#$&* course PHY 121 Question: `qQuery principles of physics and general college physics 7.02. Delivery truck 18 blocks north, 10 blocks east, 16 blocks south. What is the final displacement from the origin?YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: `aThe final position of the truck is 2 blocks south and 10 blocks east. This is equivalent to a displacement of +10 blocks in the x direction and -2 blocks in the y direction. The distance is therefore sqrt( (10 blocks)^2 + (-2 blocks)^2 ) = sqrt( 100 blocks^2 + 4 blocks^2) = sqrt(104 blocks^2) = sqrt(104) * sqrt(blocks^2) = 10.2 blocks. The direction makes and angle of theta = arcTan(-2 blocks / (10 blocks) ) = arcTan(-.2) = -12 degrees with the positive x axis, as measured counterclockwise from that axis. This puts the displacement at an angle of 12 degrees in the clockwise direction from that axis, or 12 degrees south of east. STUDENT QUESTION: Why don’t we add 180 to the angle since the y is negative? INSTRUCTOR RESPONSE: We add 180 degrees when the x component is negative, not when the y component is negative. You that 168 degrees is in the second quadrant, where the y component is positive. The arctan gives us -12 degrees, which is in the fourth quadrant (where the y component is negative and the x component positive, consistent with the given information). We often want an angle between 0 and 360 deg; when the vector is in the fourth quadrant, so that the angle is negative, we can always add 360 degrees to get an equivalent angle (called a 'coterminal' angle, 'coterminal' meaning 'ending at the same point'). In this case the angle could be expressed as -12 degrees or -12 degrees = 360 degrees = 348 degrees. Either angle specifies a vector at 12 degrees below horizontal. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery principles of physics and general college physics 7.18: Diver leaves cliff traveling in the horizontal direction at 1.8 m/s, hits the water 3.0 sec later. How high is the cliff and how far from the base does he land? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 0 * 3.0 sec + .5 * 9.8 m/s^2 * (3.0 sec)^2 = 44 m for the height of the cliff 1.8 m/s * 3.0 s = 5.4 meters from the base confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe diver's initial vertical velocity is zero, since the initial velocity is horizontal. Vertical velocity is characterized by the acceleration of gravity at 9.8 m/s^2 in the downward direction. We will choose downward as the positive direction, so the vertical motion has v0 = 0, constant acceleration 9.8 m/s^2 and time interval `dt = 3.0 seconds. The third equation of uniformly accelerated motion tells us that the vertical displacement is therefore vertical motion: `ds = v0 `dt + .5 a `dt^2 = 0 * 3.0 sec + .5 * 9.8 m/s^2 * (3.0 sec)^2 = 0 + 44 m = 44 m. The cliff is therefore 44 m high. The horizontal motion is characterized 0 net force in this direction, resulting in horizontal acceleration zero. This results in uniform horizontal velocity so in the horizontal direction v0 = vf = vAve. Since v0 = 1.8 m/s, vAve = 1.8 m/s and we have horizontal motion: `ds = vAve * `dt = 1.8 m/s * 3.0 s = 5.4 meters. STUDENT COMMENT/QUESTION Why do we not calculate the magnitude for this problem, I know the number are identical but it seems that this would tell us how far from the base the diver traveled? I understand how to calculate the magnitude using the pythagorean theorem and the directions using arc tan, but I am not quite clear on why and when it is neccessary. ? INSTRUCTOR RESPONSE The diver doesn't travel a straight-line path. His path is part of a parabola. It would be possible to calculate the distance traveled along his parabolic arc. However that would require calculus (beyond the scope of your course) and while it would be an interesting exercise, it wouldn't contribute much to understanding the physics of the situation. What you did calculate using the Pythagorean theorem is the magnitude of the displacement from start to finish, i.e. the straight-line distance from start to finish. The diver's displacement is a vector with a magnitude (which you calculated) and and angle (which you could easily have calculated using the arcTangent). However this vector is not in the direction of any force or acceleration involved in the problem, and it's not required to answer any of the questions posed by this situation. So in this case the displacement not particularly important for the physics of the situation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGen phy 3.13 A 44 N at 28 deg, B 26.5 N at 56 deg, C 31.0 N at 270 deg. Give your solution to the problem. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The solution given here is for a previous edition, in which the forces are Force A of 66 at 28 deg Force B of B 40 at 56 deg Force C of 46.8 at 270 deg These forces are very close to 2/3 as great as the forces given in the current edition, and all correct results will therefore be very close to 2/3 as great as those given here. Calculations to the nearest whole number: A has x and y components Ax = 66 cos(28 deg) = 58 and Ay = 66 sin(28 deg) = 31 Bhas x and y components Bx = 40 cos(124 deg) = -22 and By = 40 sin(124 deg) = 33 C has x and y components Cx = 46.8 cos(270 deg) = 0 and Cy = 46.8 sin(270 deg) = -47 A - B + C therefore has components Rx = Ax-Bx+Cx = 58 - (-22) + 0 = 80 and Ry = Ay - By + Cy = 31-33-47=-49, which places it is the fourth quadrant and gives it magnitude `sqrt(Rx^2 + Ry^2) = `sqrt(80^2 + (-49)^2) = 94 at angle tan^-1(Ry / Rx) = tan^-1(-49/53) = -32 deg or 360 deg - 32 deg = 328 deg. Thus A - B + C has magnitude 93 at angle 328 deg. B-2A has components Rx = Bx - 2 Ax = -22 - 2 ( 58 ) = -139 and Ry = By - 2 Ay = 33 - 2(31) = -29, placing the resultant in the third quadrant and giving it magnitude `sqrt( (-139)^2 + (-29)^2 ) = 142 at angle tan^-1(Ry / Rx) or tan^-1(Ry / Rx) + 180 deg. Since x < 0 this gives us angle tan^-1(-29 / -139) + 180 deg = 11 deg + 180 deg = 191 deg. Thus B - 2 A has magnitude 142 at angle 191 deg. Note that the 180 deg is added because the angle is in the third quadrant and the inverse tangent gives angles only in the first or fourth quandrant ( when the x coordinate is negative we'll be in the second or third quadrant and must add 180 deg). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qUniv. 3.58. (This problem has apparently been eliminated from recent editions, due to the now policitally incorrect nature of the device being thrown. The problem is a very good one and has been edited to eliminate politically incorrect references). Good guys in a car at 90 km/hr are following bad guys driving their car, which at a certain instant is 15.8 m in front of them and moving at a constant 110 km/hr; an electronic jamming device is thrown by the good guys at 45 deg above horizontal, as they observe it. This device must land in the bad guy's car. With what speed must the device be thrown relative to the good guys, and with what speed relative to the ground? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The device is thrown at velocity v0 at 45 deg, giving it v0y = .71 v0 and v0x = .71 v0. The device will return to its original vertical position so we have `dsy = 0. Using `dsy = v0y `dt + .5 g `dt^2 with `dsy = 0 and assuming the upward direction to be positive we obtain v0y `dt + .5 (-g) `dt^2 = 0 so that `dt = 0 or `dt = - 2 * v0y / (-g) = 2 * 71 v0 / g. In time `dt the horizontal displacement relative to the car will be `dsx = v0x `dt + ax `dt; since acceleration ax in the x direction and v0x = .71 v0 is zero we have `dsx = .71 v0 * `dt. We also know that relative to the first car the second is moving at 20 km / hr = 20,000 m / (3600 sec) = 5.55 m/s, approx.; since its initial position is 15.8 m in front of the first car we have `dsx = 15.8 m + 5.55 m/s * `dt. To keep the equations symbolic we use x0Relative and vRelative for the relative initial position and velocity of the second car with respect to the first. We thus have three equations: `dt = 2 * .71 v0 / g = 1.42 v0 / g. `dsx = .71 v0 * `dt `dsx = x0Relative + v0Relative * `dt. This gives us three equations in the variables v0, `dt and `dsx, which we reduce to two by substituting the expression -2 to obtain: `dsx = .71 v0 * 1.42 v0 / g = v0^2 / g `dsx = x0Relative + v0Relative * 1.42 v0 / g. Setting the right-hand sides equal we have v0^2 / g = x0Relative + v0Relative * 1.42 v0 / g, or v0^2 - v0Relative * 1.42 v0 - g * x0Relative = 0. We get v0 = [1.42 v0Relative +-sqrt( (1.42 v0Relative)^2 - 4 * (-g * x0Relative) ) ] / 2 = [1.42 * v0Relative +-sqrt( (1.42 * v0Relative)^2 + 4 * g * x0Relative) ] / 2. Substituting 5.55 m/s for v0Relative and 15.8 m for x0Relative we get [1.42 * 5.55 m/s +-sqrt( (1.42 *5.55 m/s)^2 + 4 * 9.8 m/s^2 *15.8 m) ] / 2 = 17 m/s or -9.1 m/s, approx.. We conclude that the initial velocity with respect to the first case must be 17 m/s. Checking this we see that the device will have initial x and y velocities 7.1 * 17 m/s = 12 m/s, approx., and will therefore stay aloft for 2 * 12 m/s / (9.8 m/s^2) = 2.4 sec, approx.. It will therefore travel 2.4 sec * 12 m/s = 28 m, approx. in the horizontal direction relative to the first car. During this time the second car will travel about 5.55 m/s * 2.4 sec = 13 m, approx., resulting in relative position 15.8 m + 13 m = 28.8 m with respect to the first. This is reasonably close to the 28 m obtained from the motion of the projectile. Correcting for roundoff errors will result in precise agreement. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!