital letters all graphs meeting the stated condition Write none if no graphs satisfy the condition a Velocity is constant A F b Acceleration is constant A B E F c Total displacement is zero over the time interval shown B d Velocity reverses direction B e Velocity increases over the whole time interval shown 2 Four projectiles are launched with different initial speeds so that they reach the same maximum height as shown Rank the trajectories in order of increasing initial speed Be sure to note if any of the initial speeds are equal Use less than symbols and equals signs to indicate rankings C E A B C D x B C A D Copyright c 2019 University of Georgia 4 Physics 1111 Exam 1 2 30 A Solutions 8 February 2019 Last Name II Order Choice Questions Explanations 1 15 points total 3 per part 0 5 per each item correctly listed or not listed a For 1 D motion velocity is the slope of position vs time so a constant slope of x vs t means a constant velocity An acceleration of zero also gives constant velocity b Constant velocity also means constant zero acceleration Additionally a constant slope of v vs t indicates …