A computer screen is 18 cm wide and 12cm high. The x is the horizontal and y is the vertical. \n\n\n1)A target moves from the origin from the bottom left hand corner of the screen. Its position vector t seconds is r T = 7.5ti  + (12t – 5t^2) j         0≤ t ≤ T\na)Prove that the target reaches bottom right hand corner of the screen and calculate the value of T, and the time taken to reach the point.
216882
If you assume that at the bottom left corner the coordinates are i = 0, j= 0. and at the other end (18,0).\nT has to be:\nT = 18/7.5\n\nFor Y coordinate, it must be zero:\n12t - 5t^2 = 0\nt = 0, or 12/5\n\nso 12/5 = 18/7.5 which is at the bottom right (18,0) and proves the question.
117140
Let's break the area into many vertical rectangular strips between x=1 and x=2, each strip so narrow that we can assume that the top of the strip is horizontal. Thus each strip will be a very narrow and tall rectangle.\n\nLet's say the number of vertical strips we use is n. The first strip will have its left-hand side at x = 1, the next will have its left-hand side at x = 1 + 1/n, the third will have its left-hand side at x = 1 + 2/n, and in general the kth strip will have its left-hand side at x = 1 + (k-1)/n.\n\nNow, what is the y value at each of these x points? Using f(x)=x² we get, for the kth point,\n\ny = (1 + (k-1)/n)² = 1 + 2(k-1)/n + ((k-1)/n)² \n\nNow the area of each strip is equal to its base times its height. The base of each strip is 1/n. So the area of the kth strip is\n\nA = 1/n * (1 + 2(k-1)/n + ((k-1)/n)²)\n\nNow we have to add the areas of all these strips from k = 1 to k = n and rearrange the terms.\n\n.....k=n\nA = Σ (1/n + 2(k-1)/n² + (k-1)²/n^3)\n.....k=1   \n\n.....k=n\nA = Σ (1/n + (2k - 2)/n² + (k² - 2k + 1)/n^3)\n.....k=1\n\nGiving all fractions the common denominator of n^3,\n\n.....k=n\nA = Σ ((n² + 2kn - 2n + k² - 2k + 1)/n^3)\n.....k=1\n\n.....k=n\nA = Σ ((n² - 2n + 1)/n^3  + (2kn + k² - 2k)/n^3)\n.....k=1\n\nThe first term in the summation sign appears n times in the sum, so we can take it out by multiplying it by n. \n\nSum going from k=1 to k=n:\nA =  (n² - 2n + 1)/n² + Σ (2kn + k² - 2k)/n^3\n\nNow let's clear the parentheses in the summation:\n\nSum going from k=1 to k=n:\nA =  (n² - 2n + 1)/n² + Σ 2k/n² + Σ k²/n^3 - Σ 2k/n^3)\n\nThe first summation contains k=1, 2, 3, ... n, so we use the formula for the sum of the arithmetic series:\n\nS = n(n+1)/2\n\nso the first summation becomes\n\nn(n+1)/n² = (n² + n)/n² = 1 + 1/n\n\nThe last summation also contains k=1, 2, 3, ... n, so we use the same formula for the sum of a series, and we get\n\nn(n+1)/n^3 = (n² + n)/n^3 = 1/n + 1/n²\n\nFor the middle summation, we have to compute the sum of k² for k=1, 2, 3, ... n, or that is the sum of 1, 4, 9, 16, ... n².\n\nThe site\n\nhttp://mathworld.wolfram.com/PowerSum.html\n\ngives the formula for the sum of the squares of a series as\n\nS = (2n^3 + 3n² + n)/6\n\nso the middle summation becomes\n\n1/3 + 1/2n + 1/6n²\n\nNow, we are almost done.\n\nLet's let the number of strips get very large, go to infinity as a limit. Let's see what each term becomes:\n\n(n² - 2n + 1)/n² becomes 1 as n gets very large,\n1 + 1/n becomes 1 as n gets very large,\n1/n + 1/n² becomes 0 as n gets very large, and\n1/3 + 1/2n + 1/6n² becomes 1/3 as n gets very large.\n\nThe area under the curve is 1 + 1 + 1/3, or 7/3.\n\nYou can see why integral calculus was invented. Using integral calculus, this problem can be solved in two steps. The integral of f(x)=x² is x^3/3. Inserting the limits of x=1 and x=2 gives\n\nA = 2^3/3 - 1^3/3 = 7/3
343421
horizontal component is force*cos(20 degrees), so its \n9cos(20 degrees).\nvertical component is 9sin(20 degrees)
611131
you have not given the distance on screen D\nso i took D=5cm\n \nso displacement y=m*wavelenght*D/d\n                             =0.4152 cm\n\nand angle q=4.7469 degree\n\nok if you don understand then go on this link and put the values
286380
Just solve for h:\nV = πr^2h \n\nDivide both sides by  πr^2\nV / πr^2 = h\n\nSwapping sides:\nh = V / πr^2\n\nSo if V = 121 and r = 3\nh = 121 / 9π\nh = 4.27949958 cm\n\nSorry, but I can't graph the function for you.  It would be a 3D graph with 3 variables (V, h, r).  For example for V = 0, h would be 0 for all values.  Similarly for r = 0, h would be 0 for all values.  Other than that, plug in coordinates for V and r and then graph h on the vertical (z) dimension.\n\nFor the value of h, however it is approx. 4.28 cm.
470009
Use the certical component to find the time in the air and since the horizontal velocity does not change; Dx = vx * time\n\nDx = v^2 / g\n\nheight = v^2 / (2g)\n\nHope this helps without actually solving it for you.
461081
I finished the problem.  The answer is s = 0.48517m.  It was hard, although now that I've worked it, it doesn't seem as difficult anymore.  Had to do some serious thinking though.\n\nThe collar A will tend to slide <down> the bar CD.  The vector CD (from C to D) is [-0.2, -0.3, 0.25] with magnitude 0.43875 and direction cosines [-0.45584, -0.68376, 0.56980].  The weight of collar A, directed downward, is 8g = 78.4N and is represented as [0, -78.4, 0].\n\nWe don't have to worry about the collar being supported by bar CD (the normal force).  The component of the collar directed down the bar is the inner (scalar, dot) product of the weight vector and the direction cosines.  This force is 78.4 x .68376 = 53.60706N.\n\nLet E be the point on the bar CD such that BE is perpendicular to CD.  Then BEA is a right triangle, and by the Pythagorean Theorem, the force along BE is sqrt(150^2 - 53.60706^2) = 140.09384N.  (I'm using numbers stored in the calculator and showing five decimal digits.  All this can be rounded off at the end.)\n\nSince the force triangle BEA is similar to the distance triangle BEA, we just have to get one of the legs to find our distance s = AB.  We will get the distance BE.\n\nTriangle BEA is in the plane defined by the points B, C, and D.  Vector CB = [-0.4, 0.2, 0.15].  The vector cross product CB x CD (see above) is normal to the plane defined by points B, C, and D.  CB x CD = [0.095, 0.07, 0.16].\n\nNow, the cross product CD x (CB x CD) lies in the plane defined by B, C, and D; is perpendicular to CD; and is parallel to the line segment BE.\n\nCD x (CB x CD) = [-0.0655, 0.05575, 0.0145] with magnitude 0.08723 and direction cosines [-0.75091, 0.63914, 0.16623].\n\nNow we can get parametric equations of the lines CEAD and BE, and find the point of intersection E.  Starting with point C and using vector CD, the first line is\n\nx = 0.4 - 0.2t\ny = 0.3 - 0.3t\nz = 0 + 0.25t\n\nStarting from point B and using the direction cosines from vector BE, the second line is\n\nx = 0 + 0.75091u\ny = 0.5 - 0.63914u\nz = 0.15 - 0.16623u\n\nSetting like terms equal to each other, we have three equations in two unknowns.  The three equations are not linearly independent.  Solving the x and y equations for t and u, we get\n\nt = 0.29871 and u = 0.45313\n\n(and these results are confirmed in the z equations).  Plugging these results into the x, y, and z equations, we find that the intersection point E is (0.34026, 0.21039, 0.07468).\n\nWe know the coordinates of point B, so by the Pythagorean Theorem, the distance BE is 0.45313m.\n\nComparing the force and distance vectors in similar triangles BEA, we have\n\ns/150 = 0.45313 / 140.09384\n\nand the length s of the cable AB is 0.48517m.  (Answer)\n\nA difficult, but fun, problem.
383811
a=(delta V)/(delta t)\n\ndelta V = V_final - V_initial = 0 - 3.2\ndelta t = 0.01 sec\n\na=3.2/0.01=320 m/s2\n\nforce=mxa=0.084x320=26.88 N
63705
1. Dy = 1/2 A*t^2 + Vy*t, Vy = 0 (speed is horizontal),\nso t = sqrt(66m/(1/2*9.8m/s^2)) = 3.67 s.\n   Dx = Vx*t = 6 m/s * 3.67 s = 22 m.\n\n2.(a)  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso t = sqrt(1.150m/(1/2*g)) = 0.484 s\n  (b)  Dx = Vx*t, so Vx = Dx/t = 0.385m/0.484s = 0.795 m/s\n\n3.  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso Dy = 1/2*9.8m/s*(2.9s)^2 = 41.2 m\n\n   Dx = Vx*t = 2.4 m/s * 2.9 s = 6.96 m\n\n4. Dy = 0, Dx = 97 m, a = 50 degrees\n   Dx = Vx*t = V*cos(a)*t, \nso t = Dx/(V*cos(a))\n   -1/2*g*t^2 + Vyt = Dy = 0 ( Vy = V*sin(a))\ndivide by t, so -1/2*g*t + V*sin(a) = 0, or\nt = V*sin(a)/(1/2*g)).  Equate the two expressions for t:\nV*sin(a)/(1/2*g)=Dx/(V*cos(a)), multiply both sides by V\nV^2*sin(a)/(1/2*g) = Dx/cos(a)\nV= sqrt(Dx/cos(a)/sin(a)*(1/2*g))\nV = 31.1 m/s
353782
a) they give you everything you need.  v0 = 32, s0 = 0, so\n\ns(t) = -16t² + 32t\n\nb)\n\ns(1) = -16 + 32 = 16\n\nc)  The ball hits the ground when s(t) = 0 (the second time).\n\n0 = -16t² + 32t = -16t(t-2) so t=0 or t-2 = 0, t=2.\n\nSo the ball hits the ground when t=2.\n\nd)  The maximum height is attained at the vertex of the parabola.\n\ny - 16 = -16(t² - 2t + 1)\ny - 16 = -16(t - 1)²\n\nSo the vertex is (1, 16), that is, at time t = 1, when the height of the ball is 16.
64315
Let V = speed (the magnitude of the initial velocity)\na = angle, t = time, \nVx = horizontal (x) component of velocity\nVy = vertical (y) component of velocity\nAx = acceleration in x direction\nAy = acceleration in y direction\nDx = net distance traveled in x direction\nDy = net distance traveled in y direction\n\nVx = Vcos(a)\nVy = Vsin(a)\n\n  Dx = 1/2 Ax*t^2 + Vx*t\nbut Ax = 0, so Dx = Vx*t, t = Dx/Vx\n\n  Dy = 1/2 Ay*t^2 + Vy*t\n(Ay = -g, the acceleration of gravity, negative for downward)\n -1/2*g*Dx^2/(Vcos(a))^2 + Vsin(a)*Dx/(Vcos(a)) = Dy\nMultiply by V^2, and collect terms with V^2 on the left:\n(Dx*tan(a)-Dy) V^2 = 1/2*g*(Dx)^2/(cos(a))^2\nV = sqrt(1/2*g*(Dx)^2/(cos(a))^2/(Dx*tan(a)-Dy))\nsubstitute a = 30 degrees, Dx = 6.05 m, Dy = 3.05-2.2 = 0.85 m, g = 9.8 m/s^2\nV = 9.51 m/s\nFor a = 60 degrees,\nV = 8.63 m/s
352452
[on] [y=] [(] [x,t,o,n] [+] [2] [)] [/] [x,t,o,n] [enter] [5] [-] [2] [x,t,o,n] [enter] [(-)] [2] [x,t,o,n] [-] [3] [enter] [window] [(-)] [5] [enter] [5] [enter] [1] [enter] [(-)] [5] [enter] [5] [enter] [graph]\n\nYou should have on screen a vertical asymptote at x=0, horizontal asymptote at y=1, a curve going to negative infinity towards 0 on the left, a curve going to positive infinity towards 0 on the right, on the left tangent line y=-2x-3 touches point(-1,-1), and on the right tangent line y=5-2x touches point(1,3).
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x =70m, t = 7s, Vf = 15m/s\na.)\nx = 1/2(Vi + Vf)t\nVi = initial velocity\nVf = final velocity\n\n70m = 1/2(Vi + 15m/s)(7s)\n10m/s = 1/2Vi + 7.5m/s\n2.5m/s = 1/2Vi\nVi = 5m/s\nb.)\nVf = Vo + at\n15m/s = 5m/s + 7s(a)\n10m/s = 7s(a)\na = 1.43m/s/s\nbasic kinematics\nyou can't use distance = rate x time because there is acceleration and the speed is not constant.
651316
Set 16t^2 +32t -225=0\n\nUse quadratic formula gives t = 2.9sec
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a)\nRate = Distance/time = (24000 ft)/(20 min) = 1200 ft/min\n\nb)\nTo write A(t), you know that the max. value is 24000 and the plane descends at a constant, linear rate of 1200 ft/min.  Thus, you can write a linear equation of this model:\n\nA(t) = 24000 - 1200t, where t = time (in minutes)\n\nPlugging in 0 (at initial time), the height is 24000 feet, which matches the fact in the problem.\n\nc)\nThough linear functions have infinite ranges and domains, this problem is limited to the realism of the problem.  The range is only limited from the maximium value (24000 ft) to the ground (0 ft).  You can't have a negative value for altitude because that would not make sense in the problem.  The domain is limited to the time in air (20 min).  You can't have a negative time and time after 20 min is irrelevant because the plane has already landed.  Summarizing:\n\nDomain: 0 <= t <= 20\nRange: 0 <= A(t) <= 24000\n\n--------\n\nHope this helps
457823
u can also solve this by takin s= -100m\nn by the formula S= ut + 1/2 gt^2\ng = 9.8m/s^2. u = -6m/s .......4 the 2nd part\nand \n\nfind time of the 1st half by the formula \n=> v = u + at....[ where u = 6m/s, v = -6m/s(velocity in oposit direction)]\n\nALSO REMEMBER THAT WEN A TRAJECTORY AGAIN REACHES ITS STARTIN PT.  IT HAS THE SAME MAGNITUDE OF VELOCITY\n\n-ve sign is given to all dat is below the level or in the direction below the startin point\ncombine these 2................ n voila..\nwith time u can master this easy.... n logical method
32330
First Question:\nVertical and horizontal motion are independent from each other.  Use d=(v.i)(t)+(1/2)(a)(t)^2 to find the time that the dart is in the air.  d is distance(0.32m), v.i is the initial velocity(0.00m/s vertically since it was thrown exactly horizontally), a is acceleration (9.8m/s^2: gravity) and t is the time in seconds.  Now that you have the time, use that in the equation x=vt, where x is the horizontal distance (range) that you are asked to solve for, v is the velocity(11.8m/s), and t is the time that you just found.  The answer to the problem is the x variable in this equation.\n\nSecond Question:\nI have attempted this problem for about 40 minutes and I have tried every possible combination of variables and equations, but there is not enough information to solve it.  Therefore, I have to say that it is an impossible problem.  If you figure out how to solve it, please, send me an e-mail at the address I give later and tell me how to do it.\n\nThird Question:\nFirst, find the distance that the basket is from the launch point by subtracting the two.  Then, use (v.f)^2=(v.i)^2+(2)(a)(d), where v.f is the final velocity, v.i is the initial velocity, a is the acceleration (9.8m/s^2), and d is the distance which you just found.  To  get the ball to the basket, it only needs to go the exact distance to the basket, so the final velocity is 0.00m/s.  Use the previous equation to find the intial vertical velocity.  The use v.iy=(v)sin(angle), where v.iy is the initial vertical velocity and v is the initial angular velocity, to find the velocity the ball must be launched at for each angle.\n\nPlease send me an e-mail at bge1234@yahoo.com after you turn in your work and get the solution so that I can see how your teacher did it.  In particular, tell me how to do the second question.  My ways SHOULD work as long as you don't make any calculatory errors.  I hope this helps you.