Question.5587 - Chapter 3 Kinematics in Two Dimensions: VectorsAs we start Chapter 3, and continue upcoming chapters through the rest of the semester, a certain format will be used. The format has posted lectures such as this, given in about 3-5 pages. The posted lectures have been “boiled down” to contain all the concepts and skills you will need to succeed. Thus I highly suggest using these lectures on their own, while referring to either the recommended Giancoli text, or the optional Zero Cost Textbook (ZTC) as you may find useful. Comparing the Giancoli and ZTC texts has the latter with near-double the page count and less useful figures. For this reason, the numbering of sections, figures and examples in my lectures follows Giancoli. The ZTC text has the same order but “more spread out and less detailed”. So, please use the content in my posted lectures to succeed. Homework is the final 2 pages of this posting.This lecture begins with scalars and vectors, which were covered concisely in the Week 2 lecture. These present an “x and y column format” for vectors, such as used in engineering. The use of the column format is compact and gives reliable answers “every time”. It’s use continues thru this semester, and in the next 2B electromagnetics course. Giancoli takes 10 pages to cover it, and the ZTC text more. Starting with “scalar” and “vector” definitions, a scalar is a quantity with an amount only, and a vector is a quantity that has both amount and direction. Easiest example is 60 mph on a speedometer (scalar quantity), vs. 60 mph going due east on a freeway (vector quantity). Speed is a scalar, and speed with direction is a vector. Last week discussed an example of scalars vs. vectors in terms of doing laps on a ¼ mile track.The basic vector approach is to first differentiate vectors vs. scalars, and then set up the vector calculations in a 2 column format. Left and right columns are x and y coordinates, respectively. The figure describes how to do vector calculations using x and y components. Trig functions are used, with the buttons on your calculator (gray highlighting). Put your calculator in degrees mode!So long as the angle is counter-clockwise (CCW) from the +x axis, cosine goes with x, and sine goes with y. These are button pushes on your calculator. The figure to the right shows this set up. The angle is the Greek symbol ÆŸ “theta”. The top 2 equations get the x and y components from the vector’s amount A and angle ÆŸ. Otherwise if we already have the x and y components, the bottom 2 equations give the vector’s amount A and its angle ÆŸ. The square root equation is the Pythagoras equation. We use the top two equations to set up the columns in our vector calculations. The Example is “3 miles, at a +40 degree angle”. Here east is along +x, and north is along +y. “How far east, and how far north?” The columns are x = A cosÆŸ y = A sinÆŸ, putting in the numbers, x = 3 cos 400 y = 3 sin 400 x = 2.298 y = 1.928 and the units are miles. Next is adding vectors. Maintain your columns as done last week in the Example, by adding 2 vectors. Get x and y components of the first vector in two columns, then beneath, do these columns with x and y for the second. Adding the vectors simply means adding the x components, separately, and then adding the y components. These final x and y results get combined into a final answer, by using Pythagoras and then arctan (2nd tan), on the final summed x and y values.We will now blend in Giancolis, using the above column approach. Example 3-2 Displacement, Giancolis p. 56. Add the two subsequent vectors, 22 km north, and 47 km 600 “south of east” which is a −600 angle as shown Figure 3-15a. Using columns,Add the separate x’s and y’s, and final x = 23.5, final y = −18.7. Then A = (23.52+18.72)^.5=30.0 km, ÆŸ=tan-1(-18.7/23.5)=−38.50Note that this column approach does in 5 lines, what Giancolis takes twice as many lines to do (in a style that’s hard to see).Projectile Motion (Giancolis Section 3-5, pp. 58-64) We call “free fall” motion anything that goes straight up and down, under the downward acceleration of gravity. We call “projectile” motion anything that moves horizontally while being pulled down by the acceleration of gravity. Last week’s lecture gave an introduction to free fall. We are now focusing on projectile motion which is 2 dimensional, x and y, for horizontal and vertical components of motion. The 2 figures on the next page show these two types of projectile trajectories we need to consider. One goes “out and off the edge of a cliff” and the second is like a kicked football on a level playing field. Each has a time of flight, from initial launch to final impact.Doing 2 D projectile problems has you needing to keep the x and y components of the motion entirely separate, until the end of the problem. There is a basic reason to keep these x and y components separate. There is no acceleration horizontally along x, while there is a continual downward acceleration along y (due to gravity). We need separate “x” and “y” equations to do the analysis. This brings up “notation” where x is distance along the x coordinate and y is the distance along the y coordinate on the trajectory. Distance is final position – initial position, and a subscript zero is used to denote initial values of each. So overall x distance = x – x0, and overall y distance = y – y0. Now we write out last week’s Column 2 equations in terms of x and y, separately. We introduce subscripts on the velocity “v”, because it has separate values in the x and y directions and these are vx and vy. Each has a separate initial value vx0 and vy0.The x-equations have no acceleration. Also the trajectory has an angle in terms of vx and vy, where θ = tan-1(vy/vx).1. x = x0 + vx0 t and 2. vx = vx0 along x, and, 3. y = y0 +vy0 t + ½ a t2 and 4. vy = vy0 + at along y (initial x velocity vx0, and initial y velocity vy0 each have entirely different values). Here, vx0 = v0 cosθ, vy0 = v0 sinθ. We also have R = (vo2 / g) sin(2 θ).Let’s explain their use, starting with the horizontal launch off a cliff shown below in the figure. It’s a horizontal launch at t = 0, here the vx0 is the launch speed off the edge and note that there is no vertical component of velocity at t = 0 meaning vy0 = 0. A good rule-of-thumb in all projectile problems is to start with the vertical part of the problem and get the time-of-flight “t”.Example 3-5 Driving Off a Cliff (p. 61) The motorcycle drives horizontally off the cliff 50 m high, and lands 90 m from the base of the cliff. How fast was it going? We start with the vertical part of the problem, the y coordinate. We want to get the time of flight from this. First notice the green arrows, vertical ones pointing down and horizontal ones to the right These are velocity vector x,y components. The vertical ones keep stretching longer as the trajectory continues, and indeed it is picking up larger speeds downward along the way. But the horizontal arrows stay the same length, as there is no acceleration or speed change horizontally. Back to the vertical arrows, there is no down arrow just as it leaves the edge of the cliff. This is because its vertical speed there, initially, is vy0 = 0.We are first doing the vertical part of the problem to get the time of flight. Here use the y components, as follows: We start with symbols and values. y0 = height of the cliff, y0 = 50 m. Final y = height at impact, y = 0. Initial vertical velocity vy0 = 0, and final vy is unknown. Acceleration is downward, so a = -9.8 m/s2 (note the minus). We want to select an equation to find the time t, and we choose, y = y0 +vy0 t + ½ a t2 with vy0 = 0, so 0 = 50 + 0 – ½ (9.8) t2. Solving for t, t = 3.194 s. Now use the time t for the horizontal part of the problem. Here x0 = 0 (the base of the cliff) and x = the horizontal distance covered, x = 90 m. The “vx” is just the horizontal speed that never changes, and this is the speed that the problem requests. We choose the x = x0 + vx0 t equation, and this equation is 90 = 0 + vx0 (3.194), so vx0 = vx = 28.17 m/s, final answer, constant horizontal speed along x.Example 3-6 Kicked Football (p, 62). This is the second type of projectile problem. It launches and lands on a level surface. We follow the same basic approach, doing the vertical part first to get the time of flight. Then we do the horizontal part to solve other questions that may be asked. Consider first the green arrows. The vertical ones decrease to nothing as the ball reaches the apex, and then pick up downward as the ball heads down. The vertical velocity at the apex is zero, note this. The horizontal arrows are all same length meaning same horizontal speed throughout the trajectory. Here the ball is kicked at an angle of 370 at a speed of 20 m/s. It asks maximum height, total travel time, and how far down the field it goes. Note the initial speed is at the 370 angle.Starting with the vertical y component part of the problem, we see that at time t =0 when it is kicked, the height y0 = 0. The initial vertical velocity is vy0, and our vector work makes this easy, y goes with sine, so vy0 = 20 sin 370 = 12.04 m/s. Now we split the vertical part in two halves, the “going up” half and the “going down” half. Both times are the same, “going up” time equals the “going down” time, and this insight makes this type of projectile problem much easier. On the “going up half”, the final vertical velocity vy = 0, right? So choose an equation to get the time t, per our basic approach that first uses the vertical part to get the time. This is vy = vy0 + at, and 0 = 12.04 – 9.8 t, giving t = 1.228 s for the time “going up”. The time “going down” equals this, so the total time is double. This gives an overall time of 2 — 1.228 = 2.456 s. Don’t forget that a is negative, as it is in the -y direction. This problem asks another vertical part question, what is the maximum height? This is for the “going up” part where t = 1.228 s. The y = y0 +vy0 t + ½ a t2 equation is a good choice, giving y = 0 + 12.04 (1.228) + ½ (-9.8) 1.2282 = 7.396 m. We used the time “going up” to get the height, but we use the overall time to finish the problem, getting the total horizontal distance. We will use x = x0 + vx0 t where vx = vx0 is the horizontal velocity, just as in the cliff problem above. Horizontal goes with cosine, so vx0 = 20 cos 370 = 15.97 m/s (and this never changes). Finally, x = x0 + vx0 t so x = 0 + 15.97 (2.456) = 39.2 m.Also for this second trajectory example, is a useful equation that gives the maximum range “R” for a given launch angle.This is if launch and impact points are at the same height, and is R = v02/g sin(2θ). Here this is, R = (202/9.8) sin(2—37) = 39.24 m. This checks with the above result, to 3 significant figures. Note that “g” is used here, always +9.8 m/s2 at sea level (whereas “a” would have been −9.8 m/s2). This equation gives a maximum range “R” when θ = 450 (because sin(2—450) = sin900 =1, which is the sin function’s maximum value).This finishes our Chapter 3 coverage. Its format is narrative lecture notes. Background material comes from the recommended Giancoli text, and the optional ZTC text (Chapter 3 in each). The lecture presented above is complete, giving necessary concepts, figures, equations, and worked examples. There is weekly homework, and Graded Homework, given at the end of the lecture. The overall homework is numbered below, 1 – 24. Even numbered homework is the Graded Homework for you to turn in, 20 points maximum. Please turn in answers only for these; you need not show your work.Odd numbers are not to turn in, all have worked answers—with methods taken directly from the lecture. Maximize your success in this class by working through all examples “off to the side”, and then the odd numbered homework.There is a retry opportunity if turned in by 8 pm Thursday, otherwise Friday midnight is the firm deadline. Retries encourage an early start, better comprehension, and improved scores.Graded Homework is even numbers, please submit these in Assignments with answers onlyOverall Homework is 1 – 24, answers to odd problems follow to increase your understanding.Questions (narrative, possibly with appropriate equations).A car goes due east at 40 mph, and a second car goes north at 40 mph. Are their velocities equal? Explain.2. Is it true that a car cannot be accelerating, if its speedometer indicates a steady 60 mph? Explain.3. Which of the following is a vector: a person’s height, the altitude on Mt. Everest, the age of the Earth, the boiling point of water, the cost of this book, the Earth’s population, the acceleration of gravity?4. During baseball practice, the batter hits a very high fly ball, and then runs in a straight line to catch it. Which displacement was greater, the player’s or the ball’s? 5. What do vectors and scalars have in common? How do they differ?6. Does a car’s odometer measure a scalar or a vector quantity? What about the speedometer?7. Explain why a vector cannot have a component greater than its own magnitude.8. You are riding in an enclosed train car moving at 45 mph. If you throw a baseball straight up, where will the baseball land? (a) Ahead of you. (b) Behind you. (c) Down into your hand. (d) Need more information.9. A clod of dirt falls from the bed of a moving truck. It strikes the ground directly below the end of the truck. What is the direction of its velocity relative to the truck just before it hits? Explain.Problems Vector Calculations Problem 11 Problem 14 Problem 1610. A car is driven 225 km west and then 98 km southwest (45 ° between south and west). What is the displacement of the car from the point of origin (magnitude and direction)? 11. Find the north and east components of the displacement for the hikers shown in the figure above.12. A delivery truck travels 21 blocks north, 16 blocks east, and 26 blocks south. What is its final displacement from the origin? Assume the blocks are equal length.13. If Vx = 9.80 units and Vy = –6.40 units, determine the magnitude and direction of .14. An airplane is traveling 835 km/h in a direction 41.5 ° west of north (see figure above). (a) Find the components of the velocity vector in the northerly and westerly directions. (b) How far north and how far west has the plane traveled after 1.75 h?15. Explain why it is not possible to add a scalar to a vector.16. Three vectors are shown in the figure above. Their magnitudes are given in arbitrary units. Determine the sum of the three vectors. Give the result in terms of (a) components, and the (b) magnitude and angle with the +x axis.Projectile Motion (all without air resistance)17. A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 300 above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and y distances, from where the projectile was launched, to where it strikes the target?18. A daring person leaps horizontally from a 7.5-m-high rock with a speed of 3.0 m/s into deep water (about 25 ft and 7 mph). How far from the base of the rock will this person “splash in”?19. A battleship cannon fires a shell a maximum distance of 32.0 km. What is its initial velocity and maximum height? 20. A diver running 2.5 m/s dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below. How high was the cliff and how far from its base did the diver hit the water?21. Gun sights are adjusted to aim high to compensate for the effect of gravity, making the gun accurate only for that specific range. (a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? The bullet’s muzzle velocity is 275 m/s.22. A ball is thrown horizontally from the roof of a building 7.5 m tall and lands 9.5 m from the base. What was the ball’s initial speed?23. A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball?24. A projectile is fired with an initial speed of 36.6 m/s at an angle of 42.2 ° above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range), (d) the speed of the projectile 1.50 s after firing, and (e) the projectile’s speed at the apex of the trajectory. Answers to Odd Questions and Problems1. Equal magnitudes, but different directions (the first at 00 and the second at +900). Velocity is a vector. Vectors are defined in terms of both magnitude and direction, so the 2 vectors are different.3. magnitudedirection scalarvector Person’s heightyesupnoyes Mt. Everest altitudeyesupnoyes Earth’s ageyesnoyesno Water boiling pointyesnoyesno Book costyes (zero, OER)noyesno Earth’s populationyesnoyesno Acceleration of gravityyesdownnoyes5. Scalars and vectors have magnitudes. Only vectors have a geometric direction.7. Consider the Pythagoras formula , A = magnitude, and x and y are the components. Both x2 and y2 are always ≥ 0. If both = 0 then there is no vector. Otherwise one can be = 0 and the other not, and here, A = the magnitude of the nonzero component. If both are nonzero, then A must be > the largest of x and y. Thus A always ≥ the largest of x and y. Then the largest of x and y always ≤ A, and the statement is proved.9. It drops straight down from the moving truck, so its velocity is “vertical down” relative to the truck. Yet viewed from the side, by a stationary observer, its velocity is slanted going down, in the truck’s Direction of the truck’s motion.11. north component 3.21 km, east component 3.83 km (north is along y, and east along x), from, North, y = 5 sin(400) East, x = 5 cos(400)13. V = (vx2 + vy2)^.5 (using exponent .5 is the quick way to get the square root, from vx2 + vy2 answer. V = (9.82 + 6.42)^.5 = 11.70 units, note that we do not have to use – numbers in Pythagoras Θ = tan-1(vy/vx) = tan-1(-6.4/9.8) = −33.10, do have to use the – in the angle. And also must double check this angle from your calculator--these can only give angles in the first and fourth quadrants. So we double check the actual x and y, x is + and y is −, putting the angle in the fourth quadrant. The angle of −33.10 is in the fourth quadrant, consistent with vx = +9.8 and vy = −6.4, so it’s OK. But if the x and y had indicated the 2nd or 3rd quadrants, then you must automatically add +1800 to the this answer, to have the angle correct where positive angles are always measured from the +x axis.15. A scalar has magnitude only and no direction specified, while a vector has both magnitude and direction. You would not know the direction of the scalar, to properly add it to the vector.17. First get vxo = 50 cos 300 = 43.3 m/s and vy0 = 50 sin 300 = 25.0 m/s. Then x = x0 + vx0 t = 0 + 43.3(3) = 129.9 m and y = y0 +vy0 t + ½ a t2 = 0 + 25(3) + .5 (-9.8) 32 = 30.9 m.19. First use R = v02/g sin(2θ), set R = 3200 = v02/9.8 sin(2—450), knowing that 450 gives maximum range. Solving gives v0 = 177.1 m/s. Next get the max height, a “vertical question”, by getting the time first. Start this time question with vy0 = 177.1 sin(450) = 125.2 m/s, then use vy = vy0 + at knowing that vy = 0 at the apex or max height. Thus, 0 = 125.2 – 9.8 t, and t = 12.78 s. Finally get the height for this “time to max height” from y = y0 +vy0 t + ½ a t2, or, y = 0 + 125.2 (12.78) + .5 (-9.8) 12.782 = 799.7 m.21. Here v0 = 275 m/s, x0 = 0 and at x = 100 m, the height y is the same as y0. Use x = R = v02/g sin(2θ), 100 = 2752/9.8 sin(2θ), so θ = .371250, a shallow shooting angle. We need to get the time to the 150 m distance (along x) and first get vxo = 275 cos.371250 = 274.99 m/s. Use x = x0 + vx0 t or 150 = 0+274.88 t, so t = .5455 s. Finally use y = y0 +vy0 t + ½ a t2 to get the y coordinate--vy0 = 275 sin.371250 = 1.78 m/s. So y = 0 + 1.78 (.5455) + .5(-9.8) .54552 = −.487 m.23. We have vx0 = 16 m/s and vy0 = 12 m/s. (a) kicked, and hits the ground implies equal launch and impact heights. Here, symmetry in going up, and going down, implies the final vy = −12 m/s. Its final velocity uses Pythagoras for this vy and vx, or v = (162 + 122)^.5 = 20 m/s. (b) We use the vertical direction to get time. To the apex tup has vy0 = 12 and vy = 0, and vy = vy0 + at, 0 = 12 + (-9.8) t, so t = 1.224 s. It is in the air for double this time (tup = tdown, the total t = tup + tdown) so total time is 2— 1.224 = 2.448 s. (c) To get maximum height, use y = y0 +vy0 t + ½ a t2 and t = tup, thus height = y = 0 + 12 (1.224) + .5 (-9.8) 1.2242 = 7.35 m.

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