Answers to Homework (and Hints) - AS.171.101(06 Bin Ren 0 Spreadsheet(See Q00 HINTS Refer to Section 2.4 and Section 2.5 in the textbook(Fundamentals of

Answers to Homework (and Hints) - AS.171.101(06 Bin Ren 0...

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Unformatted text preview: AS.171.101 (06), 08/29/2014, Bin Ren 0. Spreadsheet: (See Q00.pdf) HINTS: Refer to Section 2.4 and Section 2.5 in the textbook (Fundamentals of Physics by Halliday). For your information, the reason why the ball is under a constant acceleration is that the force the ball received is constant along its line of motion. If you need my help in this problem, feel free to go to my office hour. Q01. (Sub) Two submarines on a secret mission need to arrive at a place in the middle of the ocean at the same time. They start out at the same time from positions equally distant from the rendezvous point. They travel at different velocities but both go in a straight line. The first submarine travels at an average velocity of 20 km/hr for the first 500 km, 40 km/hr for the next 500 km, 30 km/hr for the next 500 km and 50 km/hr for the final 500 km. The second submarine is required to travel at a constant velocity; what should be the magnitude of that velocity? ANS/HINTS: (Key: Average Speed) If the 2nd submarine wants to reach the same destination as the 1st submarine at the same time, it should cover the same distance within the time that the 1st submarine uses. Then, if we can get the distance and time that the 1st sub takes, and since the 2nd sub travels at a constant speed, we can get the average speed of 1st sub, which should be equal to the speed of the 2nd sub. Therefore, ! = ! = !! !! = !""!!""!!""!!"" !"" !"" !"" !"" ! ! ! !" !" !" !" = ⋯, (Do It Yourself) where ! is the speed of the 2nd sub, and ! denotes the average speed of the 1st sub, ! and ! is the distance and time the 1st sub traveled when they meet. Please get the answer yourself. Caution: a. Explanation of what the numbers mean in the above equation: there are 1 AS.171.101 (06), 08/29/2014, Bin Ren four terms in right side of the above equation, both in the numerator and denominator, and they are for the four stages of the travelling of the 1st sub, correspondingly. b. You’d better write the equation step-­‐by-­‐step, which would help you get the most credits if your result is wrong, the specific step should be like: 1. The distance that the 1st submarine travels is, ! = !! + !! + !! + !! = 500 + 500 + 500 + 500 = ⋯. (eq1.1) 2. The time that the 1st submarine travels is, ! = !! + !! + !! + !! = !"" !" + !"" !" + !"" !" + !"" !" = ⋯. (eq1.2) 3. If the 2nd submarine wants to arrive at the same point with the 1st submarine at the same time, it should have an average speed that is the same as that of the 1st submarine, which should be: ! = ! = !! !! = ⋯, (eq1.3) The reason why you should split the answers into steps is that it will help the readers catch up with what you are thinking in a logical way, and even though your answer might be wrong in equations 1, 2, or 3, the readers, or at least me, would understand that your reasoning is correct and give you more credits than the circumstance that you only put a equation like this: = !""!!""!!""! !"" !"" !"" !"" ! ! ! !" !" !" !" (Be careful: the last 50 is incorrect, which should be 500). c. In addition, you’d better number the equations, just like what I did in the parentheses – in this way, you can easily refer to any equations you want. Q02. (Canoe) A person in a canoe drifts down a river with the current at 1.5 mile per hour. A second person in another canoe begins at the same time as the first from 10 miles downstream and paddles upstream until the two canoes meet. This person can paddle fast enough to travel at an average velocity of 2.5 miles per hour relative to the shore when they go against this river current. When the two canoes meet, they will come to shore. You decide to go to that spot ahead of time and wait. Where 2 AS.171.101 (06), 08/29/2014, Bin Ren should you wait? ANS/HINTS: (Key: Relative Motion) A→ ←B (river flows →) Fig 2 Lets say the river flows towards right, and A is the person drifting downstream, and B is the one paddling upstream, sketched in Figure 2. If we set the direction that the river flows to be the reference direction (that is to say, any vector that has the same direction of the reference direction will have a positive magnitude, otherwise if it is on the opposite direction, negative.), then A would have a positive velocity, and B’s would be negative. =! ! !"#$%&'" =! ! ! !!! = ⋯ (eq2.1) Be careful, in eq. 2.1, ! is subtracted by ! , which is the definition of relative motion/speed. Since ! is positive, and ! negative, the result should be positive. And, you should calculate the location you wait (DIY). Q03. (Rocket) You have a summer job working for a university research group investigating the causes of the ozone depletion in the atmosphere. The plan is to collect data on the chemical composition of the atmosphere as a function of the distance from the ground using a mass spectrometer located in the nose cone of a rocket fired vertically. To make sure the delicate instruments survive the launch, your task is to determine the acceleration of the rocket before it uses up its fuel. The rocket is launched straight up with a constant acceleration until the fuel is gone 30 seconds later. To collect enough data, the total flight time must be 5.0 minutes before the rocket crashes into the ground. ANS/HINTS: The knowledge required is no more than that of Example 2.12 in the textbook. You can solve the question using the method, but here I want to introduce the v-­‐t graph (refer to Figure 2.11 in the textbook for details). 3 AS.171.101 (06), 08/29/2014, Bin Ren Let the direction of pointing vertically up be the direction of reference, then at first the rocket’s speed will increase (because of acceleration a), continues for ! = 30 sec ; when the fuel runs out, the rocket will continue going up but decelerates because of gravity, lasts for time period ! ; then at the highest point, its speed becomes zero, then accelerates downwards because of gravity (at this time, the velocity would become negative), at last, it will hit the ground, the time is ! and ! . See Figure 3.1 for details. A B C D Fig 3.1 The rocket will go up under acceleration , and reach the height of ℎ! at the end of ! : ! ! ℎ! = ! ! (eq3.1), which is the area of the first triangle (A), with a velocity of ! = !"# = ! (eq3.2). When the fuel runs out, the rocket will move only under the gravity. In ! and ! , the motion would be “symmetric”, see the right panel of Figure 3.1, thus they will have the same absolute value of velocity when the heights are the same, but with opposite directions. While in ! , the rocket will begin falling with a speed of !! = ! , (eq3.3) end with !! = !! + ! , (eq3.4),1 covering a distance of (just the area of trapezoid D) 1 The acceleration due to gravity should be negative, since it is pointing down. 4 AS.171.101 (06), 08/29/2014, Bin Ren ! ! = ! ! !! + !! ×! = ! ! + ! + ! ! . (eq3.5) From the right panel of Figure 3.1, the distance in ! and ! should be same, making Eq3.1 and Eq3.5 equal to each other: ! ! ! ! ! = ! ! + ! + ! ! , ! ! = 2! + ! ! . (eq3.6) Substituting ! = − ! − ! − ! = − ! − 2 !!! ! , (eq3.7) where T is the total time (300 sec), and t1 = 30s, into Eq3.6, you will get the relationship between acceleration and . (DIY) !" FYI: = !" . Note: With the help of v-­‐t diagram (left panel of Fig3.1), one can just calculate the area 5 ...
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