Velocity distance problem?
914719
After trying to solve this multiple times, I've come to a simple conclusion.  The dog is constantly moving at a velocity of 4.5(m/s).  Distance is time*velocity.  The dog is in motion until the lady reaches the end.  If we use an inversion of that formula, we get time=distance/velocity.  \n\n4000(m)/2.5(m/s) = 1600 seconds.  \n       (time it takes her to reach the river)\n1600(s)*4.5(m/s)=7200 meters or 7.2km \n       (that time multiplied by the dogs velocity to figure distance)\n\n     So, after failed attempts using calculus and other math that is unnecessary, we've come to the answer of 7.2km of distance covered by the dog.  Hope this helps you out!
250483
Try using a non dimensional correlation for the Bio number or possible Prandtl number. I can't remember off the top of my head, but one of them should relate your Mach number to pressure or viscosity allowing you to have only one unknown. Another idea would be to model the situation with a known boundary layer pattern and find where the boundary layers meet. Check the solution and see if it is reasonable.
45821
are you sure that you want acceleration instead of velocity??\n\nwell, the acceleration is going to be 0.  since the speed is going to be the same the whole time, there won't be any acceleration.
497479
When calculating with error, remember to MAXIMIZE error wherever possible. For example, since V1 and V2 are in the denominator of their respective terms, error should be subtracted to maximize the overall terms. Check me on this, but I found an average velocity of (19.5 ± 0.5) m/s.
678647
The water starts from one end and reaches the other end in 30 minutes\n\nThe distance traveled is 1700 meter. Time taken 30 minutes = 1800 second.\n\nThe velocity is distance traveled per second. Therefore, it is 1700 /1800 m/s =0.94m/s
443173
This problem can be solved using v^2=2ax, where v is the velocity, a is the acceleration due to gravity, and x is the distance (this is a simplified version of a more complex equation, but this equation will work for this scenario).\n\na=32.15 ft/ s^2\nx= 30 ft\n\nSo, v^2=2(32.15ft/s^2)(30ft)\n\nOnce you solve and convert to miles per hour, you get approximately 29.9 mph.
1135961
in order to solve this problem, you need to work part (b) first before (a)\n\nb): you know gravity (g=32.2 ft/sec) acts opposite the rocket, so calculate, using:\n\nv_final = v_initial - 2*g*y_1\n\nyou know that the final velocity is zero (max height) and you are given the inital velocity. the negative sign is to account for the opposite direction of gravity. move the equation around to solve for y_1 (height from tower to max. height). then add 100 feet (to account for the tower) and you have your answer.\n\na): since you know your maximum height (from part b), substitute that answer for h. [*NOTE: remember to subtract 100 feet from your answer in part b, because t=0 at the top of the tower] You now have a quadratic equation and solve for t using the quadratic formula (you should know what that is). your answer will be the answer!\n\nThis prob is easy only because it worries upto the max height. if we had to worry about the falling down part, well thats another story.\n\n** i already worked out this prob, so if you want to ask more questions/verify your answer, just post an additional response and ill modify my question to answer it. GL!
790665
starting from the very beginning\nv ≡ ds/dt\na ≡ d²s/dt² = dv/dt\ndv = adt\nif a is constant,\n∫dv = a∫dt\nv = at + C\ncall C v0, or initial velocity:\nv = v0 + at\nv = 0 +(9.80662 m/sec^2)(2 sec)\nv = 19.61324 m/s
337691
if the question is: If it takes 3 hours to travel 24 miles downstream and 6 hours to travel UPstream, how fast is the current? then,\nlets call the travelling speed Vt and the current speed Vc\ndownstream we have Vt+Vc=24miles/3hours=8mph\nupstream we have Vt-Vc=24/6=4mph\nto solve the system of two equations we can add the equations and get 2Vt=12, Vt=6mph\nVc=8-6=6-4=2mph\nThe current is 2mph fast.
864633
He's right. You take the derivative of the equation and plug in the values for t (just consult any book on differential calculus for details on this)\n\nSo h'(t)= -3.92t+10 (the derivative)\n\nWhen t=2.3, the velocity will then be (-3.92)(2.3)+10, or .984 m/s\n\nWhen t=2.7, the velocity will then be (-3.92)(2.7)+10, or -.584 m/s (.584 m/s in the opposite direction)\n\nNot bad for a high school sophomore who taught himself differential calculus, eh?
889350
to compute for the maximum height, the vertical component of the velocity is needed.\nSo Vy = 49sin30\n      Vy = 24.5m/s\n\nto solve for the max. height use the formula\nVy = sqrt(2*g*h)\nh=(V^2)/(2*g) = [(24.5m/s)^2]/(2*9.81m/s2)\nhmax = 30.594 m
980357
The answer has to be a speed!  It involves seeing that dy/dt = 1/2 x^(-1/2) dx/dt in this case.  Also, the distance from the origin is given by s^2 = x^2 + y^2; to get ds/dt (the required result), differentiate that wrt "t".  Put this all together yourself; it's the only way to learn rather than simply parrot.\n\nWithout a cast-iron guarantee, I get the value 27/(4 x sqrt 5) \ncm/s.\n\nLive long and prosper.
782037
I solved the problem by stating that P would travel distance d in time t\nd=20*t+.5*.3*t*t\n\nand Q would travel d+20 in time t\nt is the time to collision\n\nd+20=30*t\nd=30*t-20\nSubtract the equations and solve for t:\n\n0=.15*t*t-10t+20\nt=2.064 seconds\n(the second root is superfluous)\nd=41.9m\n\nWhich proves collision\n\nj
626127
V = (v + u) / (1 + uv/c²)\n\nTry that. The velocities are quite high so relativistic. Haven't tried it yet though.\n\nNevermind, the velocites aren't nearly high enough to be relativistic. If it was km/s it would work.
1135263
The height of the trajectory is 1 m (3.05-2.05)\n\nThe horizontal distance is 6.02 m\n\nNow you have ucos(25)t = 6.02\n\nand usin(25)t-0.5*9.8*t^2 = 1\n\nSubstitute t = 6.02/ucos(25) in the 2nd eqn and solv.
807345
Kinetic energy is m v^2 / 2.\nTo find v, you need to find the magnitude of the vector that represents the object's velocity. (Actually, the above expression requires the SQUARE of the velocity, which will save us a step in the solution.)\n\nUsing vector algebra and the Pythagorean theorem, the velocity is sqrt (6^2 + 2^2). Don't bother to take the square root and you'll have v^2, which you can use in the equation for kinetic energy.\n\nThe velocity changes by 8i+4j. Let's assume that this change occurs over a period of 1 second. (It turns out that the amount of work done does not depend on the time period, so this assumption doesn't affect the answer that we will calculate.)\n\nTo change the velocity by 8i+4j, we need to apply a force of 24 newtons (8 x 3) in the i direction, and 12 newtons (4 x 3) in the j direction for 1 second.\n\nThe amount of work done by each of these forces (considered separately) is equal to the force times the distance over which it acts. The 24 newton force in the i direction acts over a distance of 10 meters (since the average i-direction velocity during this 1-second period is 10 m/s; you calculate that by averaging the initial and final velocities, 6 m/s and (6+8) m/s). So the work done is 240 newton-meters, or 240 joules.\n\nIn the j direction, the force of 12 newtons acts over a distance of 1 meter, for total work done of 12 joules. (Note that the average velocity in the j direction is 1 meter/second, found by averaging -2 m/s and (-2+4) m/s.)\n\nSo the total work done is 252 joules. (I'm a little uncomfortable about treating these separately and adding them up. If I had a physics book here, I'd read it to decide whether this is the right treatment. If you have a physics book, I suggest you do that.)\n\nP.S. I have assumed that the velocity changes BY an amount of (8i+4j). If instead it changes TO (8i+4j), then you need to figure out the amount BY which the velocity has changed, and plus those i and j values in where I used 8 and 4.
581110
You have to decide the velocity. You need to keep Re<2000. You know the diameter of the pipe, u need to know the coefficient of viscocity. Velocity range is to be calculated by putting all these quantities in the equation for Re, keeping Re between 0 and 2000. Then the velocity is to be controlled in that range.