Physics 1111 Exam #1 2:30-B Solutions8 February 2019 Name:Physics 1111Exam #1 2:30-B SolutionsInstructions:This is a closed-book, closed-notes exam. You will be given a separate sheet of formulas andnumerical data that you may consult. Nothing written on the formula sheet will begraded.There is space after each question to show your work; if you need more space, you may usethe back of the page, or request more paper. Please clearly indicate where your work foreach 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 startso 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 forbid-den. 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 forParts 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 agreeto abide by them. You also certify that you will personally uphold the university’s standardsof academic honesty for this exam, and will not tolerate any violations of these standards byothers. Unsigned exams will not be graded.Signature:UGA ID (81x) #: Row/Seat:Copyrightc 2019 University of Georgia. Unauthorized duplication or distribution prohibited.Physics 1111 Exam #1 2:30-B Solutions8 February 2019 Last Name:Part I II III IV B Total Score GradeScore /12 /27 /40 /21 /100 /100I: Multiple-Choice Questions (12 points)For each question below, choose the single best response and write the correspondingcapital 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 speedv, and Marie can paddle at a speed 2v in still water. In what general direction shouldshe head to cross the river to the east bank in the least amount of time?DA. The three directions take the same amount of time.B. North of East.C. South of East.D. Due East.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?CA. At the 10-yard line, just after the football leaves the kicker’s foot.B. At the 20-yard line.C. At the 30-yard line.D. At the 40-yard line.E. At the 50-yard line, just before the football lands.F. The speed is the same throughout.3. A car’s brakes produce a constant-magnitude acceleration. For some initial speed, thestopping distance of the car is d. If the initial speed doubles, what will be the newstopping distance?BA. 8dB. 4dC. 2dD. dE. d/2Copyrightc 2019 University of Georgia. 2Physics 1111 Exam #1 2:30-B Solutions8 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 componentof the kayak’s velocity relative to the shore. The river will only contribute a Southerlycomponent, and will not influence the motion across the river. The component pointedacross the river will come entirely from the velocity of the kayak through the water. In orderto maximize the Easterly component of that velocity, we should point the kayak straightacross the river.2. Since we ignore air resistance, the horizontal speed of the ball will remain constant through-out the flight. Thus, the minimum overall speed will occur when the vertical component issmallest. This happens at the top of the motion, when the vertical velocity is zero. Sincethe football comes back to the same height at which it started, the top of the motion ishalfway between the starting and ending points.3. This is constant-acceleration motion, which means we can use the constant-accelerationequations to answer this. First, let’s look at the original case. We want to relate thedistance traveled to the initial speed. We also know that our final speed will be zero inboth cases, and the acceleration will also be the same in both cases. We have an equationthat relates these four quantities together: v2f= v2i+ 2a(∆x). Let’s apply that to thiscase, letting v be the initial speed, and setting the forward direction to be positive, so theacceleration will be in the negative direction:0 = v2− 2|a|d =⇒ d =v22|a|.Now, we can use the same relationship for the second case to find the new distance, D.0 = (2v)2− 2|a|D =⇒ D =(2v)22|a|= 4v22|a|= 4d.Copyrightc 2019 University of Georgia. 3Physics 1111 Exam #1 2:30-B Solutions8 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 CtatvtxD E FtatvtxIn the space provided, identify in capital letters all graphs meeting the stated con-dition. Write “none” if no graphs satisfy the condition.(a) Velocity is constant. C, D(b) Acceleration is constant. B, C, D, E(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. A, E2. Four projectiles are launched with different ini-tial speeds so that they reach the same maximumheight, as shown. Rank the trajectories in order ofincreasing initial speed. Be sure to note if any ofthe initial speeds are equal. Use less-than symbolsand equals signs (<, =) to indicate rankings.C <B <A <DxA BCDCopyrightc 2019 University of Georgia. 4Physics 1111 Exam #1 2:30-B Solutions8 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 xvs. t means a constant velocity. An acceleration of zero also gives constant velocity.(b) Constant velocity also means constant (zero) acceleration. Additionally, a constantslope